Chp. 8 - Chapter 8 Conservation of Linear Momentum...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 8 Conservation of Linear Momentum Conceptual Problems 1 • [SSM] Show that if two particles have equal kinetic energies, the magnitudes of their momenta are equal only if they have the same mass. Determine the Concept The kinetic energy of a particle, as a function of its momentum, is given by K = p 2 2m . The kinetic energy of the particles is given by: 2 p12 p2 K1 = and K 2 = 2m1 2m 2 Equate these kinetic energies to obtain: p12 p2 =2 2m1 2m2 Because the magnitudes of their momenta are equal: 1 1 = and m1 = m2 m1 m2 2 • Particle A has twice the (magnitude) momentum and four times the kinetic energy of particle B. A also has four times the kinetic energy of B. What is the ratio of their masses (the mass of particle A to that of particle B)? Explain your reasoning. Determine the Concept The kinetic energy of a particle, as a function of its momentum, is given by K = p 2 2m . The kinetic energy of particle A is given by: KA = 2 pA p2 ⇒ mA = A 2 mA 2K A The kinetic energy of particle B is given by: KB = 2 pB p2 ⇒ mB = B 2 mB 2K B Divide the first of these equations by the second and simplify to obtain: 2 pA 2 2 mA 2 K A K B p A K B ⎛ p A ⎞ ⎟ ⎜ = 2= = 2 mB K A pB K A ⎜ pB ⎟ pB ⎠ ⎝ 2K B 721 722 Chapter 8 Because particle A has twice the (magnitude) momentum of particle B and four times as much kinetic energy: 2 mA K ⎛ 2p ⎞ = B ⎜ B⎟ = 1 mB 4 K B ⎜ p B ⎟ ⎠ ⎝ 3 • Using SI units, show that the units of momentum squared divided by those of mass is equivalent to the joule. Determine the Concept The SI units of momentum are kg⋅m/s. Express the ratio of the square of the units of momentum to the units of mass: m⎞ ⎛ ⎜ kg ⋅ ⎟ s⎠ ⎝ kg 2 Simplify to obtain: 2 m⎞ ⎛ m2 ⎜ kg ⋅ ⎟ kg 2 ⋅ 2 2 s⎠ ⎝ s = kg ⋅ m = ⎛ kg ⋅ m ⎞ ⋅ m = N ⋅ m = J = ⎟ ⎜ s2 ⎠ s2 ⎝ kg kg True or false: 4 • (a) The total linear momentum of a system may be conserved even when the mechanical energy of the system is not. For the total linear momentum of a system to be conserved, there must be no external forces acting on the system. The velocity of the center of mass of a system changes only when there is a net external force on the system. (b) ( c) (a) True. Consider the collision of two objects of equal mass traveling in opposite directions with the same speed. Assume that they collide inelastically. The mechanical energy of the system is not conserved (it is transformed into other forms of energy), but the momentum of the system is the same after the collision as before the collision; that is, zero. Therefore, for any inelastic collision, the momentum of a system may be conserved even when mechanical energy is not. (b) False. The net external force must be zero if the linear momentum of the system is to be conserved. (c) True. This non-zero net force accelerates the center of mass. Hence its velocity changes. Conservation of Linear Momentum 723 5 • If a bullet is fired due west, explain how conservation of linear momentum enables you to predict that the recoil of the rifle be exactly due east. Is kinetic energy conserved here? Determine the Concept The momentum of the bullet-gun system is initially zero. After firing, the bullet’s momentum is directed west. Momentum conservation requires that the system’s total momentum does not change, so the gun’s momentum must be directed east. Kinetic energy is not conserved. Some of the initial chemical energy of the gun powder, for example, is transformed into thermal energy and sound energy. 6 • A child jumps from a small boat to a dock. Why does she have to jump with more effort than she would need if she were jumping through an identical displacement, but from a boulder to a tree stump? Determine the Concept When she jumps from a boat to a dock, she must, in order for momentum to be conserved, give the boat a recoil momentum, that is, her forward momentum must be the same as the boat’s backward momentum. When she jumps through an identical displacement from a boulder to a tree stump, the mass of the boulder plus Earth is so large that the momentum she imparts to them is essentially zero. 7 •• [SSM] Much early research in rocket motion was done by Robert Goddard, physics professor at Clark College in Worcester, Massachusetts. A quotation from a 1920 editorial in the New York Times illustrates the public opinion of his work: ″That Professor Goddard with his ′chair′ at Clark College and the countenance of the Smithsonian Institution does not know the relation between action and reaction, and the need to have something better than a vacuum against which to react—to say that would be absurd. Of course, he only seems to lack the knowledge ladled out daily in high schools.″ The belief that a rocket needs something to push against was a prevalent misconception before rockets in space were commonplace. Explain why that belief is wrong. Determine the Concept In a way, the rocket does need something to push upon. It pushes the exhaust in one direction, and the exhaust pushes it in the opposite direction. However, the rocket does not push against the air. Another way to look at this is conservation of momentum. The momentum of the exhaust is equal and opposite to the momentum of the rocket, so momentum is conserved. 8 • Two identical bowling balls are moving with the same center-of-mass velocity, but one just slides down the alley without rotating, whereas the other rolls down the alley. Which ball has more kinetic energy? 2 Determine the Concept The kinetic energy of the sliding ball is 1 mvcm . The 2 2 kinetic energy of the rolling ball is 1 mvcm + K rel , where K rel is its kinetic energy 2 724 Chapter 8 relative to its center of mass. Because the bowling balls are identical and have the same velocity, the rolling ball has more kinetic energy. There is no problem here because the relationship K = p 2 2m is between the center of mass kinetic energy of the ball and its linear momentum. 9 • A philosopher tells you, ″Changing motion of objects is impossible. Forces always come in equal but opposite pairs. Therefore, all forces cancel out. Because forces cancel, the momenta of objects can never be changed.″ Answer his argument. Determine the Concept Think of someone pushing a box across a floor. Her push on the box is equal but opposite to the push of the box on her, but the action and reaction forces act on different objects. Newton’s second law is that the sum of the forces acting on the box equals the rate of change of momentum of the box. This sum does not include the force of the box on her. 10 • A moving object collides with a stationary object. Is it possible for both objects to be at rest immediately after the collision? (Assume any external forces acting on this two-object system are negligibly small.) Is it possible for one object to be at rest immediately after the collision? Explain. Determine the Concept It’s not possible for both to remain at rest after the collision, as that wouldn't satisfy the requirement that momentum is conserved. It is possible for one to remain at rest. This is what happens for a one-dimensional collision of two identical particles colliding elastically. 11 • Several researchers in physics education claim that part of the cause of physical misconceptions amongst students comes from special effects they observe in cartoons and movies. Using the conservation of linear momentum, how would you explain to a class of high school physics students what is conceptually wrong with a superhero hovering at rest in midair while tossing massive objects such as cars at villains? Does this action violate conservation of energy as well? Explain. Determine the Concept Hovering in midair while tossing objects violates the conservation of linear momentum! To throw something forward requires being pushed backward. Superheroes are not depicted as experiencing this backward motion that is predicted by conservation of linear momentum. This action does not violate the conservation of energy. 12 •• A struggling physics student asks ″If only external forces can cause the center of mass of a system of particles to accelerate, how can a car move? Doesn’t the car’s engine supply the force needed to accelerate the car? ″ Explain what external agent produces the force that accelerates the car, and explain how the engine makes that agent do so. Conservation of Linear Momentum 725 Determine the Concept There is only one force which can cause the car to move forward−the friction of the road! The car’s engine causes the tires to rotate, but if the road were frictionless (as is closely approximated by icy conditions) the wheels would simply spin without the car moving anywhere. Because of friction, the car’s tire pushes backwards against the road and the frictional force acting on the tire pushes it forward. This may seem odd, as we tend to think of friction as being a retarding force only, but it is true. 13 •• When we push on the brake pedal to slow down a car, a brake pad is pressed against the rotor so that the friction of the pad slows the rotation of the rotor and thus the rotation of the wheel. However, the friction of the pad against the rotor can’t be the force that slows the car down, because it is an internal force—both the rotor and the wheel are parts of the car, so any forces between them are internal, not external, forces. What external agent exerts the force that slows down the car? Give a detailed explanation of how this force operates. Determine the Concept The frictional force by the road on the tire causes the car to slow. Normally the wheel is rotating at just the right speed so both the road and the tread in contact with the road are moving backward at the same speed relative to the car. By stepping on the brake pedal, you slow the rotation rate of the wheel. The tread in contact with the road is no longer moving as fast, relative to the car, as the road. To oppose the tendency to skid, the tread exerts a forward frictional force on the road and the road exerts an equal and opposite force on the tread. 14 • Explain why a circus performer falling into a safety net can survive unharmed, while a circus performer falling from the same height onto the hard concrete floor suffers serious injury or death. Base your explanation on the impulse-momentum theorem. Determine the Concept Because Δp = FΔt is constant, the safety net reduces the force acting on the performer by increasing the time Δt during which the slowing force acts. 15 •• [SSM] In Problem 14, estimate the ratio of the collision time with the safety net to the collision time with the concrete for the performer falling from a height of 25 m. Hint: Use the procedure outlined in Step 4 of the ProblemSolving Strategy located in Section 8-3. Determine the Concept The stopping time for the performer is the ratio of the distance traveled during stopping to the average speed during stopping. 726 Chapter 8 Letting dnet be the distance the net gives on impact, dconcrete the distance the concrete gives, and vav, with net and vav,without net the average speeds during stopping, express the ratio of the impact times: Assuming constant acceleration, the average speed of the performer during stopping is given by: d net r= v Δt net = av, with net d concrete Δt concrete vav, without net (1) vf + v 2 or, because vf = 0 in both cases, vav = 1 v 2 vav = where v is the impact speed. Substituting in equation (1) and simplifying yields: Assuming that the net gives about 1 m and concrete about 0.1 mm yields: r= r= d net 1 2v d concrete 1 2v = d net d concrete 1m ≈ 10 4 0.1 mm 16 •• (a) Why does a drinking glass survive a fall onto a carpet but not onto a concrete floor? (b) On many race tracks, dangerous curves are surrounded by massive bails of hay. Explain how this setup reduces the chances of car damage and driver injury. Determine the Concept In both (a) and (b), longer impulse times (Impulse = FavΔt) are the result of collisions with a carpet and bails of hay. The average force on a drinking glass or a car is reduced (nothing can be done about the impulse, or change in linear momentum, during a collision but increasing the impulse time decreases the average force acting on an object) and the likelihood of breakage, damage or injury is reduced. 17 • True or false: (a) Following any perfectly inelastic collision, the kinetic energy of the system is zero after the collision in all inertial reference frames. (b) For a head-on elastic collision, the relative speed of recession equals the relative speed of approach. (c) During a perfectly inelastic head-on collision with one object initially at rest, only some of the system’s kinetic energy is dissipated. Conservation of Linear Momentum 727 (d) After a perfectly inelastic head-on collision along the east-west direction, the two objects are observed to be moving west. The initial total system momentum was, therefore, to the west. (a) False. Following a perfectly inelastic collision, the colliding bodies stick together but may or may not continue moving, depending on the momentum each brings to the collision. (b) True. For a head-on elastic collision both kinetic energy and momentum are conserved and the relative speeds of approach and recession are equal. (c) True. This is the definition of an inelastic collision. (d) True. The linear momentum of the system before the collision must be in the same direction as the linear momentum of the system after the collision. 18 •• Under what conditions can all the initial kinetic energy of an isolated system consisting of two colliding objects be lost in a collision? Explain how this result can be, and yet the momentum of the system can be conserved. Determine the Concept If the collision is perfectly inelastic, the bodies stick together and neither will be moving after the collision. Therefore, the final kinetic energy will be zero and all of it will have been lost (that is, transformed into some other form of energy). Momentum is conserved because in an isolated system the net external force is zero. 19 •• Consider a perfectly inelastic collision of two objects of equal mass. (a) Is the loss of kinetic energy greater if the two objects are moving in opposite directions, each moving at speed v/2, or if one of the two objects is initially at rest and the other has an initial speed of v? (b) In which of these situations is the percentage loss in kinetic energy the greatest? Determine the Concept We can find the loss of kinetic energy in these two collisions by finding the initial and final kinetic energies. We’ll use conservation of momentum to find the final velocities of the two masses in each perfectly elastic collision. (a) Letting V represent the velocity of the masses after their perfectly inelastic collision, use conservation of momentum to determine V: pbefore = pafter or mv − mv = 2mV ⇒ V = 0 728 Chapter 8 Express the loss of kinetic energy for the case in which the two objects have oppositely directed velocities of magnitude v/2: ⎛ ⎛ v ⎞2 ⎞ ΔK = K f − K i = 0 − 2⎜ 1 m⎜ ⎟ ⎟ ⎜2 ⎝2⎠ ⎟ ⎝ ⎠ 2 = − 1 mv 4 Letting V represent the velocity of the masses after their perfectly inelastic collision, use conservation of momentum to determine V: pbefore = pafter Express the loss of kinetic energy for the case in which the one object is initially at rest and the other has an initial velocity v: or mv = 2mV ⇒ V = 1 v 2 ΔK = K f − K i 2 ⎛v⎞ = (2m )⎜ ⎟ − 1 mv 2 = − 1 mv 2 2 4 ⎝2⎠ 1 2 The loss of kinetic energy is the same in both cases. (b) Express the percentage loss for the case in which the two objects have oppositely directed velocities of magnitude v/2: 2 1 ΔK 4 mv = 1 2 = 100% K before 4 mv Express the percentage loss for the case in which the one object is initially at rest and the other has an initial velocity v: 2 1 ΔK 4 mv = = 50% K before 1 mv 2 2 The percentage loss is greatest for the case in which the two objects have oppositely directed velocities of magnitude v/2. 20 •• A double-barreled pea shooter is shown in Figure 8-41. Air is blown into the left end of the pea shooter, and identical peas A and B are positioned inside each straw as shown. If the pea shooter is held horizontally while the peas are shot off, which pea, A or B, will travel farther after leaving the straw? Explain. Base your explanation on the impulse–momentum theorem. Determine the Concept Pea A will travel farther. Both peas are acted on by the same force, but pea A is acted on by that force for a longer time. By the impulsemomentum theorem, its momentum (and, hence, speed) will be higher than pea B’s speed on leaving the shooter. Conservation of Linear Momentum 729 21 •• A particle of mass m1 traveling with a speed v makes a head-on elastic collision with a stationary particle of mass m2. In which scenario will the largest amount of energy be imparted to the particle of mass m2? (a) m2 < m1, (b) m2 = m1, (c) m2 > m1, (d) None of the above. Determine the Concept Refer to the particles as particle 1 and particle 2. Let the direction particle 1 is moving before the collision be the positive x direction. We’ll use both conservation of momentum and conservation of mechanical energy to obtain an expression for the velocity of particle 2 after the collision. Finally, we’ll examine the ratio of the final kinetic energy of particle 2 to that of particle 1 to determine the condition under which there is maximum energy transfer from particle 1 to particle 2. Use conservation of momentum to obtain one relation for the final velocities: Use conservation of mechanical energy to set the velocity of recession equal to the negative of the velocity of approach: m1v1,i = m1v1,f + m2 v 2,f (1) v 2,f − v1,f = −(v 2,i − v1,i ) = v1,i (2) To eliminate v1,f, solve equation (2) for v1,f, and substitute the result in equation (1): v1,f = v 2,f + v1,i Solve for v2,f to obtain: v 2,f = Express the ratio R of K2,f to K1,i in terms of m1 and m2: m1v1, i = m1 (v2, f − v1,i ) + m2v2, f 2 R= = Differentiate this ratio with respect to m2, set the derivative equal to zero, and obtain the quadratic equation: 2m1 v1,i m1 + m2 − K 2,f K1,i ⎛ 2m1 ⎞ 2 m2 ⎜ ⎜ m + m ⎟ v1,i ⎟ 2⎠ ⎝1 = 2 1 2 m1v1,i 1 2 m2 4m12 m1 (m1 + m2 )2 2 m2 +1 = 0 m12 730 Chapter 8 Solve this equation for m2 to determine its value for maximum energy transfer: (b ) m 2 = m1 is correct because all of particle 1’s kinetic energy is transferred to particle 2 when m2 = m1 . 22 •• Suppose you are in charge of an accident-reconstruction team which has reconstructed an accident in which a car was ″rear-ended″ causing the two cars to lock bumpers and skid to a halt. During the trial, you are on the stand as an expert witness for the prosecution and the defense lawyer claims that you wrongly neglected friction and the force of gravity during the fraction of a second while the cars collided. Defend your report. Why were you correct in ignoring these forces? You did not ignore these two forces in your skid analysis both before and after the collision. Can you explain to the jury why you did not ignore these two forces during the pre- and post-collision skids? Determine the Concept You only used conservation of linear momentum for a fraction of a second of actual contact between the cars. Over that short time, friction and other external forces can be neglected. In the long run, over the duration of the accident, they cannot. 23 •• Nozzles for a garden hose are often made with a right-angle shape as shown in Figure 8-42. If you open the nozzle and spray water out, you will find that the nozzle presses against your hand with a pretty strong force—much stronger than if you used a nozzle not bent into a right angle. Why is this situation true? Determine the Concept The water is changing direction when it rounds the corner in the nozzle. Therefore, the nozzle must exert a force on the stream of water to change its momentum, and from Newton’s third law, the water exerts an equal but opposite force on the nozzle. This requires a net force in the direction of the momentum change. Conceptual Problems from Optional Sections 24 •• Describe a perfectly inelastic head-on collision between two stunt cars as viewed in the center-of-mass reference frame. Determine the Concept In the center-of-mass reference frame the two objects approach with equal but opposite momenta and remain at rest after the collision. Conservation of Linear Momentum 731 25 •• One air-hockey puck is initially at rest. An identical air-hockey puck collides with it, striking it with a glancing blow. Assume the collision was elastic and neglect any rotational motion of the pucks. Describe the collision in the center-of-mass frame of the pucks. Determine the Concept In the center-of-mass frame the two velocities are equal and opposite, both before and after the collision. In addition, the speed of each puck is the same before and after the collision. The direction of the velocity of each puck changes by some angle during the collision. 26 •• A baton with one end more massive than the other is tossed at an angle into the air. (a) Describe the trajectory of the center of mass of the baton in the reference frame of the ground. (b) Describe the motion of the two ends of the baton in the center-of-mass frame of the baton. Determine the Concept (a) In the center-of-mass frame of the ground, the center of mass moves in a parabolic arc. (b) Relative to the center of mass, each end of the baton would describe a circular path. The more massive end of the baton would travel in the circle with the smaller radius because it is closer to the location of the center of mass. 27 •• Describe the forces acting on a descending Lunar lander as it fires its retrorockets to slow it down for a safe landing. (Assume the lander’s mass loss during the rocket firing is not negligible.) Determine the Concept The forces acting on a descending Lunar lander are the downward force of lunar gravity and the upward thrust provided by the rocket engines. 28 •• A railroad car rolling along by itself is passing by a grain elevator, which is dumping grain into it at a constant rate. (a) Does momentum conservation imply that the railroad car should be slowing down as it passes the grain elevator? Assume that the track is frictionless and perfectly level and that the grain is falling vertically. (b) If the car is slowing down, this situation implies that there is some external force acting on the car to slow it down. Where does this force come from? (c) After passing the elevator, the railroad car springs a leak, and grain starts leaking out of a vertical hole in its floor at a constant rate. Should the car speed up as it loses mass? Determine the Concept We can apply conservation of linear momentum and Newton’s laws of motion to each of these scenarios. (a) Yes, the car should slow down. An easy way of seeing this is to imagine a "packet" of grain being dumped into the car all at once: This is a completely inelastic collision, with the packet having an initial horizontal velocity of 0. After 732 Chapter 8 the collision, it is moving with the same horizontal velocity that the car does, so the car must slow down. (b) When the packet of grain lands in the car, it initially has a horizontal velocity of 0, so it must be accelerated to come to the same speed as the car of the train. Therefore, the train must exert a force on it to accelerate it. By Newton’s third law, the grain exerts an equal but opposite force on the car, slowing it down. In general, this is a frictional force which causes the grain to come to the same speed as the car. (c) No it dos not speed up. Imagine a packet of grain being "dumped" out of the railroad car. This can be treated as a collision, too. It has the same horizontal speed as the railroad car when it leaks out, so the train car doesn’t have to speed up or slow down to conserve momentum. 29 ••• [SSM] To show that even really intelligent people can make mistakes, consider the following problem which was asked of a freshman class at Caltech on an exam (paraphrased): A sailboat is sitting in the water on a windless day. In order to make the boat move, a misguided sailor sets up a fan in the back of the boat to blow into the sails to make the boat move forward. Explain why the boat won’t move. The idea was that the net force of the wind pushing the sail forward would be counteracted by the force pushing the fan back (Newton’s third law). However, as one of the students pointed out to his professor, the sailboat could in fact move forward. Why is that? Determine the Concept Think of the sail facing the fan (like the sail on a square rigger might), and think of the stream of air molecules hitting the sail. Imagine that they bounce off the sail elastically−their net change in momentum is then roughly twice the change in momentum that they experienced going through the fan. Thus the change in momentum of the air is backward, so to conserve momentum of the air-fan-boat system, the change in momentum of the fan-boat system will be forward. Estimation and Approximation 30 •• A 2000-kg car traveling at 90 km/h crashes into an immovable concrete wall. (a) Estimate the time of collision, assuming that the center of the car travels halfway to the wall with constant acceleration. (Use any plausible length for the car.) (b) Estimate the average force exerted by the wall on the car. Picture the Problem We can estimate the time of collision from the average speed of the car and the distance traveled by the center of the car during the collision. We’ll assume a car length of 6.0 m. We can calculate the average force exerted by the wall on the car from the car’s change in momentum and its stopping time. Conservation of Linear Momentum 733 ( 1 Lcar ) = 2 (a) Relate the stopping time to the assumption that the center of the car travels halfway to the wall with constant deceleration: Δt = d stopping Because a is constant, the average speed of the car is given by: vav = vi + v f 2 Substitute numerical values and evaluate vav: km 1h 1000 m × × h 3600 s km vav = 2 = 12.5 m/s Substitute numerical values in equation (1) and evaluate Δt: Δt = vav = 1 2 vav 1 4 Lcar (1) vav 0 + 90 1 4 (6.0 m ) 12.5 m/s = 0.120 s = 0.12 s (b) Relate the average force exerted by the wall on the car to the car’s change in momentum: ⎛ Fav = Δp = Δt 1h 1000 m ⎞ ⎟ × h 3600 s km ⎟ ⎠ = 4.2 × 10 5 N 0.120 s (2000 kg )⎜ 90 km × ⎜ ⎝ 31 •• In hand-pumped railcar races, a speed of 32.0 km/h has been achieved by teams of four people. A car that has a mass equal to 350 kg is moving at that speed toward a river when Carlos, the chief pumper, notices that the bridge ahead is out. All four people (each with a mass of 75.0 kg) simultaneously jump backward off the car with a velocity that has a horizontal component of 4.00 m/s relative to the car. The car proceeds off the bank and falls into the water a horizontal distance of 25.0 m from the bank. (a) Estimate the time of the fall of the railcar. (b) What is the horizontal component of the velocity of the pumpers when they hit the ground? Picture the Problem Let the direction the railcar is moving be the +x direction and the system include Earth, the pumpers, and the railcar. We’ll also denote the railcar with the letter c and the pumpers with the letter p. We’ll use conservation of linear momentum to relate the center of mass frame velocities of the car and the pumpers and then transform to the Earth frame of reference to find the time of fall of the car. 734 Chapter 8 (a) Relate the time of fall of the railcar to the horizontal distance it travels and its horizontal speed as it leaves the bank: Δt = Δx vc (1) r r pi = pf Use conservation of momentum to find the speed of the car relative to the speed of its center of mass: or mc uc + mp u p = 0 Relate uc to up and solve for uc: u p − uc = −4.00 m/s and u p = uc − 4.00 m/s Substitute for up to obtain: mc uc + mp (u c − 4.00 m/s) = 0 4.00 m/s m 1+ c mp Solving for uc yields: uc = Substitute numerical values and evaluate uc: uc = Relate the speed of the car to its speed relative to the center of mass of the system: vc = u c + vcm 4.00 m/s = 1.846 m/s 350 kg 1+ 4(75.0 kg ) Substitute numerical values and evaluate vc: vc = 1.846 m⎛ km ⎞ ⎛ 1 h ⎞⎛ 1000 m ⎞ ⎟⎜ + ⎜ 32.0 ⎟⎜ ⎟ = 10.73 m/s s⎝ h ⎠ ⎜ 3600 s ⎟⎝ km ⎠ ⎠ ⎝ 25.0 m = 2.33 s 10.73 m/s Substitute numerical values in equation (1) and evaluate Δt: Δt = (b) The horizontal velocity of the pumpers when they hit the ground is: vp = vc − u p = 10.73 m/s − 4.00 m/s = 6.7 m/s Conservation of Linear Momentum 735 32 •• A wooden block and a gun are firmly fixed to opposite ends of a long glider mounted on a frictionless air track (Figure 8-43). The block and gun are a distance L apart. The system is initially at rest. The gun is fired and the bullet leaves the gun with a velocity vb and impacts the block, becoming imbedded in it. The mass of the bullet is mb and the mass of the gun–glider–block system is mp. (a) What is the velocity of the glider immediately after the bullet leaves the gun? (b) What is the velocity of the glider immediately after the bullet comes to rest in the block? (c) How far does the glider move while the bullet is in transit between the gun and the block? Picture the Problem Let the system include Earth, platform, gun, bullet, and r block. Then Fnet,ext = 0 and momentum is conserved within the system. Choose a coordinate system in which the +x direction is the direction of the bullet and let b and p denote the bullet and platform, respectively. r r pbefore = pafter (a) Apply conservation of linear momentum to the system just before and just after the bullet leaves the gun: or r r 0 = pbullet + pglider r r Substitute for pbullet and pglider to r ˆ 0 = m b v b i + mp v p obtain: r Solving for v p yields: (b) Apply conservation of momentum to the system just before the bullet leaves the gun and just after it comes to rest in the block: r m ˆ v p = − b vb i mp r r p before = pafter or r r 0 = pglider ⇒ vglider = 0 (c) Express the distance Δs traveled by the glider: Δs = vp Δt Express the velocity of the bullet relative to the glider: vrel = v b − vp = vb + ⎛m = ⎜1 + b ⎜m p ⎝ mb vb mp ⎞ m + mb ⎟v b = p vb ⎟ mp ⎠ 736 Chapter 8 Relate the time of flight Δt to L and vrel: Δt = L v rel Substitute and simplify to find the distance Δs moved by the glider in time Δt: ⎞ ⎛ ⎟ ⎜ ⎛ mb ⎞⎛ L ⎞ ⎛ mb ⎞ ⎜ L ⎟ ⎜ ⎟ Δs = vp Δt = ⎜ vb ⎟⎜ ⎜ v ⎟ = ⎜ m vb ⎟ ⎜ mp + mb ⎟ = ⎟ ⎜m ⎟ ⎝ p ⎠⎝ rel ⎠ ⎝ p ⎠ ⎜ vb ⎟ ⎟ ⎜m p ⎠ ⎝ ⎛ mb ⎞ ⎜ ⎟L ⎜m +m ⎟ p b⎠ ⎝ Conservation of Linear Momentum 33 • [SSM] Tyrone, an 85-kg teenager, runs off the end of a horizontal pier and lands on a free-floating 150-kg raft that was initially at rest. After he lands on the raft, the raft, with him on it, moves away from the pier at 2.0 m/s. What was Tyrone’s speed as he ran off the end of the pier? Picture the Problem Let the system include Earth, the raft, and Tyrone and apply conservation of linear momentum to find Tyrone’s speed when he ran off the end of the pier. Apply conservation of linear momentum to the system consisting of the raft and Tyrone to obtain: Because the raft is initially at rest: r r r Δpsystem = ΔpTyrone + Δpraft = 0 or, because the motion is onedimensional, pf,Tyrone − pi , Tyrone + pf,raft − pi,raft = 0 pf,Tyrone − pi , Tyrone + pf,raft = 0 Use the definition of linear momentum to obtain: mTyrone vf,Tyrone − mTyrone vi,Tyrone + mraft vf,raft = 0 Solve for vi,Tyrone to obtain: vi,Tyrone = Letting v represent the common final speed of the raft and Tyrone yields: ⎛ ⎞ m vi,Tyrone = ⎜1 + raft ⎟v ⎜m ⎟ Tyrone ⎠ ⎝ mraft vf,raft + vf,Tyrone mTyrone Conservation of Linear Momentum 737 Substitute numerical values and evaluate vi,Tyrone : ⎛ 150 kg ⎞ ⎟ (2.0 m/s ) vi, Tyrone = ⎜1 + ⎜ 85 kg ⎟ ⎝ ⎠ = 5.5 m/s 34 •• A 55-kg woman contestant on a reality television show is at rest at the south end of a horizontal 150-kg raft that is floating in crocodile-infested waters. She and the raft are initially at rest. She needs to jump from the raft to a platform that is several meters off the north end of the raft. She takes a running start. When she reaches the north end of the raft she is running at 5.0 m/s relative to the raft. At that instant, what is her velocity relative to the water? Picture the Problem Let the system include the woman, the raft, and Earth. Then the net external force is zero and linear momentum is conserved as she jumps off the raft. Let the direction the woman is running be the +x direction. Apply conservation of linear momentum to the system: r Solving for vraft yields: r ∑m v ii r r = mwoman v woman + mraft v raft = 0 r r mwoman v woman v raft = − mraft Substituting numerical values gives: ⎛ 55 kg ⎞ r r ⎛ 55 ⎞ r v raft = −⎜ ⎜ 150 kg ⎟v woman = −⎜ 150 ⎟v woman ⎟ ⎝ ⎠ ⎝ ⎠ It is given that: r r ˆ v woman − v raft = (5.0 m/s ) i r Substituting for v raft yields: r Solve for v woman to obtain: r ⎛ 55 v woman + ⎜ ⎝ 150 ⎞r ˆ ⎟v woman = (5.0 m/s ) i ⎠ ⎛ ⎞ ⎜ 5.0 m/s ⎟ r ˆ ⎟i = v woman = ⎜ 55 ⎟ ⎜ 1+ ⎜ ⎟ ⎝ 150 ⎠ (3.7 m/s) iˆ 35 • A 5.0-kg object and a 10-kg object, both resting on a frictionless table, are connected by a massless compressed spring. The spring is released and the objects fly off in opposite directions. The 5.0-kg object has a velocity of 8.0 m/s to the left. What is the velocity of the 10-kg object? 738 Chapter 8 Picture the Problem If we include Earth in our system, then the net external force is zero and linear momentum is conserved as the spring delivers its energy to the two objects. Choose a coordinate system in which the +x direction is to the right. Apply conservation of linear momentum to the system: r Solving for v10 yields: Substitute numerical values and r evaluate v10 : r ∑m v ii r r = m5 v 5 + m10 v10 = 0 ⎛ m ⎞r r v10 = ⎜ − 5 ⎟v5 ⎜m⎟ 10 ⎠ ⎝ ⎡ (5.0 kg )(− 8.0 m/s ) ⎤ ˆ r v10 = ⎢− ⎥i 10 kg ⎣ ⎦ ˆ = (4.0 m/s ) i or 4.0 m/s to the right. 36 • Figure 8-44 shows the behavior of a projectile just after it has broken up into three pieces. What was the speed of the projectile the instant before it broke up? (a) v3. (b) v3/3. (c) v3/4. (d) 4v3. (e) (v1 + v2 + v3)/4. Picture the Problem This is an explosion-like event in which linear momentum is conserved. Thus we can equate the initial and final momenta in the x direction and the initial and final momenta in the y direction. Choose a coordinate system in the +x direction is to the right and the +y direction is upward. Equate the momenta in the y direction before and after the explosion: ∑p y,i = ∑ py,f = mv2 − 2mv1 = m(2v1 ) − 2mv1 = 0 We can conclude that the momentum was entirely in the x direction before the particle exploded. ∑p = ∑ p x,f Equate the momenta in the x direction before and after the explosion: or 4mvprojectile = mv3 Solving for vprojectile yields: vprojectile = 1 v3 and 4 x,i (c ) is correct. Conservation of Linear Momentum 739 37 • A shell of mass m and speed v explodes into two identical fragments. If the shell was moving horizontally with respect to Earth, and one of the r fragments is subsequently moving vertically with speed v, find the velocity v ′ of the other fragment immediately following the explosion. Picture the Problem Choose the direction the shell is moving just before the explosion to be the positive x direction and apply conservation of momentum. Use conservation of momentum to relate the masses of the fragments to their velocities: r r pi = pf or r r ˆj ˆ mvi = 1 mvˆ + 1 mv ' ⇒ v ' = 2vi − vˆ j2 2 38 •• During this week’s physics lab, the experimental setup consists of two gliders on a horizontal frictionless air track (see Figure 8-45). Each glider supports a strong magnet centered on top of it, and the magnets are oriented so they attract each other. The mass of glider 1 and its magnet is 0.100 kg and the mass of glider 2 and its magnet is 0.200 kg. You and your lab partners take the origin to be at the left end of the track and to center glider 1 at x1 = 0.100 m and glider 2 at x2 = 1.600 m. Glider 1 is 10.0 cm long, while glider 2 is 20.0 cm long and each glider has its center of mass at its geometric center. When the two are released from rest, they will move toward each other and stick. (a) Predict the position of the center of each glider when they first touch. (b) Predict the velocity the two gliders will continue to move with after they stick. Explain the reasoning behind this prediction for your lab partners. Picture the Problem Because no external forces act on either glider, the center of mass of the two-glider system can’t move. We can use the data concerning the masses and separation of the gliders initially to calculate its location and then apply the definition of the center of mass a second time to relate the positions x1 and x2 of the centers of the carts when they first touch. We can also use the separation of the centers of the gliders when they touch to obtain a second equation in x1 and x2 that we can solve simultaneously with the equation obtained from the location of the center of mass. (a) The x coordinate of the center of mass of the 2-glider system is given by: xcm = m1 x1 + m2 x2 m1 + m2 Substitute numerical values and evaluate xcm: xcm = (0.100 kg )(0.100 m ) + (0.200 kg )(1.600 m ) = 1.10 m 0.100 kg + 0.200 kg from the left end of the air track. 740 Chapter 8 Because the location of the center of mass has not moved when two gliders first touch: 1.10 m = m1 X 1 + m2 X 2 m1 + m2 Substitute numerical values and simplify to obtain: 1.10 m = 1 X 1 + 2 X 2 3 3 Also, when they first touch, their centers are separated by half their combined lengths: X 2 − X1 = Thus we have: X 1 + 2 X 2 = 1.10 m 3 and X 2 − X 1 = 0.150 m Solving these equations simultaneously yields: X 1 = 1.00 m and X 2 = 1.15 m 1 2 (10.0 cm + 20.0 cm ) = 0.150 m 1 3 (b) Because the momentum of the system was zero initially, it must be zero just before the collision and after the collision in which the gliders stick together. Hence their velocity after the collision must be 0 . 39 •• Bored, a boy shoots his pellet gun at a piece of cheese that sits on a massive block of ice. On one particular shot, his 1.2 g pellet gets stuck in the cheese, causing it to slide 25 cm before coming to a stop. If the muzzle velocity of the gun is known to be 65 m/s, and the cheese has a mass of 120 g, what is the coefficient of friction between the cheese and ice? Picture the Problem Let the system consist of the pellet and the cheese. Then we can apply the conservation of linear momentum and the conservation of energy with friction to this inelastic collision to find the coefficient of friction between the cheese and the ice. Apply conservation of linear momentum to the system to obtain: Because the cheese is initially at rest: r r r Δpsystem = Δppellet + Δpcheese = 0 or, because the motion is onedimensional, pf,pellet − pi,pellet + pf,cheese − pi,cheese = 0 pf,pellet − pi,pellet + pf,cheeset = 0 Conservation of Linear Momentum 741 Letting m represent the mass of the pellet, M the mass of the cheese, and v the common final speed of the pellet and the cheese, use the definition of linear momentum to obtain: mv − mvi,pellet + Mv = 0 Solving for v yields: v= Apply the conservation of energy with friction to the system to obtain: Wext = ΔEmech + ΔE therm Because ΔUg = Kf = 0, and ΔE therm = fΔs (where Δs is the m vi,pellet m+M (1) or, because Wext = 0, ΔK + ΔU g + ΔE therm = 0 − 1 (m + M )v 2 + fΔs = 0 2 distance the cheese slides on the ice): f is given by: f = μ k (m + M )g Substituting for f yields: − 1 (m + M )v 2 + μ k (m + M )gΔs = 0 2 Substitute for v from equation (1) to obtain: 2 ⎛m ⎞ − (m + M )⎜ vi,pellet ⎟ + μ k (m + M )gΔs = 0 ⎝m+M ⎠ 1 2 Solving for μk yields: 1 ⎛ mvi,pellet ⎞ ⎟ ⎜ μk = 2 gΔs ⎜ m + M ⎟ ⎠ ⎝ 2 Substitute numerical values and evaluate μk: ⎛ (0.0012 kg )(65 m/s ) ⎞ 1 ⎟ = 0.084 ⎜ μk = 2 2(9.81 m/s )(0.25 m ) ⎜ 0.0012 kg + 0.120 kg ⎟ ⎠ ⎝ 2 40 ••• A wedge of mass M, as shown in Figure 8-46, is placed on a frictionless, horizontal surface, and a block of mass m is placed on the wedge, whose surface is also frictionless. The center of mass of the block moves 742 Chapter 8 downward a distance h, as the block slides from its initial position to the horizontal floor. (a) What are the speeds of the block and of the wedge, as they separate from each other and each go their own way? (b) Check your calculation plausibility by considering the limiting case when M >>m. Picture the Problem Let the system include Earth, the block, and the wedge and apply conservation of energy and conservation of linear momentum. Wext = ΔK + ΔU (a) Apply conservation of energy with no frictional forces to the system to obtain: or, because Wext = 0, ΔK + ΔU = 0 Substituting for ΔK and ΔU yields: Kf − Ki + U f −Ui = 0 Because Ki = Uf = 0: Kf −Ui = 0 Letting ″b″ refer to the block and ″w″ to the wedge yields: K b,f + K w,f − U b,i = 0 Substitute for Kb,f, Kw,f, and Ub,i to obtain: 1 2 Applying conservation of linear momentum to the system yields: 2 2 mvb + 1 Mvw − mgh = 0 2 (1) r r r Δpsys = Δpb + Δpw = 0 or r r r r pb,f − pb,i + pw,f − pw,i = 0 r r Because pb,i = pw,i = 0 : r r pb,f + pw,f = 0 r r Substituting for pb,f and pw,f yields: ˆ ˆ − mvb i + Mvw i = 0 or − mvb + Mvw = 0 Solve for vw to obtain: Substituting for vw in equation (1) yields: Solve for vb to obtain: vw = m vb M (2) 2 1 2 ⎛m ⎞ 2 mvb + 1 M ⎜ vb ⎟ − mgh = 0 2 ⎝M ⎠ vb = 2 ghM M +m (3) Conservation of Linear Momentum 743 Substitute for vb in equation (2) and simplify to obtain: vw = m M 2 ghm 2 M (M + m ) = (b) Rewriting equation (3) by dividing the numerator and denominator of the radicand by M yields: vb = 2 ghM M +m (4) 2 gh m 1+ M When M >> m: vb = 2 gh Rewriting equation (4) by dividing the numerator and denominator of the radicand by M yields: ⎛m⎞ 2 gh⎜ ⎟ ⎝M ⎠ vw = m 1+ M When M >> m: vw = 0 2 These results are exactly what we would expect in this case: The physics is that of a block sliding down a fixed wedge incline with no movement of the incline. Kinetic Energy of a System of Particles 41 •• [SSM] A 3.0-kg block is traveling to the right (the +x direction) at 5.0 m/s, and a second 3.0-kg block is traveling to the left at 2.0 m/s. (a) Find the total kinetic energy of the two blocks. (b) Find the velocity of the center of mass of the two-block system. (c) Find the velocity of each block relative to the center of mass. (d) Find the kinetic energy of the blocks relative to the center of mass. (e) Show that your answer for Part (a) is greater than your answer for Part (d) by an amount equal to the kinetic energy associated with the motion of the center of mass. Picture the Problem Choose a coordinate system in which the +x direction is to the right. Use the expression for the total momentum of a system to find the velocity of the center of mass and the definition of relative velocity to express the sum of the kinetic energies relative to the center of mass. 744 Chapter 8 (a) The total kinetic energy is the sum of the kinetic energies of the blocks: 2 K = K1 + K 2 = 1 m1v12 + 1 m2 v2 2 2 Substitute numerical values and evaluate K: K= 1 2 (3.0 kg )(5.0 m/s)2 + 1 (3.0 kg )(2.0 m/s )2 = 43.5 J = 2 (b) Relate the velocity of the center of mass of the system to its total momentum: r Solving for vcm yields: 44 J r r r Mv cm = m1v1 + m2 v 2 r r r m1v1 + m2 v 2 vcm = m1 + m2 r Substitute numerical values and evaluate vcm : r (3.0 kg )(5.0 m/s) iˆ + (3.0 kg )(− 2.0 m/s ) iˆ = (1.5 m/s) iˆ vcm = 3.0 kg + 3.0 kg (c) The velocity of an object relative to the center of mass is given by: Substitute numerical values to obtain: r rr v rel = v − v cm r ˆ ˆ v1,rel = (5.0 m/s ) i − (1.5 m/s ) i (3.5 m/s ) iˆ r ˆ ˆ v 2,rel = (− 2.0 m/s ) i − (1.5 m/s ) i = = (d) Express the sum of the kinetic energies relative to the center of mass: (− 3.5 m/s) iˆ 2 K rel = K1,rel + K 2,rel = 1 m1v12,rel + 1 m2v2,rel 2 2 Substitute numerical values and evaluate K rel : K rel = (e) K cm is given by: 1 2 (3.0 kg )(3.5 m/s)2 + 1 (3.0 kg )(− 3.5 m/s)2 = 2 2 K cm = 1 mtot vcm 2 37 J Conservation of Linear Momentum 745 Substitute numerical values and evaluate K cm : K cm = 1 2 (6.0 kg )(1.5 m/s)2 = 6.75 J = 43.5 J − 36.75 J = K − K rel 42 •• Repeat Problem 41 with the second 3.0-kg block replaced by a 5.0-kg block moving to the right at 3.0 m/s. Picture the Problem Choose a coordinate system in which the +x direction is to the right. Use the expression for the total momentum of a system to find the velocity of the center of mass and the definition of relative velocity to express the sum of the kinetic energies relative to the center of mass. (a) The total kinetic energy is the sum of the kinetic energies of the blocks: 2 K = K1 + K 2 = 1 m1v12 + 1 m2 v2 2 2 Substitute numerical values and evaluate K: K= 1 2 (3.0 kg )(5.0 m/s)2 + 1 (5.0 kg )(3.0 m/s)2 = 60.0 J = 2 (b) Relate the velocity of the center of mass of the system to its total momentum: r Solving for v cm yields: 60 J r r r Mv cm = m1v1 + m2 v 2 r r r m1v1 + m2 v 2 v cm = m1 + m2 r Substitute numerical values and evaluate vcm : r (3.0 kg )(5.0 m/s) iˆ + (5.0 kg ) (3.0 m/s) iˆ = (3.75 m/s) iˆ = v cm = 3.0 kg + 5.0 kg (c) The velocity of an object relative to the center of mass is given by: r rr v rel = v − v cm (3.8 m/s) iˆ 746 Chapter 8 Substitute numerical values to obtain: r ˆ ˆ v1,rel = (5.0 m/s ) i − (3.75 m/s ) i ˆ = (1.3 m/s ) i r ˆ ˆ v 2,rel = (3.0 m/s ) i − (3.75 m/s ) i ˆ = (− 0.75 m/s ) i (− 0.8 m/s) iˆ = (d) Express the sum of the kinetic energies relative to the center of mass: 2 K rel = K1,rel + K 2,rel = 1 m1v12,rel + 1 m2v2,rel 2 2 Substitute numerical values and evaluate K rel : K rel = 1 2 (3.0 kg )(1.25 m/s )2 + 1 (5.0 kg )(− 0.75 m/s)2 = 3.75 J = 2 (e) K cm is given by: 2 K cm = 1 mtot vcm 2 Substitute numerical values and evaluate K cm : K cm = 4J 1 2 (8.0 kg )(3.75 m/s)2 ≈ 56.3 J ≈ K − K rel Impulse and Average Force 43 • [SSM] You kick a soccer ball whose mass is 0.43 kg. The ball leaves your foot with an initial speed of 25 m/s. (a) What is the magnitude of the impulse associated with the force of your foot on the ball? (b) If your foot is in contact with the ball for 8.0 ms, what is the magnitude of the average force exerted by your foot on the ball? Picture the Problem The impulse imparted to the ball by the kicker equals the change in the ball’s momentum. The impulse is also the product of the average force exerted on the ball by the kicker and the time during which the average force acts. (a) Relate the magnitude of the impulse delivered to the ball to its change in momentum: Substitute numerical values and evaluate I: r I = Δp = pf − pi or, because vi = 0, I = mvf I = (0.43 kg )(25 m/s ) = 10.8 N ⋅ s = 11 N ⋅ s Conservation of Linear Momentum 747 (b) The impulse delivered to the ball as a function of the average force acting on it is given by: I = Fav Δt ⇒ Fav = Substitute numerical values and evaluate Fav : Fav = I Δt 10.8 N ⋅ s = 1.3 kN 0.0080 s 44 • A 0.30-kg brick is dropped from a height of 8.0 m. It hits the ground and comes to rest. (a) What is the impulse exerted by the ground on the brick during the collision? (b) If it takes 0.0013 s from the time the brick first touches the ground until it comes to rest, what is the average force exerted by the ground on the brick during impact? Picture the Problem The impulse exerted by the ground on the brick equals the change in momentum of the brick and is also the product of the average force exerted by the ground on the brick and the time during which the average force acts. (a) Express the magnitude of the impulse exerted by the ground on the brick: I = Δpbrick = pf,brick − pi,brick Because pf,brick = 0: I = pi,brick = mbrick v Use conservation of energy to determine the speed of the brick at impact: ΔK + ΔU = 0 or Because Uf = Ki = 0: Kf Ui = 0 and 2 1 2 gh 2 mbrick v − mbrick gh = 0 ⇒ v = Substitute in equation (1) to obtain: I = mbrick 2 gh Substitute numerical values and evaluate I: I = (0.30 kg ) 2 9.81m/s 2 (8.0 m ) (c) The average force acting on the brick is: Fav = (1) Kf − Ki + U f − U i = 0 ( = 3.76 N ⋅ s = 3.8 N ⋅ s I Δt ) 748 Chapter 8 Substitute numerical values and evaluate Fav: Fav = 3.76 N ⋅ s = 2.9 kN 0.0013 s 45 • A meteorite that has a mass equal to 30.8 tonne (1 tonne = 1000 kg) is exhibited in the American Museum of Natural History in New York City. Suppose that the kinetic energy of the meteorite as it hit the ground was 617 MJ. Find the magnitude of the impulse I experienced by the meteorite up to the time its kinetic energy was halved (which took about t = 3.0 s). Find also the average force F exerted on the meteorite during this time interval. Picture the Problem The impulse exerted by the ground on the meteorite equals the change in momentum of the meteorite and is also the product of the average force exerted by the ground on the meteorite and the time during which the average force acts. Express the magnitude of the impulse exerted by the ground on the meteorite: Relate the kinetic energy of the meteorite to its initial momentum and solve for its initial momentum: r I = Δpmeteorite = pf − pi Ki = pi2 ⇒ pi = 2mK i 2m Express the ratio of the initial and final kinetic energies of the meteorite: pi2 p Ki p2 = 2m = i2 = 2 ⇒ p f = i 2 Kf pf pf 2 2m Substitute in our expression for I and simplify: I= pi ⎞ ⎛1 − pi = pi ⎜ − 1⎟ 2 ⎝2 ⎠ ⎞ ⎛1 = 2mK i ⎜ − 1⎟ ⎝2 ⎠ Because our interest is in its magnitude, substitute numerical values and evaluate the absolute value of I: I= ⎛1 ⎞ 2 30.8 × 10 3 kg 617 × 10 6 J ⎜ − 1⎟ = 1.81 MN ⋅ s ⎝2 ⎠ ( The average force acting on the meteorite is: )( ) Fav = I Δt Conservation of Linear Momentum 749 Substitute numerical values and evaluate Fav: Fav = 1.81 MN ⋅ s = 0.60 MN 3.0 s 46 •• A 0.15-kg baseball traveling horizontally is hit by a bat and its direction exactly reversed. Its velocity changes from +20 m/s to –20 m/s. (a) What is the magnitude of the impulse delivered by the bat to the ball? (b) If the baseball is in contact with the bat for 1.3 ms, what is the average force exerted by the bat on the ball? Picture the Problem The impulse exerted by the bat on the ball equals the change in momentum of the ball and is also the product of the average force exerted by the bat on the ball and the time during which the bat and ball were in contact. (a) Express the impulse exerted by the bat on the ball in terms of the change in momentum of the ball: Substitute for m and v and r evaluate I : r r rr I = Δpball = pf − pi ˆ ˆ ˆ = mv i − − mv i = 2mv i f ( i ) where v = vf = vi I = 2(0.15 kg )(20 m/s ) = 6.00 N ⋅ s = 6.0 N ⋅ s (b) The average force acting on the ball is: Fav = I Δt Substitute numerical values and evaluate Fav: Fav = 6.00 N ⋅ s = 4.6 kN 1.3 ms 47 •• A 60-g handball moving with a speed of 5.0 m/s strikes the wall at an angle of 40º with the normal, and then bounces off with the same speed at the same angle with the normal. It is in contact with the wall for 2.0 ms. What is the average force exerted by the ball on the wall? 750 Chapter 8 Picture the Problem The figure shows the handball just before and immediately after its collision with the wall. Choose a coordinate system in which the +x direction is to the right. The wall changes the momentum of the ball by exerting a force on it during the ball’s collision with it. The reaction to this force is the force the ball exerts on the wall. Because these action and reaction forces are equal in magnitude, we can find the average force exerted on the ball by finding the change in momentum of the ball. Using Newton’s 3rd law, relate the average force exerted by the ball on the wall to the average force exerted by the wall on the ball: Relate the average force exerted by the wall on the ball to its change in momentum: r r Fav on wall = − Fav on ball and Fav on wall = Fav on ball (1) r r r Δp mΔv Fav on ball = = Δt Δt r Express Δv in terms of its components: r r r Δv = Δv x + Δv y r or, because Δv y = vf , y ˆ − vi , y ˆ and j j r r vf , y = vi , y , Δv = Δv x r Express Δv x for the ball: r ˆ ˆ Δv x = vf , x i − vi , x i or, because vi, x = v cosθ and vf, x = −v cosθ , r ˆ ˆ ˆ Δv x = −v cos θ i − v cos θ i = −2v cos θ i Substituting in the expression for r Fav on ball yields: r r 2mv cosθ ˆ mΔv =− Fav on ball = i Δt Δt r The magnitude of Fav on ball is: Fav on ball = 2mv cosθ Δt Conservation of Linear Momentum 751 2(0.060 kg )(5.0 m/s )cos40° 2.0 ms = 0.23 kN Substitute numerical values and evaluate Fav on ball: Fav on ball = Substitute in equation (1) to obtain: Fav on wall = 0.23 kN 48 •• You throw a 150-g ball straight up to a height of 40.0 m. (a) Use a reasonable value for the displacement of the ball while it is in your hand to estimate the time the ball is in your hand while you are throwing it. (b) Calculate the average force exerted by your hand while you are throwing it. (Is it OK to neglect the gravitational force on the ball while it is being thrown?) Picture the Problem The pictorial representation shows the ball during the interval of time during which you are exerting a force on it to accelerate it upward. The average force you exert can be determined from the change in momentum of the ball. The change in the velocity of the ball can be found by applying conservation of mechanical energy to the ball-Earth system once it has left your hand. t 2 = t1 + Δ t y2 = d v2 = ? 2 r Fav r r Fg = m g t1 = 0 1 m y1 = 0 v1 = 0 Δy (a) Relate the time the ball is in your hand to its average speed while it is in your hand and the displacement of your hand: Δt = Letting Ug = 0 at the initial elevation of your hand, use conservation of mechanical energy to relate the initial kinetic energy of the ball to its potential energy when it is at its highest point: ΔK + ΔU = 0 or, because Kf = Ui = 0, − Ki + U f = 0 Substitute for Kf and Ui and solve for v2: 2 − 1 mv2 + mgh = 0 ⇒ v2 = 2 gh 2 Because vav, in your hand = 1 v2 : 2 Δt = vav, in your hand Δy 2Δy = v2 2 gh 1 2 752 Chapter 8 2(0.70 m ) Assuming the displacement of your hand is 0.70 m as you throw the ball straight up, substitute numerical values and evaluate Δt: Δt = (b) Relate the average force exerted by your hand on the ball to the change in momentum of the ball: Δp p 2 − p1 = Δt Δt or, because v1 = p1 = 0, mv2 Fav = Δt Substitute for v2 to obtain: Substitute numerical values and evaluate Fav: ( ) 2 9.81 m/s 2 (40 m ) = 50.0 ms = 50 ms Fav = Fav = Fav = m 2 gh Δt (0.15 kg ) ( ) 2 9.81m/s 2 (40 m ) 50.0 ms = 84.04 N = 84 N Express the ratio of the gravitational force on the ball to the average force acting on it: Substitute numerical values and evaluate Fg/Fav: Fg Fav Fg Fav = mg Fav (0.15 kg )(9.81m/s2 ) < 2% = 84.04 N Because the gravitational force acting on the ball is less than 2% of the average force exerted by your hand on the ball, it is reasonable to have neglected the gravitational force. 49 •• A 0.060-kg handball is thrown straight toward a wall with a speed of 10 m/s. It rebounds straight backward at a speed of 8.0 m/s. (a) What impulse is exerted on the wall? (b) If the ball is in contact with the wall for 3.0 ms, what average force is exerted on the wall by the ball? (c) The rebounding ball is caught by a player who brings it to rest. During the process, her hand moves back 0.50 m. What is the impulse received by the player? (d) What average force was exerted on the player by the ball? Picture the Problem Choose a coordinate system in which the direction the ball is moving after its collision with the wall is the +x direction. The impulse delivered to the wall or received by the player equals the change in the momentum of the ball during these two collisions. We can find the average forces from the rate of change in the momentum of the ball. Conservation of Linear Momentum 753 (a) The impulse delivered to the wall is the change in momentum of the handball: Substitute numerical values and r evaluate I : r r r r I = Δp = mv f − mv i r ˆ I = (0.060 kg )(8.0 m/s ) i [ ˆ − − (0.060 kg )(10 m/s )i ˆ ˆ = (1.08 N ⋅ s ) i = (1.1 N ⋅ s ) i or 1.1 N⋅s directed into the wall. (b) Fav is the rate of change of the ball’s momentum: Fav = Δp Δt Substitute numerical values and evaluate Fav: Fav = 1.08 N ⋅ s = 360 N 0.0030 s = 0.36 kN, into the wall. (c) The impulse received by the player from the change in momentum of the ball is given by: Substitute numerical values and evaluate I: (d) Relate Fav to the change in the ball’s momentum: Express the stopping time in terms of the average speed vav of the ball and its stopping distance d: I = Δp ball = mΔv I = (0.060 kg )(8.0 m/s ) = 0.480 N ⋅ s = 0.48 N ⋅ s, away from the wall. Fav = Δt = Δp ball Δt d vav Substitute for Δt and simplify to obtain: Fav = Substitute numerical values and evaluate Fav : Fav = vav Δpball d (4.0 m/s)(0.480 N ⋅ s ) 0.50 m = 3.8 N, away from the wall. 754 Chapter 8 50 •• A spherical 0.34-kg orange, 2.0 cm in radius, is dropped from the top of a 35 m-tall building. After striking the pavement, the shape of the orange is a 0.50 cm thick pancake. Neglect air resistance and assume that the collision is completely inelastic. (a) How much time did the orange take to completely ″squish″ to a stop? (b) What average force did the pavement exert on the orange during the collision? Picture the Problem The following pictorial representation shows the orange moving with speed v, just before impact, after falling from a height of 35 m. Let the system be the orange and let the zero of gravitational potential energy be at the center of mass of the squished orange. The external forces are gravity, acting on the orange throughout its fall, and the normal force exerted by the ground that acts on the orange as it is squished. We can find the squishing time from the displacement of the center-of-mass of the orange as it stops and its average speed during this period of (assumed) constant acceleration. We can use the impulsemomentum theorem to find the average force exerted by the ground on the orange as it slowed to a stop. y CM just before collision R CM of squished orange Ug = 0 d w 0 (a) Express the stopping time for the orange in terms of its average speed and the distance traveled by its center of mass: R− 1w d 2 = (1) vav vav where w is the thickness of the pancaked orange. In order to find vav, apply the conservation of mechanical energy to the free-fall portion of the orange’s motion: K f − K i + U g,f − U g,i = 0 Substituting for Kf, Ug,f, and Ug,i yields: Solving for v yields: Δt = or, because Ki = 0, K f + U g,f − U g,i = 0 1 2 mv 2 + mgd − mg (h − d ) = 0 d⎞ ⎛ v = 2 gh⎜1 − 2 ⎟ h⎠ ⎝ or, because d << h, v ≈ 2 gh Conservation of Linear Momentum 755 Assuming constant acceleration as the orange squishes: vav = 1 v = 2 Substituting for vav in equation (1) and simplifying yields: Δt = Substitute numerical values and evaluate Δt: Δt = 1 2 2 gh 2(R − 1 w) 2 2 gh 2[2.0 cm − 1 (0.50 cm )] 2 ( ) 2 9.81 m/s 2 (35 m ) = 1.336 × 10 −3 s = 1.3 ms (b) Apply the impulse-momentum theorem to the squishing orange to obtain: r r rr Fav Δt = Δp = pf − pi or, because pf = 0, Fav Δt = pi ⇒ Fav = In Part (a) we showed that vi = v ≈ 2 gh . Therefore: Substitute numerical values and evaluate Fav : Fav = Fav = pi mvi = Δt Δt m 2 gh Δt (0.34 kg ) ( ) 2 9.81 m/s 2 (35 m ) 1.336 × 10 −3 s ≈ 6.7 kN 51 •• The pole-vault landing pad at an Olympic competition contains what is essentially a bag of air that compresses from its ″resting″ height of 1.2 m down to 0.20 m as the vaulter is slowed to a stop. (a) What is the time interval during which a vaulter who has just cleared a height of 6.40 m slows to a stop? (b) What is the time interval if instead the vaulter is brought to rest by a 20 cm layer of sawdust that compresses to 5.0 cm when he lands? (c) Qualitatively discuss the difference in the average force the vaulter experiences from the two different landing pads. That is, which landing pad exerts the least force on the vaulter and why? Picture the Problem The pictorial representation shows the vaulter just before impact on the landing pad after falling from a height of 6.40 m. In order to determine the time interval during which the vaulter stops, we have to know his momentum change and the average net force acting on him. With knowledge of these quantities, we can use the impulse-momentum equation, Fnet Δt = Δp . We can determine the average force by noting that as the vaulter comes to a stop on the landing pad, work is done on him by the airbag. 756 Chapter 8 y 1.2 m r v d 0.20 m (a) Use the impulse-momentum theorem to relate the stopping time to the average force acting on the vaulter: Use the work-kinetic energy theorem to obtain: Substituting for Fnet in equation (1) yields: Solve for Δt to obtain: Use K = p2 to obtain: 2m Rationalizing the denominator of this expression and simplifying yields: Fnet Δt = Δp Wnet = Fnet d = (Fairbag − mg )d = ΔK where d is the distance the vaulter moves while being decelerated. (F airbag − mg )Δt = Δp or ΔK Δt = Δp d dΔp d ( pf − pi ) = ΔK Kf − Ki or, because Kf = pf = 0, d (− pi ) pi d Δt = = Ki − Ki Δt = Δt = 2mpi d 2md 2md = = 2 pi pi 2mK i Δt = d 2m Ki (1) Conservation of Linear Momentum 757 K i = mgΔy Ki is equal to the change in the gravitational potential energy of the vaulter as he falls a distance Δy before hitting the airbag: Substituting for Ki in equation (1) and simplifying yields: Δt = d 2 gΔy Substitute numerical values and evaluate Δt: Δt = (1.2 m − 0.2 m ) 2 = 0.198 s = 0.20 s 9.81 m/s (6.4 m − 1.2 m ) ( 2 ) (b) In this case, d1 = 20 cm, d2 = 5.0 cm. Substitute numerical values and evaluate Δt: Δt = (0.20 m − 0.05 m ) 2 = 27.2 ms = 27 ms 9.81 m/s (6.4 m − 0.20 m ) ( 2 ) (c) Because the collision time is much shorter for the sawdust landing, the average force exerted on the vaulter by the airbag is much less than the average force the sawdust exerts on him. 52 ••• Great limestone caverns have been formed by dripping water. (a) If water droplets of 0.030 mL fall from a height of 5.0 m at a rate of 10 droplets per minute, what is the average force exerted on the limestone floor by the droplets of water during a 1.0-min period? (Assume the water does not accumulate on the floor.) (b) Compare this force to the weight of one water droplet. Picture the Problem The average force exerted on the limestone by the droplets of water equals the rate at which momentum is being delivered to the floor. We’re given the number of droplets that arrive per minute and can use conservation of mechanical energy to determine their velocity as they reach the floor. (a) Letting N represent the number of droplets that fall in a time interval ∆t, relate Fav to the change in the droplet’s momentum: Δpdroplets N mv (1) Δt Δt where v is their speed after falling 5.0 m from rest. Fav = = 758 Chapter 8 The mass of the droplets is the product of their density and volume: m = ρV Letting Ug = 0 at the point of impact of the droplets, use conservation of mechanical energy to relate their speed at impact to their fall distance: ΔK + ΔU = 0 or Kf − Ki + U f − U i = 0 Because Ki = Uf = 0: 1 2 Substitute for m and v in equation (1) to obtain: Fav = Substitute numerical values and evaluate Fav: Fav = 4.95 × 10 −5 N = 50 μN (b) Express the ratio of the weight of a droplet to Fav: w mg = Fav Fav Substitute numerical values and evaluate w/Fav: w 3 × 10 −5 kg 9.81m/s 2 = ≈6 Fav 4.95 × 10 −5 N mv 2 − mgh = 0 ⇒ v = 2 gh N ρV 2 gh Δt ( )( ) Collisions in One Dimension 53 • [SSM] A 2000-kg car traveling to the right at 30 m/s is chasing a second car of the same mass that is traveling in the same direction at 10 m/s. (a) If the two cars collide and stick together, what is their speed just after the collision? (b) What fraction of the initial kinetic energy of the cars is lost during this collision? Where does it go? Picture the Problem We can apply conservation of linear momentum to this perfectly inelastic collision to find the after-collision speed of the two cars. The ratio of the transformed kinetic energy to kinetic energy before the collision is the fraction of kinetic energy lost in the collision. (a) Letting V be the velocity of the two cars after their collision, apply conservation of linear momentum to their perfectly inelastic collision: pinitial = pfinal or mv1 + mv2 = (m + m )V ⇒ V = v1 + v2 2 Conservation of Linear Momentum 759 30 m/s + 10 m/s = 20 m/s 2 Substitute numerical values and evaluate V: V= (b) The ratio of the kinetic energy that is lost to the kinetic energy of the two cars before the collision is: K − K initial K final ΔK = final = −1 K initial K initial K initial Substitute for the kinetic energies and simplify to obtain: 1 (2m )V 2 ΔK = 1 2 2 1 2 −1 K initial 2 mv1 + 2 mv2 = Substitute numerical values and evaluate ΔK/Kinitial: 2V 2 −1 2 v12 + v2 ΔK 2(20 m/s ) = −1 K initial (30 m/s )2 + (10 m/s )2 = −0.20 2 Twenty percent of the initial kinetic energy is transformed into thermal energy, acoustical energy, and the deformation of the materials from which the car is constructed. 54 • An 85-kg running back moving at 7.0 m/s makes a perfectly inelastic head-on collision with a 105-kg linebacker who is initially at rest. What is the speed of the players just after their collision? Picture the Problem We can apply conservation of linear momentum to this perfectly inelastic collision to find the after-collision speed of the two players. Letting the subscript 1 refer to the running back and the subscript 2 refer to the linebacker, apply conservation of momentum to their perfectly inelastic collision: pi = p f or Substitute numerical values and evaluate V: V= m1v1 = (m1 + m2 )V ⇒ V = m1 v1 m1 + m2 85 kg (7.0 m/s ) = 3.1m/s 85 kg + 105 kg 55 • A 5.0-kg object with a speed of 4.0 m/s collides head-on with a 10-kg object moving toward it with a speed of 3.0 m/s. The 10-kg object stops dead after the collision. (a) What is the post-collision speed of the 5.0-kg object? (b) Is the collision elastic? 760 Chapter 8 Picture the Problem We can apply conservation of linear momentum to this collision to find the post-collision speed of the 5.0-kg object. Let the direction the 5.0-kg object is moving before the collision be the positive direction. We can decide whether the collision was elastic by examining the initial and final kinetic energies of the system. (a) Letting the subscript 5 refer to the 5.0-kg object and the subscript 10 refer to the 10-kg object, apply conservation of momentum to obtain: pi = p f or m5vi,5 − m10 vi,10 = m5vf,5 Solve for vf,5: vf,5 = Substitute numerical values and evaluate vf,5: vf,5 = m5vi,5 − m10 vi,10 m5 (5.0 kg )(4.0 m/s) − (10 kg )(3.0 m/s) 5 kg = − 2.0 m/s where the minus sign means that the 5.0-kg object is moving to the left after the collision. (b) Evaluate ΔK for the collision: ΔK = K f − K i = 1 2 (5.0 kg )(2.0 m/s)2 − [1 (5.0 kg )(4.0 m/s)2 + 1 (10 kg )(3.0 m/s)2 ] 2 2 = −75 J Because ΔK ≠ 0, the collision was not elastic. 56 • A small superball of mass m moves with speed v to the right toward a much more massive bat that is moving to the left with speed v. Find the speed of the ball after it makes an elastic head-on collision with the bat. Conservation of Linear Momentum 761 Picture the Problem The pictorial representation shows the ball and bat just before and just after their collision. Take the direction the bat is moving to be the positive direction. Because the collision is elastic, we can equate the speeds of recession and approach, with the approximation that vi,bat ≈ vf,bat to find vf,ball. Express the speed of approach of the bat and ball: Because the mass of the bat is much greater than that of the ball: vf, bat − vf, ball = −(vi, bat − vi, ball ) vi,bat ≈ vf,bat Substitute to obtain: vf, bat − vf, ball = −(vf, bat − vi, ball ) Solve for and evaluate vf,ball: vf,ball = vf,bat + (vf,bat − vi,ball ) = −vi,ball + 2vf,bat = v + 2v = 3v 57 •• A proton that has a mass m and is moving at 300 m/s undergoes a head-on elastic collision with a stationary carbon nucleus of mass 12m. Find the velocity of the proton and the carbon nucleus after the collision. Picture the Problem Let the direction the proton is moving before the collision be the +x direction. We can use both conservation of momentum and conservation of mechanical energy to obtain an expression for velocities of the proton and the carbon nucleus after the collision. Use conservation of linear momentum to obtain one relation for the final velocities: Use conservation of mechanical energy to set the velocity of recession equal to the negative of the velocity of approach: mp v p,i = mp v p,f + mnuc v nuc,f (1) v nuc,f − v p,f = −(v nuc,i − v p,i ) = v p,i (2) 762 Chapter 8 To eliminate vnuc,f, solve equation (2) for vnuc,f, and substitute the result in equation (1): Solving for vp,f yields: v nuc,f = v p,i + v p,f and mp v p,i = mp v p,f + mnuc (v p,i + v p,f ) vp,f = mp − mnuc mp + mnuc vp,i 11 m − 12m vp,i = − vp,i 13 m + 12m Substituting for mp and mnuc and simplifying yields: vp,f = Substitute the numerical value of vp,i and evaluate vp,f: 11 (300 m/s) = − 254 m/s 13 where the minus sign tells us that the velocity of the proton was reversed in the collision. Solving equation (2) for vnuc,f yields: vnuc,f = vp,i + vp,f Substitute numerical values and evaluate vnuc,f: vnuc,f = 300 m/s − 254 m/s vp,f = − = 46 m/s, forward 58 •• A 3.0-kg block moving at 4.0 m/s makes a head-on elastic collision with a stationary block of mass 2.0 kg. Use conservation of momentum and the fact that the relative speed of recession equals the relative speed of approach to find the velocity of each block after the collision. Check your answer by calculating the initial and final kinetic energies of each block. Picture the Problem We can use conservation of momentum and the definition of an elastic collision to obtain two equations in v2f and v3f that we can solve simultaneously. Use conservation of momentum to obtain one relation for the final velocities: Use conservation of mechanical energy to set the speed of recession equal to the negative of the speed of approach: m3 v3i = m3 v3f + m2 v 2f (1) v 2f − v3f = −(v 2i − v3i ) = v3i (2) Conservation of Linear Momentum 763 Solve equation (2) for v3f , substitute in equation (1) to eliminate v3f, and solve for v2f to obtain: v2f = 2m3v3i m2 + m3 Substitute numerical values and evaluate v2f: v2f = 2(3.0 kg )(4.0 m/s ) = 4.80 m/s 2.0 kg + 3.0 kg = 4.8 m/s v3f = v2f − v3i = 4.8 m/s − 4.0 m/s Use equation (2) to find v3f: = 0.8 m/s Evaluate Ki and Kf: 2 K i = K 3i = 1 m3 v3i = 2 1 2 (3.0 kg )(4.0 m/s)2 = 24 J and 2 2 K f = K 3f + K 2f = 1 m3 v3f + 1 m2 v2f 2 2 = 1 2 (3.0 kg )(0.8 m/s)2 + 1 (2.0 kg )(4.8 m/s)2 = 24 J 2 Because Ki = Kf, we can conclude that the values obtained for v2f and v3f are consistent with the collision having been elastic. 59 •• A block of mass m1 = 2.0 kg slides along a frictionless table with a speed of 10 m/s. Directly in front of it, and moving in the same direction with a speed of 3.0 m/s, is a block of mass m2 = 5.0 kg. A massless spring that has a force constant k = 1120 N/m is attached to the second block as in Figure 8-47. (a) What is the velocity of the center of mass of the system? (b) During the collision, the spring is compressed by a maximum amount Δx. What is the value of Δx? (c) The blocks will eventually separate again. What are the final velocities of the two blocks measured in the reference frame of the table, after they separate? Picture the Problem We can find the velocity of the center of mass from the definition of the total momentum of the system. We’ll use conservation of energy to find the maximum compression of the spring and express the initial (i.e., before collision) and final (i.e., at separation) velocities. Finally, we’ll transform the velocities from the center-of-mass frame of reference to the table frame of reference. Let the direction the block of mass m1 is moving be the +x direction. 764 Chapter 8 (a) Use the definition of the total momentum of a system to relate the initial momenta to the velocity of the center of mass: r r r P = ∑ mi vi = Mv cm i or r r r m1v1i + m2v 2i = (m1 + m2 ) vcm r vcm = Solving for vcm gives: m1 r m2 r v1i + v2i m1 + m2 m1 + m2 r Substitute numerical values and evaluate vcm : r ⎞ ⎞ ⎛ 2.0 kg 3.0 kg ˆ⎛ ˆ vcm = ⎜ ⎟ ⎜ ⎟ ⎜ 2.0 kg + 5.0 kg ⎟ (10 m/s ) i + ⎜ 2.0 kg + 5.0 kg ⎟ (3.0 m/s ) i ⎠ ⎝ ⎠ ⎝ ˆ ˆ = (5.00 m/s ) i = (5.0 m/s ) i (b) Find the kinetic energy of the system at maximum compression (u1 = u2 = 0): 2 K = K cm = 1 Mvcm 2 Use conservation of mechanical energy to relate the kinetic energy of the system to the potential energy stored in the spring at maximum compression: ΔK + ΔU s = 0 or K f − K i + U sf − U si = 0 Because Kf = Kcm and Usi = 0: K cm − K i + 1 k (Δx ) = 0 2 = 1 2 (7.0 kg )(5.00 m/s)2 = 87.5 J 2 Solving for Δx yields: [ 2 2 2 1 m1v12i + 1 m2 v2i − K cm m1v12i + m2 v2i − 2 K cm 2(K i − K cm ) 2 2 = = Δx = k k k Substitute numerical values and evaluate Δx: Δx = (2.0 kg )(10 m/s)2 + (5.0 kg )(3.0 m/s)2 − 2(87.5 J ) ⎤ = 1120 N/m 1120 N/m ⎥ ⎦ 0.25 m Conservation of Linear Momentum 765 (c) Find u1i, u2i, and u1f for this elastic collision: u1i = v1i − vcm = 10 m/s − 5 m/s = 5 m/s, u 2i = v2i − vcm = 3 m/s − 5 m/s = −2 m/s, and u1f = v1f − vcm = 0 − 5 m/s = −5 m/s Use conservation of mechanical energy to set the speed of recession equal to the negative of the speed of approach and solve for u2f: Substitute numerical values and evaluate u2f: Transform u1f and u2f to the table frame of reference: u2f − u1f = −(u2i − u1i ) and u 2f = −u 2i + u1i + u1f u 2f = −(− 2.0 m/s ) + 5.0 m/s − 5.0 m/s = 2.0 m/s v1f = u1f + vcm = −5.0 m/s + 5.0 m/s =0 and v2f = u2f + vcm = 2.0 m/s + 5.0 m/s = 7.0 m/s Express the velocities of the blocks to obtain: r r v1f = 0 and v 2f = (7.0 m/s) iˆ 60 •• A bullet of mass m is fired vertically from below into a thin horizontal sheet of plywood of mass M that is initially at rest, supported by a thin sheet of paper (Figure 8-48). The bullet punches through the plywood, which rises to a height H above the paper before falling back down. The bullet continues rising to a height h above the paper. (a) Express the upward velocity of the bullet and the plywood immediately after the bullet exits the plywood in terms of h and H. (b) What is the speed of the bullet? (c) What is the mechanical energy of the system before and after the inelastic collision? (d) How much mechanical energy is dissipated during the collision? Picture the Problem Let the system include Earth, the bullet, and the sheet of plywood. Then Wext = 0. Choose the zero of gravitational potential energy to be where the bullet enters the plywood. We can apply both conservation of energy and conservation of momentum to obtain the various physical quantities called for in this problem. (a) Use conservation of mechanical energy after the bullet exits the sheet of plywood to relate its exit speed to the height to which it rises: ΔK + ΔU = 0 or, because Kf = Ui = 0, 2 − 1 mvm + mgh = 0 ⇒ vm = 2 2 gh 766 Chapter 8 Proceed similarly to relate the initial velocity of the plywood to the height to which it rises: vM = 2 gH (b) Apply conservation of momentum to the collision of the bullet and the sheet of plywood: r r pi = pf or mvmi = mvm + MvM Substitute for vm and vM and solve for vmi: vmi = 2 gh + M m 2 gH (c) Express the initial mechanical energy of the system (i.e., just before the collision): 2 Ei = 1 mvmi 2 Express the final mechanical energy of the system (that is, when the bullet and block have reached their maximum heights): Ef = mgh + MgH = g (mh + MH ) (d) Use the work-energy theorem with Wext = 0 to find the energy dissipated by friction in the inelastic collision: Ef − Ei + Wfriction = 0 ⎡ 2M = mg ⎢h + m ⎢ ⎣ 2 ⎛M ⎞ ⎤ hH + ⎜ ⎟ H ⎥ ⎝m⎠ ⎥ ⎦ and Wfriction = Ei − Ef ⎡hM ⎤ = gMH ⎢2 + − 1⎥ m ⎣H ⎦ 61 •• A proton of mass m is moving with initial speed v0 directly toward the center of an α particle of mass 4m, which is initially at rest. Both particles carry positive charge, so they repel each other. (The repulsive forces are sufficient to prevent the two particles from coming into direct contact.) Find the speed vα of the α particle (a) when the distance between the two particles is a minimum, and (b) later when the two particles are far apart. Picture the Problem We can find the speed of the center of mass from the definition of the total momentum of the system. We’ll use conservation of energy to find the speeds of the particles when their separation is a minimum and when they are far apart. (a) Noting that when the distance between the two particles is a minimum, both move at the same speed, namely vcm , use the Conservation of Linear Momentum 767 r r r P = ∑ mi vi = Mv cm i or mp vp = (mp + mα )vcm . definition of the total momentum of a system to relate the initial momenta to the speed of the center of mass: Solve for vcm to obtain: vcm = vα = mp vp + mα vα mp + mα mv0 + 0 = 0.2 v0 m + 4m Further simplification yields: vcm = vα = (b) Use conservation of linear momentum to obtain one relation for the final speeds: mp v0 = mpvpf + mα vαf (1) vpf − vαf = −(vpi − vαi ) = −vpi (2) Use conservation of mechanical energy to set the speed of recession equal to the negative of the speed of approach: Solve equation (2) for vpf , substitute in equation (1) to eliminate vpf, and solve for vαf: Simplifying further yields: vαf = vαf = 2 mp v 0 mp + mα 2mv0 = 0.4 v0 m + 4m 62 •• An electron collides elastically with a hydrogen atom initially at rest. Assume all the motion occurs along a straight line. What fraction of the electron’s initial kinetic energy is transferred to the atom? (Take the mass of the hydrogen atom to be 1840 times the mass of an electron.) Picture the Problem Let the numeral 1 denote the electron and the numeral 2 the hydrogen atom. We can find the final velocity of the electron and, hence, the fraction of its initial kinetic energy that is transferred to the atom, by transforming to the center-of-mass reference frame, calculating the post-collision velocity of the electron, and then transforming back to the laboratory frame of reference. 768 Chapter 8 Express f, the fraction of the electron’s initial kinetic energy that is transferred to the atom: Find the velocity of the center of mass: Find the initial velocity of the electron in the center-of-mass reference frame: f= Ki − Kf K =1− f Ki Ki ⎛v m1v12f =1− = 1 − ⎜ 1f 2 ⎜v m1v1i ⎝ 1i 1 2 1 2 ⎞ ⎟ ⎟ ⎠ 2 (1) m1v1i m1 + m2 or, because m2 = 1840m1, m1v1i 1 vcm = = v1i m1 + 1840m1 1841 vcm = u1i = v1i − vcm = v1i − 1 v1i 1841 1⎞ ⎛ = ⎜1 − ⎟v1i ⎝ 1841 ⎠ Find the post-collision velocity of the electron in the center-of-mass reference frame by reversing its velocity: ⎛1 ⎞ u1f = −u1i = ⎜ − 1⎟v1i ⎝ 1841 ⎠ To find the final velocity of the electron in the original frame, add vcm to its final velocity in the centerof-mass reference frame: ⎛2 ⎞ v1f = u1f + vcm = ⎜ − 1⎟v1i ⎝ 1841 ⎠ Substituting in equation (1) and simplifying yields: ⎛2 ⎞ f = 1− ⎜ − 1⎟ = 0.217% ⎝ 1841 ⎠ 2 63 •• [SSM] A16-g bullet is fired into the bob of a 1.5-kg ballistic pendulum (Figure 8-18). When the bob is at its maximum height, the strings make an angle of 60° with the vertical. The pendulum strings are 2.3 m long. Find the speed of the bullet prior to impact. Picture the Problem The pictorial representation shows the bullet about to imbed itself in the bob of the ballistic pendulum and then, later, when the bob plus bullet have risen to their maximum height. We can use conservation of momentum during the collision to relate the speed of the bullet to the initial speed of the bob plus bullet (V). The initial kinetic energy of the bob plus bullet is transformed into gravitational potential energy when they reach their maximum height. Hence we apply conservation of mechanical energy to relate V to the angle through which the bullet plus bob swings and then solve the momentum and energy equations simultaneously for the speed of the bullet. Conservation of Linear Momentum 769 θ L cos θ m r vb L Ug = 0 M Use conservation of momentum to relate the speed of the bullet just before impact to the initial speed of the bob-bullet: ⎛ M⎞ mvb = (m + M )V ⇒ vb = ⎜1 + ⎟V (1) m⎠ ⎝ Use conservation of energy to relate the initial kinetic energy of the bobbullet to their final potential energy: ΔK + ΔU = 0 or, because Kf = Ui = 0, − Ki + U f = 0 Substitute for Ki and Uf to obtain: − 1 (m + M )V 2 2 + (m + M )gL(1 − cosθ ) = 0 Solving for V yields: V = 2 gL(1 − cos θ ) Substitute for V in equation (1) to obtain: ⎛ M⎞ vb = ⎜1 + ⎟ 2 gL(1 − cos θ ) m⎠ ⎝ Substitute numerical values and evaluate vb: ⎛ 1.5 kg ⎞ 2 v b = ⎜1 + ⎟ ⎜ 0.016 kg ⎟ 2 9.81 m/s (2.3 m )(1 − cos60°) = 0.45 km/s ⎠ ⎝ ( ) 64 •• Show that in a one-dimensional elastic collision, if the mass and velocity of object 1 are m1 and v1i, and if the mass and velocity of object 2 are m2 and v2i, then their final velocities v1f and v2f are given by 2m 2 2 m1 m − m2 m − m1 v1f = 1 v 1i + v 2i and v2f = v1i + 2 v. m1 + m2 m1 + m2 m1 + m2 m1 + m2 2i 770 Chapter 8 Picture the Problem We can apply conservation of linear momentum and the definition of an elastic collision to obtain equations relating the initial and final velocities of the colliding objects that we can solve for v1f and v2f. Apply conservation of momentum to the elastic collision of the particles to obtain: Use conservation of mechanical energy to set the speed of recession equal to the negative of the speed of approach: Multiply equation (2) by m2 and add it to equation (1) to obtain: Solve for v1f to obtain: Multiply equation (2) by m1 and subtract it from equation (1) to obtain: Solve for v2f to obtain: m1v1f + m2 v2f = m1v1i + m2 v2i (1) v2f − v1f = −(v2i − v1i ) or v1f − v2f = v2i − v1i (2) (m1 + m2 )v1f = (m1 − m2 )v1i + 2m2v2i v1 f = m1 − m2 2m2 v1i + v2 i m1 + m2 m1 + m2 (m1 + m2 )v2f = (m2 − m1 )v2i + 2m1v1i v2 f = m − m1 2m1 v1i + 2 v 2i m1 + m2 m1 + m2 Remarks: Note that the velocities satisfy the condition that v 2f − v1f = −(v 2i − v1i ) . This verifies that the speed of recession equals the speed of approach. 65 •• Investigate the plausibility of the results of Problem 64 by calculating the final velocities in the following limits: (a) When the two masses are equal, show that the particles ″swap″ velocities: v1f = v2i i and v2f = v1i (b) If m2 >> m1, and v2i = 0 , show that v1f ≈ −v1i and v2f ≈ 0 . (c) If m1 >> m2, and v2i = 0 , show that v1f ≈ v1i and v2f ≈ 2v1i. Picture the Problem As in this problem, Problem 74 involves an elastic, onedimensional collision between two objects. Both solutions involve using the conservation of momentum equation m1v1f + m2 v2f = m1v1i + m2 v2i and the elastic collision equation v1f − v2 f = v2i − v1i . In Part (a) we can simply set the masses equal to each other and substitute in the equations in Problem 64 to show that the particles "swap" velocities. In Part (b) we can divide the numerator and denominator of the equations in Problem 64 by m2 and use the condition that m2 >> m1 to show that v1f ≈ −v1i+2v2i and v2f ≈ v2i. Conservation of Linear Momentum 771 (a) From Problem 64 we have: v1f = and 2m2 m1 − m2 v1i + v2i m1 + m2 m1 + m2 (1) v2f = Set m1 = m2 = m to obtain: 2m1 m − m1 v1i + 2 v 2i (2) m1 + m2 m1 + m2 v1f = 2m v2i = v2i m+m and v2 f = 2m v1i = v1i m+m (b) Divide the numerator and denominator of both terms in equation (1) by m2 to obtain: m1 −1 2 m2 v1f = v1i + v2i m1 m1 +1 +1 m2 m2 If m2 >> m1 and v2i = 0: v1f ≈ −v1i Divide the numerator and denominator of both terms in equation (2) by m2 to obtain: v2f = If m2 >> m1: v2f ≈ v 2i (c) Divide the numerator and denominator of equation (1) by m1 to obtain: m2 m 22 m1 m1 v1f = v1i + v m2 m2 2i 1+ 1+ m1 m1 If m1 >> m2 and v2i = 0: v1f ≈ v1i Divide the numerator and denominator of equation (2) by m1 to obtain: m2 −1 m1 2 v2f = v+ v m2 1i m2 2i 1+ 1+ m1 m1 If m1 >> m2 and v2i = 0: v2f ≈ 2v1i 2 m1 m2 m1 +1 m2 1− v1i + m1 m2 m1 +1 m2 v 2i 1− 772 Chapter 8 Remarks: Note that, in both parts of this problem, the velocities satisfy the condition that v 2f − v1f = −(v 2i − v1i ) . This verifies that the speed of recession equals the speed of approach. 66 •• A bullet of mass m1 is fired horizontally with a speed v0 into the bob of a ballistic pendulum of mass m2. The pendulum consists of a bob attached to one end of a very light rod of length L. The rod is free to rotate about a horizontal axis through its other end. The bullet is stopped in the bob. Find the minimum v0 such that the bob will swing through a complete circle. Picture the Problem Choose Ug = 0 at the bob’s equilibrium position. Momentum is conserved in the collision of the bullet with bob and the initial kinetic energy of the bob plus bullet is transformed into gravitational potential energy as it swings up to the top of the circle. If the bullet plus bob just makes it to the top of the circle with zero speed, it will swing through a complete circle. Use conservation of momentum to relate the speed of the bullet just before impact to the initial speed of the bob plus bullet: m1v0 − (m1 + m2 )V = 0 Solve for the speed of the bullet to obtain: ⎛ m⎞ v 0 = ⎜1 + 2 ⎟ V ⎜ m⎟ 1⎠ ⎝ Use conservation of mechanical energy to relate the initial kinetic energy of the bob plus bullet to their potential energy at the top of the circle: ΔK + ΔU = 0 or, because Kf = Ui = 0, − Ki + U f = 0 Substitute for Ki and Uf: − 1 (m1 + m2 )V 2 + (m1 + m2 )g (2 L ) = 0 2 Solving for V yields: V = 2 gL Substitute for V in equation (1) and simplify to obtain: ⎛ m⎞ v0 = 2⎜1 + 2 ⎟ gL ⎜ m⎟ 1⎠ ⎝ (1) 67 •• A bullet of mass m1is fired horizontally with a speed v into the bob of a ballistic pendulum of mass m2 (Figure 8-19). Find the maximum height h attained by the bob if the bullet passes through the bob and emerges with a speed v/3. Conservation of Linear Momentum 773 Picture the Problem Choose Ug = 0 at the equilibrium position of the ballistic pendulum. Momentum is conserved in the collision of the bullet with the bob and kinetic energy is transformed into gravitational potential energy as the bob swings up to its maximum height. Letting V represent the initial speed of the bob as it begins its upward swing, use conservation of momentum to relate this speed to the speeds of the bullet just before and after its collision with the bob: m1v = m1 (1 v ) + m2V ⇒ V = 3 Use conservation of energy to relate the initial kinetic energy of the bob to its potential energy at its maximum height: 2m1 v 3m2 ΔK + ΔU = 0 or, because Kf = Ui = 0, − Ki + U f = 0 Substitute for Ki and Uf: Substitute for V in the expression for h and simplify to obtain: − 1 m2V 2 + m2 gh = 0 ⇒ h = 2 V2 2g 2 ⎛ 2m1 ⎞ ⎜ ⎜ 3m v ⎟ ⎟ 2m12 v 2 h=⎝ 2 ⎠ = 2 2g 9m 2 g 68 •• A heavy wooden block rests on a flat table and a high-speed bullet is fired horizontally into the block, the bullet stopping in it. How far will the block slide before coming to a stop? The mass of the bullet is 10.5 g, the mass of the block is 10.5 kg, the bullet’s impact speed is 750 m/s, and the coefficient of kinetic friction between the block and the table is 0.220. (Assume that the bullet does not cause the block to spin.) Picture the Problem Let the mass of the bullet be m, that of the wooden block M, the pre-collision speed of the bullet v, and the post-collision speed of the block+bullet be V. We can use conservation of momentum to find the speed of the block with the bullet imbedded in it immediately after their perfectly inelastic collision. We can use Newton’s second law to find the acceleration of the sliding block and a constant-acceleration equation to find the distance the block slides. 774 Chapter 8 Before m r v Immediately After r Fn r V M r fk Sometime Later M +m r (M + m )g Δx Using a constant-acceleration equation, relate the speed of the block+bullet just after their collision to their acceleration and displacement before stopping: V2 2a because the final speed of the block+bullet is zero. Use conservation of momentum to relate the pre-collision speed of the bullet to the post-collision speed of the block+bullet: mv = (m + M )V ⇒ V = Substitute for V in the expression for Δx to obtain: 1⎛ m ⎞ Δx = − ⎜ v⎟ 2a ⎝ m + M ⎠ Apply r r ∑ F = ma to the block+bullet (see the force diagram above): 0 = V 2 + 2aΔx ⇒ Δx = − ∑F x m v m+M 2 = − f k = (m + M ) a and ∑ Fy =Fn − (m + M )g = 0 Use the definition of the coefficient of kinetic friction and equation (2) to obtain: − μ k (m + M )g = (m + M ) a Solve for a to obtain: a = −μk g Substituting for a in the expression for Δx yields: Δx = (2) f k = μ k Fn = μ k (m + M )g Substituting for fk in equation (1) yields: (1) 1⎛m ⎜ 2μk g ⎝ m + M ⎞ v⎟ ⎠ 2 Substitute numerical values and evaluate Δx: 1 Δx = 2(0.220) 9.81m/s 2 ( 2 ⎛ ⎞ 0.0105 kg ⎜ ⎜ 0.0105 kg + 10.5 kg (750 m/s )⎟ = 13.0 cm ⎟ )⎝ ⎠ Conservation of Linear Momentum 775 69 •• [SSM] A 0.425-kg ball with a speed of 1.30 m/s rolls across a level surface toward an open 0.327-kg box that is resting on its side. The ball enters the box, and the box (with the ball inside slides across the surface a distance of x = 0.520 m. What is the coefficient of kinetic friction between the box and the table? Picture the Problem The collision of the ball with the box is perfectly inelastic and we can find the speed of the box-and-ball immediately after their collision by applying conservation of momentum. If we assume that the kinetic friction force is constant, we can use a constant-acceleration equation to find the acceleration of the box and ball combination and the definition of μk to find its value. Using its definition, express the coefficient of kinetic friction of the table: μk = f k (M + m ) a a = = Fn (M + m )g g Use conservation of momentum to relate the speed of the ball just before the collision to the speed of the ball+box immediately after the collision: MV = (m + M ) v ⇒ v = Use a constant-acceleration equation to relate the sliding distance of the ball+box to its initial and final velocities and its acceleration: vf2 = vi2 + 2aΔx or, because vf = 0 and vi = v, v2 0 = v 2 + 2aΔx ⇒ a = − 2Δx Substitute for a in equation (1) to obtain: v2 μk = 2 gΔ x MV (2) m+M Use equation (2) to eliminate v: ⎞ ⎛ 2 1 ⎛ MV ⎞ 1⎜V⎟ ⎟ ⎜ μk = ⎜ ⎟= 2 gΔx ⎝ m + M ⎠ 2 gΔx ⎜ m 1 ⎟ +⎟ ⎜ ⎠ ⎝M 2 Substitute numerical values and evaluate μk: 2 (1) ⎞ ⎛ ⎟ ⎜ 1 ⎜ 1.30 m/s ⎟ = 0.0529 μk = 2(9.81 m/s 2 )(0.520 m ) ⎜ 0.327 kg 1 ⎟ +⎟ ⎜ ⎠ ⎝ 0.425 kg 776 Chapter 8 70 •• Tarzan is in the path of a pack of stampeding elephants when Jane swings in to the rescue on a rope vine, hauling him off to safety. The length of the vine is 25 m, and Jane starts her swing with the rope horizontal. If Jane’s mass is 54 kg, and Tarzan’s is 82 kg, to what height above the ground will the pair swing after she rescues him? (Assume that the rope is vertical when she grabs him.) Picture the Problem Jane’s collision with Tarzan is a perfectly inelastic collision. We can find her speed v1 just before she grabs Tarzan from conservation of energy and their speed V just after she grabs him from conservation of momentum. Their kinetic energy just after their collision will be transformed into gravitational potential energy when they have reached their greatest height h. Use conservation of energy to relate the potential energy of Jane and Tarzan at their highest point (2) to their kinetic energy immediately after Jane grabbed Tarzan: Apply conservation of linear momentum to relate Jane’s velocity just before she collides with Tarzan to their velocity just after their perfectly inelastic collision: mJ 0 2 mJ+T 1 Ug = 0 h U 2 = K1 or V2 mJ+T gh = mJ+TV ⇒ h = 2g 2 1 2 mJ v1 − mJ +TV = 0 ⇒ V = (1) mJ v1 (2) mJ+T Apply conservation of mechanical energy to relate Jane’s kinetic energy at 1 to her potential energy at 0: K1 = U 0 or 2 1 2 gL 2 mJ v1 = mJ gL ⇒ v1 = Substitute for v1 in equation (2) to obtain: V= mJ 2 gL mJ+T Substitute for V in equation (1) and simplify: ⎛m ⎞ 1 ⎛ mJ ⎞ ⎜ ⎟ 2 gL = ⎜ J ⎟ L h= ⎜m ⎟ 2 g ⎜ mJ+T ⎟ ⎝ ⎠ ⎝ J +T ⎠ Substitute numerical values and evaluate h: ⎛ ⎞ 54 kg h=⎜ ⎜ 54 kg + 82 kg ⎟ (25 m ) = 3.9 m ⎟ ⎝ ⎠ 2 2 2 Conservation of Linear Momentum 777 71 •• [SSM] Scientists estimate that the meteorite responsible for the creation of Barringer Meteorite Crater in Arizona weighed roughly 2.72 × 105 tonne (1 tonne = 1000 kg) and was traveling at a speed of 17.9 km/s. Take Earth’s orbital speed to be about 30.0 km/s. (a) What should the direction of impact be if Earth’s orbital speed is to be changed by the maximum possible amount? (b) Assuming the condition of collision in Part (a), estimate the maximum percentage change in Earth’s orbital speed as a result of this collision. (c) What mass asteroid, having a speed equal to Earth’s orbital speed, would be necessary to change Earth’s orbital speed by 1.00%? Picture the Problem Let the system include Earth and the asteroid. Choose a coordinate system in which the direction of Earth’s orbital speed is the +x direction. We can apply conservation of linear momentum to the perfectly inelastic collision of Earth and the asteroid to find the percentage change in Earth’s orbital speed as well as the mass of an asteroid that would change Earth’s orbital speed by 1.00%. Note that the following solution neglects the increase in Earth’s orbital speed due to the gravitational pull of the asteroid during descent. (a) The meteorite should impact Earth along a line exactly opposite Earth’s orbital velocity vector. (b) Express the percentage change in Earth’s orbital speed as a result of the collision: Apply the conservation of linear momentum to the system to obtain: v −v v Δv = Earth f = 1 − f vEarth vEarth vEarth (1) where vf is Earth’s orbital speed after the collision. rrr Δp = pf − pi = 0 or, because the asteroid and Earth are moving horizontally, pf, x − pi, x = 0 Because the collision is perfectly inelastic: (mEarth + masteroid )vf − (mEarth vEarth − masteroid vasteroid ) = 0 778 Chapter 8 Solving for vf yields: vf = = mEarth vEarth − masteroid vasteroid mEarth vEarth m v = − asteroid asteroid mEarth + masteroid mEarth + masteroid mEarth + masteroid vEarth m 1 + asteroid mEarth masteroid vasteroid mEarth − m 1 + asteroid mEarth Because masteroid << mEarth: masteroid vasteroid −1 ⎛ masteroid ⎞ mEarth masteroid ⎜1 + ⎟ vf ≈ vEarth − = vEarth − vasteroid ⎜ masteroid mEarth mEarth ⎟ ⎝ ⎠ 1+ mEarth −1 ⎛m ⎞ Expanding ⎜1 + asteroid ⎟ binomially ⎜ mEarth ⎟ ⎝ ⎠ yields: ⎛ masteroid ⎜1 + ⎜ mEarth ⎝ −1 ⎛m ⎞ Substitute for ⎜1 + asteroid ⎟ in the ⎜ mEarth ⎟ ⎝ ⎠ expression for vf to obtain: vf ≈ vEarth − ≈ vEarth − −1 ⎞ m ⎟ = 1 − asteroid ⎟ mEarth ⎠ + higher order terms ⎛m masteroid vasteroid ⎜1 − asteroid ⎜ mEarth mEarth ⎝ masteroid vasteroid mEarth Substitute for vf in equation (1) to obtain: Δv = 1− vEarth vEarth − masteroid masteroid vasteroid vasteroid mEarth mEarth = vEarth vEarth ⎞ ⎟ ⎟ ⎠ Conservation of Linear Momentum 779 Using data found in the appendices of your text or given in the problem statement, Δv substitute numerical values and evaluate : vEarth ⎛ 10 3 kg ⎞ ⎜ 2.72 × 10 5 tonne × ⎟ (17.9 km/s ) ⎜ 1 tonne ⎟ ⎝ ⎠ Δv 5.98 × 10 24 kg = = 2.71× 10 −15 % 30.0 km/s vEarth (c) If the asteroid is to change Earth’s orbital speed by one percent: masteroid vasteroid mEarth 1 = 100 vEarth Solve for masteroid to obtain: masteroid = vEarth mEarth 100vasteroid Substitute numerical values and evaluate masteroid: masteroid = (30.0 km/s )(5.98 × 10 24 kg ) = 100(17.9 km/s ) 1.00 × 10 23 kg Remarks: The mass of this asteroid is approximately that of the moon! 72 ••• William Tell shoots an apple from his son’s head. The speed of the 125-g arrow just before it strikes the apple is 25.0 m/s, and at the time of impact it is traveling horizontally. If the arrow sticks in the apple and the arrow/apple combination strikes the ground 8.50 m behind the son’s feet, how massive was the apple? Assume the son is 1.85 m tall. Picture the Problem Let the system include Earth, the apple, and the arrow. Choose a coordinate system in which the direction the arrow is traveling before imbedding itself in the apple is the +x direction. We can apply conservation of linear momentum to express the mass of the apple in terms of the speed of the arrow-apple combination just after the collision and then use constantacceleration equations to find this post-collision speed. 780 Chapter 8 Apply conservation of linear momentum to the system to obtain: rrr Δp = pf − pi = 0 or, because the arrow and apple are moving horizontally, pf, x − pi, x = 0 (m Because the collision is perfectly inelastic (the arrow is imbedded in the apple): arrow + mapple )v x − marrow vi,arrow = 0 Solving for mapple yields: ⎛v ⎞ mapple = marrow ⎜ i,arrow − 1⎟ ⎜v ⎟ ⎝x ⎠ (1) Using constant-acceleration equations, express the horizontal and vertical displacements of the applearrow after their collision: Δx = v x Δt (2) and Substituting for Δt in equation (2) yields: Δx = v x Δy = 1 g (Δt ) ⇒ Δt = 2 2 Substitute for vx in equation (1) to obtain: mapple 2Δy g 2Δy g ⇒ v x = Δx g 2Δy ⎞ ⎛ ⎟ ⎜ ⎟ ⎜ vi,arrow = marrow ⎜ − 1⎟ ⎟ ⎜ Δx g ⎟ ⎜ 2Δy ⎠ ⎝ Substitute numerical values and evaluate mapple: mapple ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ 25.0 m/s = (0.125 kg )⎜ − 1⎟ = 101 g 2 ⎜ (8.50 m ) 9.81 m/s ⎟ ⎜ (1.85 m ) ⎟ 2 ⎝ ⎠ Explosions and Radioactive Decay 73 •• [SSM] The beryllium isotope 8Be is unstable and decays into two α particles (mα = 6.64 × 10–27 kg) and releases 1.5 × 10–14 J of energy. Determine the velocities of the two α particles that arise from the decay of a 8Be nucleus at rest, assuming that all the energy appears as kinetic energy of the particles. Conservation of Linear Momentum 781 Picture the Problem This nuclear reaction is 4Be → 2α + 1.5 × 10−14 J. In order to conserve momentum, the alpha particles will have move in opposite directions with the same velocities. We’ll use conservation of energy to find their speeds. Letting E represent the energy released in the reaction, express conservation of energy for this process: Substitute numerical values and evaluate vα: ( ) 2 2 K α = 2 1 mα vα = E ⇒ vα = 2 E mα 1.5 × 10 −14 J vα = = 1.5 × 10 6 m/s − 27 6.64 × 10 kg 74 •• The light isotope of lithium, 5Li, is unstable and breaks up spontaneously into a proton and an α particle. During this process, 3.15 × 10–13 J of energy are released, appearing as the kinetic energy of the two decay products. Determine the velocities of the proton and the α particle that arise from the decay of a 5Li nucleus at rest. (Note: The masses of the proton and alpha particle are mp = 1.67 × 10–27 kg and mα = 4mp = 6.64 × 10–27 kg.) Picture the Problem This nuclear reaction is 5Li → α + p + 3.15 × 10−13 J. To conserve momentum, the alpha particle and proton must move in opposite directions. We’ll apply both conservation of energy and conservation of momentum to find the speeds of the proton and alpha particle. Use conservation of momentum in this process to express the alpha particle’s speed in terms of the proton’s: pi = p f = 0 and 0 = mp vp − mα vα Solve for vα and substitute for mα to obtain: vα = Letting E represent the energy released in the reaction, apply conservation of energy to the process: K p + Kα = E mp mα vp = mp 4mp vp = 1 vp 4 or 2 2 1 1 2 mp vp + 2 mα vα = E 2 mp vp + 1 mα (1 vp ) = E 2 4 Substitute for vα: 1 2 Solve for vp and substitute for mα to obtain: vp = 2 32 E 16mp + mα 782 Chapter 8 Substitute numerical values and evaluate vp: vp = ( ) 32 3.15 × 10 −13 J = 1.74 × 10 7 m/s − 27 − 27 16 1.67 × 10 kg + 6.64 × 10 kg ( ) vα = 1 vp = 4 Use the relationship between vp and vα to obtain vα: 1 4 (1.74 ×10 7 m/s ) = 4.34 × 106 m/s 75 ••• A 3.00-kg projectile is fired with an initial speed of 120 m/s at an angle of 30.0º with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 1.00 kg and 2.00 kg. At 3.60 s after the explosion the 2.00-kg fragment lands on the ground directly below the point of explosion. (a) Determine the velocity of the 1.00-kg fragment immediately after the explosion. (b) Find the distance between the point of firing and the point at which the 1.00-kg fragment strikes the ground. (c) Determine the energy released in the explosion. Picture the Problem The pictorial representation shows the projectile at its maximum elevation and is moving horizontally. It also shows the two fragments resulting from the explosion. We’ll choose the system to include the projectile and Earth so that no external forces act to change the momentum of the system during the explosion. With this choice of system we can also use conservation of energy to determine the elevation of the projectile when it explodes. We’ll also find it useful to use constant-acceleration equations in our description of the motion of the projectile and its fragments. Neglect air resistance. r y v 1 φ 1 r v3 3 h r v0 θ 0 r v2 Δx (a) Use conservation of linear momentum to relate the velocity of the projectile before its explosion to the velocities of its two parts after the explosion: 2 Ug = 0 d = Δ x + Δ x' x r r pi = pf r r r m3v 3 = m1v1 + m2 v 2 ˆ ˆ m3v3 i = m1v x1i + m1v y1 ˆ − m2 v y 2 ˆ j j Conservation of Linear Momentum 783 The only way this equality can hold is if the x and y components are equal: Express v3 in terms of v0 and substitute for the masses to obtain: m3v3 = m1v x1 and m1v y1 = m2 v y 2 v x1 = 3v3 = 3v0 cosθ = 3(120 m/s )cos30.0° = 312 m/s and v y1 = 2v y 2 Using a constant-acceleration equation with the downward direction positive, relate vy2 to the time it takes the 2.00-kg fragment to hit the ground: Solving for vy2 gives: With Ug = 0 at the launch site, apply conservation of energy to the climb of the projectile to its maximum elevation h: Solving for h yields: Substitute numerical values and evaluate h: Substitute numerical values in equation (2) and evaluate vy2: Substitute numerical values in equation (1) and evaluate vy1: r Express v1 in vector form: (1) h = v y 2 Δt + 1 g (Δt ) 2 2 h − 1 g (Δt ) 2 = Δt 2 vy 2 (2) ΔK + ΔU = 0 Because Kf = Ui = 0, − K i + U f = 0 or 2 − 1 m3v y 0 + m3 gh = 0 2 2 vy0 h= 2g = (v0 sin θ )2 2g [(120 m/s)sin30.0°] 2 h= ( 2 9.81m/s 2 ) = 183.5 m 183.5 m − 1 (9.81m/s 2 )(3.60 s ) 2 = 3.60 s = 33.31m/s 2 vy2 v y1 = 2(33.31m/s ) = 66.62 m/s r ˆ v1 = vx1i + v y1 ˆ j = (312 m/s) iˆ + (66.6 m/s) ˆ j 784 Chapter 8 (b) Express the total distance d traveled by the 1.00-kg fragment: Relate Δx to v0 and the time-toexplosion: d = Δx + Δx' (3) Δx = (v0 cosθ )(Δt exp ) (4) v0 sin θ g Using a constant-acceleration equation, express Δtexp: Δtexp = Substitute numerical values and evaluate Δtexp: Δt exp = Substitute in equation (4) and evaluate Δx: Δx = (120 m/s )(cos30.0°)(6.116 s ) = 635.6 m vy0 g = (120 m/s)sin30.0° = 6.116 s 9.81 m/s 2 Relate the distance traveled by the 1.00-kg fragment after the explosion to the time it takes it to reach the ground: Δx' = vx1Δt' Using a constant-acceleration equation, relate the time Δt′ for the 1.00-kg fragment to reach the ground to its initial speed in the y direction and the distance to the ground: Δy = v y1Δt' − 1 g (Δt' ) 2 Substitute to obtain the quadratic equation: (Δt' )2 − (13.6 s )Δt' − 37.4 s 2 = 0 Solve the quadratic equation to find Δt′: Δt′ = 15.945 s Substitute in equation (3) and evaluate d: d = Δx + Δx' = Δx + v x1Δt' 2 = 635.6 m + (312 m/s )(15.945 s ) = 5.6 km (c) Express the energy released in the explosion: Eexp = ΔK = K f − K i (5) Conservation of Linear Momentum 785 Find the kinetic energy of the fragments after the explosion: 2 K f = K1 + K 2 = 1 m1v12 + 1 m2 v2 2 2 = 1 2 (1.00 kg )[(312 m/s)2 + (66.6 m/s)2 ] 2 + 1 (2.00 kg )(33.3 m/s ) 2 = 52.0 kJ Find the kinetic energy of the projectile before the explosion: 2 K i = 1 m3 v3 = 1 m3 (v0 cosθ ) 2 2 2 = 1 2 (3.00 kg )[(120 m/s)cos 30°] 2 = 16.2 kJ Substitute in equation (5) to determine the energy released in the explosion: Eexp = K f − K i = 52.0 kJ − 16.2 kJ = 35.8 kJ 76 ••• The boron isotope 9B is unstable and disintegrates into a proton and two α particles. The total energy released as kinetic energy of the decay products is 4.4 × 10–14 J. After one such event, with the 9B nucleus at rest prior to decay, the velocity of the proton is measured as 6.0 × 106 m/s. If the two α particles have equal energies, find the magnitude and the direction of their velocities with respect to the direction of the proton. Picture the Problem This nuclear reaction is 9 B → 2α + p + 4.4×10−14 J. Assume that the proton moves in the –x direction as shown in the diagram. The sum of the kinetic energies of the decay products equals the energy released in the decay. We’ll use conservation of momentum to find the angle between the velocities of the proton and the alpha particles. Note that vα = v'α . Express the energy released to the kinetic energies of the decay products: y r vα α r vp θ p x θ α r vα ' K p + 2 Kα = Erel or 2 2 1 1 2 mp v p + 2 2 mα vα = E rel ( ) 786 Chapter 8 Solving for vα yields: vα = 2 Erel − 1 mp vp 2 mα Substitute numerical values and evaluate vα: 4.4 × 10 −14 J vα = − 6.64 × 10 − 27 kg 1 2 (1.67 ×10 − 27 )( kg 6.0 × 10 6 m/s 6.64 × 10 − 27 kg ) 2 = 1.44 × 10 6 m/s = 1.4 × 10 6 m/s Given that the boron isotope was at rest prior to the decay, use conservation of momentum to relate the momenta of the decay products: Substituting for mα to obtain: Solving for θ yields: Substitute numerical values and evaluate θ : Let θ ′ equal the angle the velocities of the alpha particles make with that of the proton: r r pf = pi = 0 or, because p xf = 0 , 2(mα vα cosθ ) − mp vp = 0 2(4mp vα cos θ ) − mp vp = 0 ⎡ vp ⎤ ⎥ ⎣ 8vα ⎦ θ = cos −1 ⎢ ⎡ 6.0 × 106 m/s ⎤ ⎥ = ±59° 6 ⎣ 8 1.44 × 10 m/s ⎦ θ = cos −1 ⎢ ( ) θ ' = ±(180° − 59°) = ± 121° Coefficient of Restitution 77 • [SSM] You are in charge of measuring the coefficient of restitution for a new alloy of steel. You convince your engineering team to accomplish this task by simply dropping a small ball onto a plate, both the ball and plate made from the experimental alloy. If the ball is dropped from a height of 3.0 m and rebounds to a height of 2.5 m, what is the coefficient of restitution? Picture the Problem The coefficient of restitution is defined as the ratio of the speed of recession to the speed of approach. These speeds can be determined from the height from which the ball was dropped and the height to which it rebounded by using conservation of mechanical energy. Conservation of Linear Momentum 787 Use its definition to relate the coefficient of restitution to the speeds of approach and recession: e= Letting Ug = 0 at the surface of the steel plate, apply conservation of energy to obtain: ΔK + U = 0 or, because Ki = Uf = 0, Kf −Ui = 0 vrec vapp 2 mvapp − mghapp = 0 Substituting for Kf and Ui yields: 1 2 Solving for vapp yields: vapp = 2 ghapp In like manner, show that: vrec = 2 ghrec Substitute in the equation for e to obtain: e= Substitute numerical values and evaluate e: e= 2 ghrec 2 ghapp = hrec happ 2.5 m = 0.91 3.0 m 78 • According to the official racquetball rules, to be acceptable for tournament play, a ball must bounce to a height of between 173 and 183 cm when dropped from a height of 254 cm at room temperature. What is the acceptable range of values for the coefficient of restitution for the racquetball–floor system? Picture the Problem The coefficient of restitution is defined as the ratio of the speed of recession to the speed of approach. These speeds can be determined from the heights from which the ball was dropped and the height to which it rebounded by using conservation of mechanical energy. Use its definition to relate the coefficient of restitution to the speeds of approach and recession: e= Letting Ug = 0 at the surface of the steel plate, the mechanical energy of the ball-Earth system is: ΔK + ΔU = 0 or, because Ki = Uf = 0, Kf −Ui = 0 Substituting for Kf and Ui yields: 1 2 vrec vapp 2 mvapp − mghapp = 0 ⇒ vapp = 2 ghapp 788 Chapter 8 In like manner, show that: vrec = 2 ghrec Substitute in the equation for e to obtain: e= Substitute numerical values and evaluate emin: emin = 173 cm = 0.825 254 cm Substitute numerical values and evaluate emax: emax = 183 cm = 0.849 254 cm 2 ghrec 2 ghapp = hrec happ and 0.825 ≤ e ≤ 0.849 79 •• A ball bounces to 80 percent of its original height. (a) What fraction of its mechanical energy is lost each time it bounces? (b) What is the coefficient of restitution of the ball–floor system? Picture the Problem Because the rebound kinetic energy is proportional to the rebound height, the percentage of mechanical energy lost in one bounce can be inferred from knowledge of the rebound height. The coefficient of restitution is defined as the ratio of the speed of recession to the speed of approach. These speeds can be determined from the heights from which an object was dropped and the height to which it rebounded by using conservation of mechanical energy. (a) We know, because the mechanical energy of the ball-Earth system is constant, that the kinetic energy of an object dropped from a given height h is proportional to h. If, for each bounce of the ball, hrec = 0.80happ, 20% of its mechanical energy is lost. (b) Use its definition to relate the coefficient of restitution to the speeds of approach and recession: e= Letting Ug = 0 at the surface from which the ball is rebounding, the mechanical energy of the ball is: ΔK + ΔU = 0 or, because Ki = Uf = 0, Kf −Ui = 0 vrec vapp (1) 2 mvapp − mghapp = 0 ⇒ vapp = 2 ghapp Substituting for Kf and Ui yields: 1 2 In like manner, show that: vrec = 2 ghrec Conservation of Linear Momentum 789 Substitute for vrec and vapp in equation (1) to obtain: Substitute for 2 ghrec e= 2 ghapp = hrec happ e = 0.80 = 0.89 hrec to obtain: happ 80 •• A 2.0-kg object moving to the right at 6.0 m/s collides head-on with a 4.0-kg object that is initially at rest. After the collision, the 2.0-kg object is moving to the left at 1.0 m/s. (a) Find the velocity of the 4.0-kg object after the collision. (b) Find the energy lost in the collision. (c) What is the coefficient of restitution for these objects? Picture the Problem Let the numerals 2 and 4 refer, respectively, to the 2.0-kg object and the 4.0-kg object. Choose a coordinate system in which the direction the 2.0-kg object is moving before the collision is the +x direction and let the system consist of Earth, the surface on which the objects slide, and the objects. Then we can use conservation of momentum to find the velocity of the recoiling 4.0-kg object. We can find the energy transformed in the collision by calculating the difference between the pre- and post-collision kinetic energies and find the coefficient of restitution from its definition. (a) Use conservation of linear momentum in one dimension to relate the initial and final momenta of the participants in the collision: r r pi = pf or m2 v 2 i = m4 v 4 f − m2 v 2 f Solve for the final velocity of the 4.0-kg object: v4 f = Substitute numerical values and evaluate v4f: v4 f = m 2 v 2 i + m2 v 2 f m4 (2.0 kg )(6.0 m/s + 1.0 m/s) 4.0 kg = 3.50 m/s = 3.5 m/s (b) Express the energy lost in terms of the kinetic energies before and after the collision: Elost = K i − K f 2 2 2 = 1 m2 v2i − (1 m2 v2f + 1 m4 v4 f ) 2 2 2 = 1 2 [m (v 2 2 2i 2 2 − v2 f ) − m4 v4 f Substitute numerical values and evaluate Elost: Elost = 1 2 [((2.0 kg ){(6.0 m/s) 2 − (1.0 m/s ) 2 })− (4.0 kg )(3.50 m/s) ] = 2 11 J 790 Chapter 8 (c) From the definition of the coefficient of restitution we have: e= vrec v4 f − v2f = vapp v2i Substitute numerical values and evaluate e: e= 3.50 m/s − (− 1.0 m/s) = 0.75 6.0 m/s 81 •• A 2.0-kg block moving to the right with speed of 5.0 m/s collides with a 3.0-kg block that is moving in the same direction at 2.0 m/s, as in Figure 8-49. After the collision, the 3.0-kg block moves to the right at 4.2 m/s. Find (a) the velocity of the 2.0-kg block after the collision and (b) the coefficient of restitution between the two blocks. Picture the Problem Let the numeral 2 refer to the 2.0-kg block and the numeral 3 to the 3.0-kg block. Choose a coordinate system in which the direction the blocks are moving before the collision is the +x direction and let the system consist of Earth, the surface on which the blocks move, and the blocks. Then we can use conservation of momentum to find the speed of the 2.0-kg block after the collision. We can find the coefficient of restitution from its definition. r r pi = p f (a) Use conservation of linear momentum in one dimension to relate the initial and final momenta of the participants in the collision: or m2 v2i + m3 v3i = m2 v2 f + m3 v3f Solve for the final speed of the 2.0kg object: v2f = m2v2i + m3v3i − m3v3f m2 Substitute numerical values and evaluate v2f: v2 f = (2.0 kg )(5.0 m/s) + (3.0 kg )(2.0 m/s − 4.2 m/s) = 1.70 m/s = 2.0 kg 1.7 m/s (b) From the definition of the coefficient of restitution we have: e= vrec v3f − v2f = vapp v2i − v3i Substitute numerical values and evaluate e: e= 4.2 m/s − 1.7 m/s = 0.83 5.0 m/s − 2.0 m/s 82 ••• To keep homerun records and distances consistent from year to year, organized baseball randomly checks the coefficient of restitution between new baseballs and wooden surfaces similar to that of an average bat. Suppose you are in charge of making sure that no ″juiced″ baseballs are produced. (a) In a random Conservation of Linear Momentum 791 test, you find one that when dropped from 2.0 m rebounds 0.25 m. What is the coefficient of restitution for this ball? (b) What is the maximum distance home run shot you would expect from this ball, neglecting any effects due to air resistance and making reasonable assumption for bat speeds and incoming pitch speeds? Is this a ″juiced″ ball, a ″normal″ ball, or a ″dead″ ball? Picture the Problem The coefficient of restitution is defined as the ratio of the speed of recession to the speed of approach. These speeds can be determined from the heights from which the ball was dropped and the height to which it rebounded by using conservation of mechanical energy. We can use the same elevation range equation to find the maximum home run you would expect from the ball with the experimental coefficient of restitution. (a) The coefficient of restitution is the ratio of the speeds of approach and recession: e= Letting Ug = 0 at the surface of from which the ball rebounds, the mechanical energy of the ball-Earth system is: ΔK + ΔU = 0 or, because Ki = Uf = 0, Kf −Ui = 0 vrec vapp (1) 2 mvapp − mghapp = 0 ⇒ vapp = 2 ghapp Substituting for Kf and Ui yields: 1 2 In like manner, show that: vrec = 2 ghrec Substitute for vrec and vapp in equation (1) and simplify to obtain: e= Substitute numerical values and evaluate e: e= (b) The ″same-elevation″ range equation is: R= 2 ghrec 2 ghapp = hrec happ 0.25 m = 0.3536 = 0.35 2.0 m 22 2 vrec sin 2θ e vapp sin 2θ = (2) g g vapp is the sum of the speed of the ball and the speed of the bat: vapp = vball + vbat Assuming that the bat travels about 1 m in 0.2 s yields: vbat = 1m = 5 m/s 0.2 s 792 Chapter 8 Assuming that the speed of the baseball thrown by the pitcher is close to 100 mi/h yields: vball ≈ 45 m/s Evaluate vapp to obtain: vapp = 45 m/s + 5 m/s = 50 m/s Assuming a 45° launch angle, substitute numerical values in equation (2) and evaluate R: R= (0.3536)2 (50 m/s)2 sin 2(45°) 9.81 m/s 2 ≈ 32 m Because home runs must travel at least 100 m in modern major league ballparks, this is a ″dead″ ball and should be tossed out. 83 •• [SSM] To make puck handling easy, hockey pucks are kept frozen until they are used in the game. (a) Explain why room temperature pucks would be more difficult to handle on the end of a stick than a frozen puck. (Hint: Hockey pucks are made of rubber.) (b) A room-temperature puck rebounds 15 cm when dropped onto a wooden surface from 100 cm. If a frozen puck has only half the coefficient of restitution of a room-temperature one, predict how high the frozen puck would rebound under the same conditions. Picture the Problem The coefficient of restitution is defined as the ratio of the speed of recession to the speed of approach. These speeds can be determined from the heights from which the ball was dropped and the height to which it rebounded by using conservation of mechanical energy. (a) At room-temperature rubber will bounce more when it hits a stick than it will at freezing temperatures. (b) The mechanical energy of the rebounding puck is constant: If the puck’s speed of recession is vrec and it rebounds to a height h, then: The coefficient of restitution is the ratio of the speeds of approach and recession: ΔK + ΔU = 0 or, because Kf = Ui = 0, − Ki + U f = 0 − mv 1 2 e= 2 rec + mgh = 0 ⇒ h = vrec ⇒ vrec = evapp vapp 2 vrec 2g (1) Conservation of Linear Momentum 793 Substitute for vrec to obtain: Letting Ug = 0 at the surface of from which the puck rebounds, the mechanical energy of the puck-Earth system is: h= 2 e 2 vapp (2) 2g ΔK + ΔU = 0 Because Ki = Uf = 0, Kf −Ui = 0 2 mvapp − mghapp = 0 ⇒ vapp = 2 ghapp Substituting for Kf and Ui yields: 1 2 In like manner, show that: vrec = 2 ghrec Substitute for vrec and vapp in equation (1) and simplify to obtain: e= Substitute numerical values and evaluate eroom temp: eroom temp = For the falling puck, vapp is given by: 2 ghrec 2 ghapp = hrec happ 15 cm = 0.3873 100 cm vapp = 2 gH where H is the height from which the puck was dropped. 2e 2 gH = e2 H 2g Substituting for vapp in equation (2) and simplifying yields: h= For the room-temperature puck: efrozen = 1 eroom temp 2 Substituting for e in equation (3) yields: 2 h = 1 eroom temp H 4 Substitute numerical values and evaluate h: h= 1 4 (0.3873)2 (100 cm) = (3) 3.8 cm Remarks: The puck that rebounds only 3.8 cm is a much ″deader″ and, therefore, much better puck. Collisions in More Than One Dimension 84 •• In Section 8-3 it was proven by using geometry that when a particle elastically collides with another particle of equal mass that is initially at rest, the 794 Chapter 8 two post-collision velocities are perpendicular. Here we examine another way of proving this result that illustrates the power of vector notation. (a) Given that r r r A = B + C , square both sides of this equation (obtain the scalar product of each rr side with itself) to show that A 2 r= B 2 + C 2 + 2 B ⋅ C . (b) Let the momentum of the initially moving particle be P and the momenta of the particles after the r r collision be p 1 and p 2. Write the vector equation for the conservation of linear momentum and square both sides (obtain the dot product of each side with itself). Compare it to the equation gotten from the elastic-collision condition (kinetic energy is conserved) and finally show that these two equations imply that rr p 1 ⋅ p 2 = 0. Picture the Problem We can use the definition of the magnitude of a vector and the definition of the scalar product to establish the result called for in (a). In Part (b) we can use the result of Part (a), the conservation of momentum, and the definition of an elastic collision (kinetic energy is conserved) to show that the particles separate at right angles. rr (a) Find the dot product of B + C with itself: r r (B + C ) ⋅ (B + C ) = B rrr Because A = B + C : r r2 rrrr A2 = B + C = B + C ⋅ B + C ( r r 2 ( )( ) rrrr Substitute for B + C ⋅ B + C to obtain: rr + C 2 + 2B ⋅ C )( ) rr A2 = B 2 + C 2 + 2 B ⋅ C (b) Apply conservation of momentum to the collision of the particles: rr r p1 + p2 = p Form the scalar product of each side of this equation with itself to obtain: rr rr ( p1 + p2 ) ⋅ ( p1 + p2 ) = Use the definition of an elastic collision to obtain: Subtract equation (1) from equation (2) to obtain: rr p⋅ p or rr 2 p12 + p2 + 2 p1 ⋅ p2 = p 2 (1) 2 p12 p2 p2 + = 2m 2m 2m or 2 p12 + p 2 = p 2 (2) rr 2 p1 ⋅ p2 = 0 or rr p1 ⋅ p2 = 0 That is, the particles move apart along paths that are at right angles to each other. Conservation of Linear Momentum 795 85 •• During a pool game, the cue ball, which has an initial speed of 5.0 m/s, makes an elastic collision with the eight ball, which is initially at rest. After the collision, the eight ball moves at an angle of 30º to the right of the original direction of the cue ball. Assume that the balls have equal masses. (a) Find the direction of motion of the cue ball immediately after the collision. (b) Find the speed of each ball immediately after the collision. Picture the Problem Let the initial direction of motion of the cue ball be the +x direction. We can apply conservation of energy to determine the angle the cue ball makes with the +x direction and the conservation of momentum to find the final velocities of the cue ball and the eight ball. (a) Use conservation of energy to relate the velocities of the collision participants before and after the collision: 1 2 2 2 2 mvci = 1 mvcf + 1 mv8 2 2 or 2 2 2 vci = vcf + v8 This Pythagorean relationship rr r tells us that v ci , v cf , and v8 form a θ cf + θ 8 = 90° right triangle. Hence: θ cf = 60° (b) Use conservation of momentum in the x direction to relate the velocities of the collision participants before and after the collision: and r r pxi = pxf or mvci = mvcf cos θ cf + mv8 cos θ 8 r r pyi = pyf Use conservation of momentum in the y direction to obtain a second equation relating the velocities of the collision participants before and after the collision: or 0 = mvcf sin θ cf + mv8 sin θ 8 Solve these equations simultaneously to obtain: vcf = 2.50 m/s and v8 = 4.33 m/s 86 •• Object A, which has a mass m and a velocity v0 ˆ collides with object i 1 ˆ . Following the collision, object B B, which has a mass 2m and a velocity 2 v0 j ˆ has a velocity of 1 v i . (a) Determine the velocity of object A after the collision. 4 0 (b) Is the collision elastic? If not, express the change in the kinetic energy in terms of m and v0. 796 Chapter 8 Picture the Problem We can find the final velocity of the object whose mass is m by using the conservation of momentum. Whether the collision was elastic can be decided by examining the difference between the initial and final kinetic energy of the interacting objects. (a) Use conservation of linear momentum to relate the initial and final velocities of the two objects: r r pi = pf or r ˆ ˆ mv0 i + 2m 1 v0 ˆ = 2m 1 v0 i + mv1f j 4 2 Simplify to obtain: r ˆ v0 i + v0 ˆ = 1 v0 i + v1f j2ˆ r Solving for v1f yields: r v1f = (b) Express the difference between the kinetic energy of the system before the collision and its kinetic energy after the collision: Substituting for the kinetic energies yields: Substitute for speeds and simplify to obtain: () 1 2 () ˆ v0 i + v0 ˆ j ΔE = K i − K f = K1i + K 2i − (K1f + K 2f ) ΔE = 1 2 (mv 2 1i 2 2 + 2mv2i − mv12f − 2mv2f ) [ 2 2 2 12 ΔE = 1 m v0 + 2(1 v0 ) − 5 v0 − 2(16 v0 ) 4 4 2 = 1 16 2 mv0 Because ΔE ≠ 0, the collision is inelastic. 87 •• [SSM] A puck of mass 5.0 kg moving at 2.0 m/s approaches an identical puck that is stationary on frictionless ice. After the collision, the first puck leaves with a speed v1 at 30º to the original line of motion; the second puck leaves with speed v2 at 60º, as in Figure 8-50. (a) Calculate v1 and v2. (b) Was the collision elastic? Picture the Problem Let the direction of motion of the puck that is moving before the collision be the +x direction. Applying conservation of momentum to the collision in both the x and y directions will lead us to two equations in the unknowns v1 and v2 that we can solve simultaneously. We can decide whether the collision was elastic by either calculating the system’s kinetic energy before and after the collision or by determining whether the angle between the final velocities is 90°. Conservation of Linear Momentum 797 (a) Use conservation of linear momentum in the x direction to obtain: p xi = p xf or mv = mv1 cos 30° + mv2 cos 60° Simplify further to obtain: v = v1 cos 30° + v2 cos 60° Use conservation of momentum in the y direction to obtain a second equation relating the velocities of the collision participants before and after the collision: p yi = p yf or 0 = mv1 sin 30° − mv2 sin 60° Simplifying further yields: 0 = v1 sin 30° − v2 sin 60° Solve equations (1) and (2) simultaneously to obtain: v1 = 1.7 m/s and v2 = 1.0 m/s (1) (2) r r (b) Because the angle between v1 and v 2 is 90°, the collision was elastic. Figure 8-51 shows the result of a collision between two objects of 88 •• unequal mass. (a) Find the speed v2 of the larger mass after the collision and the angle θ2. (b) Show that the collision is elastic. Picture the Problem Let the direction of motion of the object that is moving before the collision be the +x direction. Applying conservation of momentum to the motion in both the x and y directions will lead us to two equations in the unknowns v2 and θ2 that we can solve simultaneously. We can show that the collision was elastic by showing that the system’s kinetic energy before and after the collision is the same. (a) Use conservation of linear momentum in the x direction to relate the velocities of the collision participants before and after the collision: Use conservation of linear momentum in the y direction to obtain a second equation relating the velocities of the collision participants before and after the collision: pxi = pxf or 3mv0 = 5mv0 cos θ1 + 2mv2 cos θ 2 or 3v0 = 5v0 cos θ1 + 2v2 cos θ 2 p yi = p yf or 0 = 5mv0 sin θ1 − 2mv 2 sin θ 2 798 Chapter 8 Simplifying further yields: Note that if tanθ1 = 2, then: Substitute in the x-direction momentum equation and simplify to obtain: Substitute in the y-direction momentum equation and simplify to obtain: 0 = 5v0 sin θ1 − 2v 2 sin θ 2 cosθ 1 = 1 and sin θ 1 = 5 3v0 = 5v0 2 5 1 + 2v2 cos θ 2 5 or v0 = v2 cos θ 2 (1) 2 − 2v2 sin θ 2 5 0 = 5v0 or 0 = v0 − v2 sin θ 2 (2) Solve equations (1) and (2) simultaneously for θ2 : θ 2 = tan −1 (1) = 45.0° Substitute in equation (1) to find v2 : v2 = (b) To show that the collision was elastic, find the before-collision and after-collision kinetic energies: v0 v0 = = cosθ 2 cos 45° 2v 0 2 K i = 1 m(3v0 ) = 4.5mv0 2 2 and ( K f = 1 m 5v0 2 )+ 2 1 2 (2m )( 2v 0 ) 2 2 = 4.5mv0 Because Ki = Kf, the collision is elastic. 89 •• A 2.0-kg ball moving at 10 m/s makes an off-center collision with a 3.0-kg ball that is initially at rest. After the collision, the 2.0-kg ball is deflected at an angle of 30º from its original direction of motion and the 3.0-kg ball is moving at 4.0 m/s. Find the speed of the 2.0-kg ball and the direction of the 3.0-kg ball after the collision. Hint: sin 2 α + cos 2 α = 1 . Picture the Problem Let the direction of motion of the ball that is moving before the collision be the +x direction and use the subscripts 2 and 3 to designate the 2.0-kg and 3.0-kg balls, respectively. Applying conservation of momentum to the collision in both the x and y directions will lead us to two equations in the unknowns v 2f and θ that we can solve simultaneously. Conservation of Linear Momentum 799 y v 3f v 2i = 10 m/s m2 0 = 4. m3 v3i = 0 m3 m/s θ x 30° v2 m2 f =? m2 v2i = m2 v2f cos 30° + m3v3f sin θ Use conservation of momentum in the x and y directions to relate the speeds and directions of the balls before and after the collision: and 0 = m3v3f sin θ − m2 v2f sin 30° Solve the first of these equations for cosθ to obtain: cosθ = m2 v2i − m2 v2f cos 30° (1) m3v3f Solve the second of these equations for sinθ to obtain: sin θ = m2 v2f sin 30° m3 v3f (2) Using the hint given in the problem statement, square and add equations (1) and (2) and simplify the result to obtain the quadratic equation: 2 v2f sin 2 30° + (v2i − v2f cos 30°) = 2 Substituting numerical values and simplifying yields: 22 m3 v3f 2 m2 2 v2f + (10 m/s − 0.866v 2f ) = 144 m 2 / s 2 2 Use the quadratic formula or your graphing calculator to obtain: v2f = 5.344 m/s or 11.977 m/s Because the larger of these values corresponds to there being more kinetic energy in the system after the collision than there was before the collision: v2f = 5.3 m/s 800 Chapter 8 Solving equation (2) for θ yields: ⎡ m2 v 2f sin 30° ⎤ ⎥ m3 v3f ⎣ ⎦ θ = sin −1 ⎢ ⎡ (2.0 kg ) (5.344 m/s )sin 30° ⎤ (3.0 kg )(4.0 m/s ) ⎥ ⎣ ⎦ Substitute numerical values and evaluate θ : θ = sin −1 ⎢ = 26° 90 •• A particle has initial speed v0. It collides with a second particle with the same mass that is initially at rest. The first particle is deflected through an angle φ. Its speed after the collision is v. The second particle recoils, and its velocity makes an angle θ with the initial direction of the first particle. (a) Show v sin φ . (b) Show that if the collision is elastic, then v = v0 cos that tan θ = (v 0 − v cos φ ) φ. Picture the Problem Choose the coordinate system shown in the following diagram with the +x direction the direction of the initial approach of the projectile particle. Call V the speed of the target particle after the collision. In Part (a) we can apply conservation of momentum in the x and y directions to obtain two equations that we can solve simultaneously for tanθ. In Part (b) we can use conservation of momentum in vector form and the elastic-collision equation to show that v = v0cosφ. r v y 1 r v0 1 2 φ x θ 2 r V (a) Apply conservation of linear momentum in the x direction to obtain: v0 = v cos φ + V cos θ (1) Apply conservation of linear momentum in the y direction to obtain: v sin φ = V sin θ (2) Solve equation (1) for Vcosθ : V cos θ = v0 − v cos φ (3) Conservation of Linear Momentum 801 Divide equation (2) by equation (3) to obtain: (b) Noting that the masses of the particles are equal, apply conservation of linear momentum to obtain: V sin θ v sin φ = V cos θ v0 − v cos φ or v sin φ tan θ = v0 − v cos φ r rr v0 = v + V r v0 Draw the vector diagram representing this equation: θ φ r V Use the definition of an elastic collision to obtain: 2 v0 = v 2 + V 2 If this Pythagorean condition is to hold, the third angle of the triangle must be a right angle and, using the definition of the cosine function: r v v = v0 cos φ *Center-of-Mass Reference Frame 91 •• In the center-of-mass reference frame a particle with mass m1 and momentum p1 makes an elastic head-on collision with a second particle of mass ′ m2 and momentum p2 = –p1. After the collision its momentum is p1 . Write the total kinetic energy in terms of m1, m2, and p1 and the total final energy in terms ′ of m1, m2, and p1 , and show that p1' = ± p1 . If p1' = − p1 , the particle is merely turned around by the collision and leaves with the speed it had initially. What is the situation for the p1' = + p1 solution? Picture the Problem The total kinetic energy of a system of particles is the sum of the kinetic energy of the center of mass and the kinetic energy relative to the center of mass. The kinetic energy of a particle of mass m is related to its momentum according to K = p 2 2m . Express the total kinetic energy of the system: K = K rel + K cm (1) 802 Chapter 8 Relate the kinetic energy relative to the center of mass to the momenta of the two particles: K rel = p12 p2 p 2 (m + m2 ) + 1=1 1 2m1 2m2 2m1m2 Express the kinetic energy of the center of mass of the two particles: (2 p1 )2 = 2 p12 K cm = 2(m1 + m2 ) m1 + m2 Substitute in equation (1) and simplify to obtain: K= = In an elastic collision: p12 (m1 + m2 ) 2 p12 + m1 + m2 2m1 m2 2 p12 ⎡ m12 + 6m1 m2 + m2 ⎤ ⎥ ⎢ 2 2 ⎣ m12 m2 + m1 m2 ⎦ Ki = Kf = = Simplify to obtain: 2 p12 ⎡ m12 + 6m1m2 + m2 ⎤ ⎢ ⎥ 2 2 ⎣ m12 m2 + m1m2 ⎦ p'12 2 2 ⎡ m12 + 6m1m2 + m2 ⎤ ⎢ ⎥ 2 2 ⎣ m1 m2 + m1m2 ⎦ (p ) = ( p ) '2 1 2 ' ⇒ p1 = ± p1 1 and if p1' = + p1 , the particles do not collide. 92 •• A 3.0-kg block is traveling in the −x direction at 5.0 m/s, and a 1.0-kg block is traveling in the +x direction at 3.0 m/s. (a) Find the velocity vcm of the center of mass. (b) Subtract vcm from the velocity of each block to find the velocity of each block in the center-of-mass reference frame. (c) After they make a head-on elastic collision, the velocity of each block is reversed (in the center-ofmass frame). Find the velocity of each block in the center-of-mass frame after the collision. (d) Transform back into the original frame by adding vcm to the velocity of each block. (e) Check your result by finding the initial and final kinetic energies of the blocks in the original frame and comparing them. Picture the Problem Let the numerals 3 and 1 denote the blocks whose masses r r are 3.0 kg and 1.0 kg respectively. We can use ∑ mi v i = Mv cm to find the i velocity of the center-of-mass of the system and simply follow the directions in the problem step by step. (a) Express the total momentum of this two-particle system in terms of the velocity of its center of mass: r Solve for vcm : Conservation of Linear Momentum 803 r r r r P = ∑ mi vi = m1v1 + m3v3 i r r = Mvcm = (m1 + m3 )vcm r r r m3v3 + m1v1 vcm = m3 + m1 r Substitute numerical values and evaluate vcm : r (3.0 kg )(− 5.0 m/s) iˆ + (1.0 kg ) (3.0 m/s) iˆ = v cm = 3.0 kg + 1.0 kg (b) Find the velocity of the 3-kg block in the center of mass reference frame: Find the velocity of the 1-kg block in the center of mass reference frame: (− 3.0 m/s) iˆ rrr u3 = v3 − vcm ˆ ˆ = (− 5.0 m/s ) i − (− 3.0 m/s ) i = (− 2.0 m/s) iˆ rrr u1 = v1 − v cm ˆ ˆ = (3.0 m/s ) i − (− 3.0 m/s ) i = (c) Express the after-collision velocities of both blocks in the center of mass reference frame: (d) Transform the after-collision velocity of the 3-kg block from the center of mass reference frame to the original reference frame: Transform the after-collision velocity of the 1-kg block from the center of mass reference frame to the original reference frame: (e) Express K i in the original frame of reference: (6.0 m/s) iˆ r u'3 = (2.0 m/s) iˆ and r' u1 = (− 6.0 m/s) iˆ r' r' r v3 = u3 + v cm ˆ ˆ = (2.0 m/s ) i + (− 3.0 m/s ) i = (− 1.0 m/s ) iˆ r' r' r v1 = u1 + v cm ˆ ˆ = (− 6.0 m/s ) i + (− 3.0 m/s ) i = (− 9.0 m/s ) iˆ 2 K i = 1 m3 v3 + 1 m1v12 2 2 804 Chapter 8 Substitute numerical values and evaluate K i : Ki = 1 2 [(3.0 kg )(5.0 m/s) 2 + (1.0 kg )(3.0 m/s ) = 42 J 2 Express K i in the original frame of K f = 1 m3 v' 32 + 1 m1v'12 2 2 reference: Substitute numerical values and evaluate K f : Kf = 1 2 [(3.0 kg )(1.0 m/s) 2 + (1.0 kg )(9.0 m/s ) = 42 J 2 93 •• [SSM] Repeat Problem 92 with the second block having a mass of 5.0 kg and moving to the right at 3.0 m/s. Picture the Problem Let the numerals 3 and 5 denote the blocks whose masses r r are 3.0 kg and 5.0 kg respectively. We can use ∑ mi v i = Mv cm to find the i velocity of the center-of-mass of the system and simply follow the directions in the problem step by step. (a) Express the total momentum of this two-particle system in terms of the velocity of its center of mass: r r r r P = ∑ mi vi = m3v3 + m5v5 i r r = Mvcm = (m3 + m5 ) vcm r r r m3v3 + m5v5 vcm = m3 + m5 r Solve for vcm : r Substitute numerical values and evaluate vcm : r (3.0 kg )(− 5.0 m/s) iˆ + (5.0 kg )(3.0 m/s) iˆ = 0 v cm = 3.0 kg + 5.0 kg (b) Find the velocity of the 3.0-kg block in the center of mass reference frame: r rr ˆ u3 = v3 − vcm = (− 5.0 m/s ) i − 0 Find the velocity of the 5.0-kg block in the center of mass reference frame: r rr ˆ u5 = v5 − vcm = (3.0 m/s ) i − 0 = = (− 5.0 m/s) iˆ (3.0 m/s) iˆ Conservation of Linear Momentum 805 r ˆ u'3 = (5.0 m/s ) i (c) Express the after-collision velocities of both blocks in the center of mass reference frame: and r' u5 = (− 3.0 m/s) iˆ (d) Transform the after-collision velocity of the 3.0-kg block from the center of mass reference frame to the original reference frame: r' r r ˆ v3 = u'3 + vcm = (5.0 m/s ) i + 0 Transform the after-collision velocity of the 5.0-kg block from the center of mass reference frame to the original reference frame: r' r r ˆ v5 = u'5 + vcm = (− 3.0 m/s ) i + 0 = = (5.0 m/s) iˆ (− 3.0 m/s) iˆ 2 2 K i = 1 m3 v3 + 1 m5 v5 2 2 (e) Express K i in the original frame of reference: Substitute numerical values and evaluate K i : Ki = 1 2 [(3.0 kg )(5.0 m/s) 2 + (5.0 kg )(3.0 m/s ) = 60 J 2 K f = 1 m3 v' 32 + 1 m5 v' 52 2 2 Express K f in the original frame of reference: Substitute numerical values and evaluate K f : Kf = 1 2 [(3.0 kg )(5.0 m/s) + (5.0 kg )(3.0 m/s) ] = 2 2 60 J *Systems With Continuously Varying Mass: Rocket Propulsion 94 • A rocket burns fuel at a rate of 200 kg/s and exhausts the gas at a relative speed of 6.00 km/s relative to the rocket. Find the magnitude of the thrust of the rocket. Picture the Problem The thrust of a rocket Fth depends on the burn rate of its fuel dm/dt and the relative speed of its exhaust gases uex according to Fth = dm dt uex . 806 Chapter 8 Using its definition, relate the rocket’s thrust to the relative speed of its exhaust gases: dm uex dt Fth = Fth = (200 kg/s )(6.00 km/s ) Substitute numerical values and evaluate Fth: = 1.20 MN 95 •• A rocket has an initial mass of 30,000 kg, of which 80 percent is the fuel. It burns fuel at a rate of 200 kg/s and exhausts its gas at a relative speed of 1.80 km/s. Find (a) the thrust of the rocket, (b) the time until burnout, and (c) its speed at burnout assuming it moves straight upward near the surface of Earth. Assume that g is constant and neglect any effects of air resistance. Picture the Problem The thrust of a rocket Fth depends on the burn rate of its fuel dm/dt and the relative speed of its exhaust gases uex according to Fth = dm dt uex . The final velocity vf of a rocket depends on the relative speed of its exhaust gases uex, its payload to initial mass ratio mf/m0 and its burn time according to vf = −uex ln (mf m0 ) − gt b . (a) Using its definition, relate the rocket’s thrust to the relative speed of its exhaust gases: Fth = dm uex dt Fth = (200 kg/s )(1.80 km/s ) Substitute numerical values and evaluate Fth: = 360 kN 0.8m0 mfuel = dm / dt dm / dt (b) Relate the time to burnout to the mass of the fuel and its burn rate: tb = Substitute numerical values and evaluate t b : tb = (c) Relate the final velocity of a rocket to its initial mass, exhaust velocity, and burn time: ⎛m ⎞ vf = −uex ln⎜ f ⎟ − gtb ⎜m ⎟ ⎝ 0⎠ (0.80)(30,000 kg ) = 200 kg/s 120 s Substitute numerical values and evaluate vf: ( ) ⎛1⎞ vf = −(1.80 km/s )ln⎜ ⎟ − 9.81 m/s 2 (120 s ) = 1.72 km/s ⎝5⎠ Conservation of Linear Momentum 807 96 •• The specific impulse of a rocket propellant is defined as Isp = Fth/(Rg), where Fth is the thrust of the propellant, g the magnitude of free-fall acceleration, and R the rate at which the propellant is burned. The rate depends predominantly on the type and exact mixture of the propellant. (a) Show that the specific impulse has the dimension of time. (b) Show that uex = gIsp, where uex is the relative speed of the exhaust. (c) What is the specific impulse (in seconds) of the propellant used in the Saturn V rocket of Example 8-16. Picture the Problem We can use the dimensions of thrust, burn rate, and acceleration to show that the dimension of specific impulse is time. Combining the definitions of rocket thrust and specific impulse will lead us to uex = gI sp . M⋅L 2 [F ] = th = T =T [R][g ] M ⋅ L T T2 (a) Express the dimension of specific impulse in terms of the dimensions of Fth, R, and g: [I ] (b) From the definition of rocket thrust we have: Fth = Ruex ⇒ uex = Substitute for Fth to obtain: (c) Solve equation (1) for Isp and substitute for uex to obtain: From Example 8-21 we have: Substitute numerical values and evaluate Isp: sp uex = RgI sp I sp = Fth R Fth Rg R = gI sp (1) R = 1.384×104 kg/s and Fth = 3.4×106 N I sp = 3.4 × 10 6 N 1.384 × 10 4 kg/s 9.81 m/s 2 ( )( ) = 25 s ••• [SSM] The initial thrust-to-weight ratio τ0 of a rocket is τ0 = Fth/(m0 g), where Fth is the rocket’s thrust and m0 the initial mass of the 97 rocket, including the propellant. (a) For a rocket launched straight up from Earth’s surface, show that τ0 = 1 + (a0/g), where a0 is the initial acceleration of the rocket. For manned rocket flight, τ0 cannot be made much larger than 4 for the comfort and safety of the astronauts. (The astronauts will feel that their weight as the rocket lifts off is equal to τ0 times their normal weight.) (b) Show that the final velocity of a rocket launched from Earth’s surface, in terms of τ0 and Isp (see Problem 96) can be written as 808 Chapter 8 ⎡ ⎛m ⎞ 1 ⎛ m ⎞⎤ vf = gIsp ⎢1n⎜ 0 ⎟ − ⎜1 − f ⎟ ⎥ ⎢ ⎥ ⎣ ⎝ mf ⎠ τ 0 ⎝ m0 ⎠ ⎦ where mf is the mass of the rocket (not including the spent propellant). (c) Using a spreadsheet program or graphing calculator, graph vf as a function of the mass ratio m0/mf for Isp = 250 s and τ0 = 2 for values of the mass ratio from 2 to 10. (Note that the mass ratio cannot be less than 1.) (d) To lift a rocket into orbit, a final velocity after burnout of vf = 7.0 km/s is needed. Calculate the mass ratio required of a single stage rocket to do this, using the values of specific impulse and thrust ratio given in Part (b). For engineering reasons, it is difficult to make a rocket with a mass ratio much greater than 10. Can you see why multistage rockets are usually used to put payloads into orbit around Earth? Picture the Problem We can use the rocket equation and the definition of rocket thrust to show that τ 0 = 1 + a0 g . In Part (b) we can express the burn time tb in terms of the initial and final masses of the rocket and the rate at which the fuel burns, and then use this equation to express the rocket’s final velocity in terms of Isp, τ0, and the mass ratio m0/mf. In Part (d) we’ll need to use trial-and-error methods or a graphing calculator to solve the transcendental equation giving vf as a function of m0/mf. (a) The rocket equation is: − mg + Ru ex = ma From the definition of rocket thrust we have: Fth = Ruex Substitute for Ruex to obtain: − mg + Fth = ma Solve for Fth at takeoff: Fth = m0 g + m0 a0 Divide both sides of this equation by m0g to obtain: Fth a = 1+ 0 m0 g g Because τ 0 = Fth /( m0 g ) : (b) Use Equation 8-39 to express the final speed of a rocket that starts from rest with mass m0: Express the burn time in terms of the burn rate R (assumed constant): τ 0 = 1+ a0 g ⎛m ⎞ vf = u ex ln⎜ 0 ⎟ − gt b , ⎜m ⎟ ⎝ f⎠ where t b is the burn time. tb = m0 − mf m0 ⎛ mf ⎞ ⎜1 − ⎟ = R R ⎜ m0 ⎟ ⎝ ⎠ (1) Conservation of Linear Momentum 809 Multiply t b by one in the form gFth and simplify to obtain: gFth gFth m0 ⎛ mf ⎞ ⎜1 − ⎟ gFth R ⎜ m0 ⎟ ⎝ ⎠ gm0 Fth ⎛ mf ⎞ ⎜1 − ⎟ = Fth gR ⎜ m0 ⎟ ⎝ ⎠ tb = = I sp ⎛ mf ⎞ ⎜1 − ⎟ τ 0 ⎜ m0 ⎟ ⎝ ⎠ Substitute in equation (1): ⎛ m ⎞ gI ⎛ m ⎞ vf = uex ln⎜ 0 ⎟ − sp ⎜1 − f ⎟ ⎜m ⎟ τ ⎜ m ⎟ 0⎝ 0⎠ ⎝ f⎠ From Problem 96 we have: u ex = gI sp , where uex is the exhaust velocity of the propellant. Substitute for uex and factor to obtain: ⎛ m ⎞ gI ⎛ m ⎞ vf = gI sp ln⎜ 0 ⎟ − sp ⎜1 − f ⎟ ⎜m ⎟ τ ⎜ m ⎟ 0⎝ 0⎠ ⎝ f⎠ ⎡ ⎛ m ⎞ 1 ⎛ m ⎞⎤ = gI sp ⎢ln⎜ 0 ⎟ − ⎜1 − f ⎟⎥ ⎜⎟ ⎟ ⎜ ⎣ ⎝ mf ⎠ τ 0 ⎝ m0 ⎠⎦ (c) A spreadsheet program to calculate the final velocity of the rocket as a function of the mass ratio m0/mf is shown below. The constants used in the velocity function and the formulas used to calculate the final velocity are as follows: Cell B1 B2 B3 D9 E8 Content/Formula 250 9.81 2 D8 + 0.25 $B$2*$B$1*(LOG(D8) − (1/$B$3)*(1/D8)) 1 2 3 4 5 6 7 A B C Isp = 250 s g = 9.81 m/s2 τ0 = 2 Algebraic Form Isp g τ0 m0/mf ⎡ ⎛ m ⎞ 1 ⎛ m ⎞⎤ gI sp ⎢ln⎜ 0 ⎟ − ⎜1 − f ⎟⎥ ⎜ ⎟ ⎜ ⎟ ⎣ ⎝ mf ⎠ τ 0 ⎝ m0 ⎠ ⎦ D E mass ratio vf 2.00 1.252E+02 2.25 3.187E+02 810 Chapter 8 8 9 10 2.50 2.75 3.00 4.854E+02 6.316E+02 7.614E+02 34 35 36 37 38 39 9.00 9.25 9.50 9.75 10.00 725.00 2.204E+03 2.237E+03 2.269E+03 2.300E+03 2.330E+03 7.013E+03 A graph of final velocity as a function of mass ratio follows. 2.5 v f (km/s) 2.0 1.5 1.0 0.5 0.0 2 3 4 5 6 7 8 9 10 m 0/m f (d) Substitute the data given in part (c) in the equation derived in Part (b) to obtain: ⎛ ⎛ m ⎞ 1 ⎛ m ⎞⎞ 7.00 km/s = 9.81m/s 2 (250 s )⎜ ln⎜ 0 ⎟ − ⎜1 − f ⎟ ⎟ ⎜ ⎜ m ⎟ 2 ⎜ m ⎟⎟ 0 ⎠⎠ ⎝ ⎝ ⎝ f⎠ ( ) or 2.854 = ln x − 0.5 + Use trial-and-error methods or a graphing calculator to solve this transcendental equation for the root greater than 1: 0.5 where x = m0/mf. x x ≈ 28 , a value considerably larger than the practical limit of 10 for singlestage rockets. 98 •• The height that a model rocket launched from Earth’s surface can reach can be estimated by assuming that the burn time is short compared to the total flight time; the rocket is therefore in free-fall for most of the flight. (This estimate neglects the burn time in calculations of both time and displacement.) For a model rocket with specific impulse Isp = 100 s, mass ratio m0/mf = 1.20, and Conservation of Linear Momentum 811 initial thrust-to-weight ratio τ0 = 5.00 (these parameters are defined in Problems 96 and 97), estimate (a) the height the rocket can reach, and (b) the total flight time. (c) Justify the assumption used in the estimates by comparing the flight time from Part (b) to the time it takes to consume the fuel. Picture the Problem We can use the velocity-at-burnout equation from Problem 96 to find vf and constant-acceleration equations to approximate the maximum height the rocket will reach and its total flight time. (a) Assuming constant acceleration, relate the maximum height reached by the model rocket to its time-totop-of-trajectory: 2 h = 1 gt top 2 From Problem 96 we have: ⎛ ⎛ m ⎞ 1 ⎛ m ⎞⎞ vf = gI sp ⎜ ln⎜ 0 ⎟ − ⎜1 − f ⎟ ⎟ ⎜ ⎜ m ⎟ τ ⎜ m ⎟⎟ 0 ⎠⎠ ⎝ ⎝ ⎝ f⎠ (1) Evaluate the velocity at burnout vf for Isp = 100 s, m0/mf = 1.2, and τ = 5: ⎡ 1⎛ 1 ⎞⎤ vf = 9.81 m/s 2 (100 s ) ⎢ln (1.2) − ⎜1 − ⎟ = 146 m/s 5 ⎝ 1.2 ⎠⎥ ⎣ ⎦ ( ) Assuming that the time for the fuel to burn up is short compared to the total flight time, find the time to the top of the trajectory: t top = vf 146 m/s = = 14.9 s g 9.81 m/s 2 Substitute in equation (1) and evaluate h: h= (9.81m/s )(14.9 s ) (b) Find the total flight time from the time it took the rocket to reach its maximum height: t flight = 2t top = 2(14.9 s ) = 29.8 s (c) The fuel burn time t b is: 1 2 2 I sp ⎛ m f ⎜1 − τ ⎜ m0 ⎝ = 3.33 s tb = 2 = 1.09 km ⎞ 100 s ⎛ 1⎞ ⎟= ⎜1 − ⎟ ⎟ 5 ⎝ 1.2 ⎠ ⎠ 812 Chapter 8 Because this burn time is approximately one-fifth of the total flight time, we can’t expect the answer we obtain in Part (b) to be very accurate. It should, however, be good to about thirty percent accuracy, as the maximum distance the model rocket could possibly move in this time is 1 vf tb = 244 m , assuming constant acceleration 2 until burnout. General Problems 99 • [SSM] A 250-g model-train car traveling at 0.50 m/s links up with a 400-g car that is initially at rest. What is the speed of the cars immediately after they link up? Find the pre- and post-collision kinetic energies of the two-car system. Picture the Problem Let the direction the 250-g car is moving before the collision be the +x direction. Let the numeral 1 refer to the 250-kg car, the numeral 2 refer to the 400-kg car, and V represent the velocity of the linked cars. Let the system include Earth and the cars. We can use conservation of momentum to find their speed after they have linked together and the definition of kinetic energy to find their pre- and post-collision kinetic energies. Use conservation of momentum to relate the speeds of the cars immediately before and immediately after their collision: pix = p fx or Substitute numerical values and evaluate V: V= m1v1 = (m1 + m2 )V ⇒ V = m1v1 m1 + m2 (0.250 kg )(0.50 m/s) = 0.192 m/s 0.250 kg + 0.400 kg = 0.19 m/s Find the pre-collision kinetic energy of the cars: K pre = 1 m1v12 = 2 Find the post-collision kinetic energy of the coupled cars: K post = 1 2 (0.250 kg )(0.50 m/s)2 = 31 mJ (m1 + m2 )V 2 2 = 1 (0.250 kg + 0.400 kg )(0.192 m/s ) 2 1 2 = 12 mJ 100 • A 250-g model train car traveling at 0.50 m/s heads toward a 400-g car that is initially at rest. (a) Find the kinetic energy of the two-car system. (b) Find the velocity of each car in the center-of-mass reference frame, and use these velocities to calculate the kinetic energy of the two-car system in the center-of- Conservation of Linear Momentum 813 mass reference. (c) Find the kinetic energy associated with the motion of the center of mass of the system. (d) Compare your answer for Part (a) with the sum of your answers for Parts (b) and (c). Picture the Problem Let the direction the 250-g car is moving before the collision be the +x direction. Let the numeral 1 refer to the 250-kg car and the numeral 2 refer to the 400-g car and the system include Earth and the cars. We can use conservation of momentum to find their speed after they have linked together and the definition of kinetic energy to find their pre- and post-collision kinetic energies. (a) The pre-collision kinetic energy of the two-car system is: (b) Relate the velocity of the center of mass to the total momentum of the system: K pre = 1 m1v12 = 2 1 2 (0.250 kg )(0.50 m/s)2 = 31.3 mJ = 31 mJ r r r Psys = ∑ mi v i = mv cm i m1v1 + m2 v2 m1 + m2 Solve for vcm : vcm = Substitute numerical values and evaluate vcm : vcm = Find the initial velocity of the 250-g car relative to the velocity of the center of mass: u1 = v1 − vcm = 0.50 m/s − 0.192 m/s Find the initial velocity of the 400-g car relative to the velocity of the center of mass: u 2 = v2 − vcm = 0 m/s − 0.192 m/s Express the pre-collision kinetic energy of the system relative to the center of mass: 2 K pre,rel = 1 m1u12 + 1 m2 u 2 2 2 Substitute numerical values and evaluate K pre,rel : K pre,rel = (0.250 kg )(0.50 m/s) = 0.192 m/s 0.250 kg + 0.400 kg = 0.31m/s = − 0.19 m/s (0.250 kg )(0.308 m/s)2 2 + 1 (0.400 kg )(− 0.192 m/s ) 2 1 2 = 19 mJ 814 Chapter 8 (c) Express the kinetic energy of the center of mass: 2 K cm = 1 Mvcm 2 Substitute numerical values and evaluate Kcm: K cm = (d) Relate the pre-collision kinetic energy of the system to its precollision kinetic energy relative to the center of mass and the kinetic energy of the center of mass: K i = K i,rel + K cm 1 2 (0.650 kg )(0.192 m/s)2 = 12 mJ = 19.2 mJ + 12.0 mJ = 31.2 mJ and K i = K i,rel + K cm 101 •• A 1500-kg car traveling north at 70 km/h collides at an intersection with a 2000-kg car traveling west at 55 km/h. The two cars stick together. (a) What is the total momentum of the system before the collision? (b) Find the magnitude and direction of the velocity of the wreckage just after the collision. Picture the Problem Let east be the positive x direction and north the positive y direction. Include both cars and Earth in the system and let the numeral 1 denote the 1500-kg car and the numeral 2 the 2000-kg car. Because the net external force acting on the system is zero, momentum is conserved in this perfectly inelastic collision. (a) Express the total momentum of the system: rrr r r p = p1 + p2 = m1v1 + m2 v 2 ˆ =mv ˆ−m v i j 11 22 r Substitute numerical values and evaluate p : r ˆ p = (1500 kg ) (70 km/h ) ˆ − (2000 kg ) (55 km/h ) i j ˆ = − 1.10 × 105 kg ⋅ km/h i + 1.05 ×105 kg ⋅ km/h ˆ j ( ( = − 1.1×105 )( ˆ kg ⋅ km/h ) i + (1.1×10 (b) The velocity of the wreckage in terms of the total momentum of the system is given by: 5 ) ) j kg ⋅ km/h ˆ r rr p v f = v cm = M Conservation of Linear Momentum 815 r Substitute numerical values and evaluate v f : ( ) ( ) ˆ r − 1.10 × 105 kg ⋅ km/h i 1.05 × 105 kg ⋅ km/h ˆ j vf = + 1500 kg + 2000 kg 1500 kg + 2000 kg ˆ = −(31.4 km/h ) i + (30.0 km/h ) ˆ j (31.4 km/h )2 + (30.0 km/h )2 Find the magnitude of the velocity of the wreckage: vf = Find the direction the wreckage moves: θ = tan −1 ⎢ = 43 km/h ⎡ 30.0 km/h ⎤ ⎥ = −43.7° ⎣ − 31.4 km/h ⎦ The direction of the wreckage is 46° west of north. 102 •• A 60-kg woman stands on the back of a 6.0-m-long, 120-kg raft that is floating at rest in still water. The raft is 0.50 m from a fixed pier, as shown in Figure 8-52. (a) The woman walks to the front of the raft and stops. How far is the raft from the pier now? (b) While the woman walks, she maintains a constant speed of 3.0 m/s relative to the raft. Find the total kinetic energy of the system (woman plus raft), and compare with the kinetic energy if the woman walked at 3.0 m/s on a raft tied to the pier. (c) Where do these kinetic energies come from, and where do they go when the woman stops at the front of the raft? (d) On land, the woman puts a lead shot 6.0 m. She stands at the back of the raft, aims forward, and puts the shot so that just after it leaves her hand, it has the same velocity relative to her as it did when she threw it from the ground. Approximately, where does her shot land? Picture the Problem Take the origin to be at the initial position of the right-hand end of raft and let the positive x direction be to the left. Let ″w″ denote the woman and ″r″ the raft, d be the distance of the end of the raft from the pier after the woman has walked to its front. The raft moves to the left as the woman moves to the right; with the center of mass of the woman-raft system remaining fixed (because Fext,net = 0). The diagram shows the initial (xw,i) and final (xw,f) positions of the woman as well as the initial (xr_cm,i) and final (xr_cm,f) positions of the center of mass of the raft both before and after the woman has walked to the front of the raft. 816 Chapter 8 CM x × xr_cm,i xw i = 6 m , xC M 0 0.5 m CM x (a) Express the distance of the raft from the pier after the woman has walked to the front of the raft: × xr_cm,f xr_cm,i P I E R 0 xw f , d d = 0.50 m + x w, f Express xcm before the woman has walked to the front of the raft: xcm = Express xcm after the woman has walked to the front of the raft: xcm = (1) mw xw,i + mr xr_cm, i m w + mr mw xw,f + mr xr_cm,f m w + mr Because Fext,net = 0, the center of mass remains fixed and we can equate these two expressions for xcm to obtain: mw xw ,i + mr xr_cm,i = mw xw,f + mr xr_cm,f Solve for x w,f : xw,f = xw,i − From the figure it can be seen that x r_cm,f − xr_cm,i = x w,f . Substitute xw,f = mr (xr_cm,f − xr_cm,i ) mw mw xw,i m w + mr x w,f for x r_cm,f − x r_cm,i to obtain: (60 kg )(6.0 m ) = 2.0 m Substitute numerical values and evaluate x w,f : x w,f = Substitute in equation (1) to obtain: d = 2.0 m + 0.50 m = 2.5 m (b) Express the total kinetic energy of the system: 2 K tot = 1 mw vw + 1 mr vr2 2 2 60 kg + 120 kg Conservation of Linear Momentum 817 Noting that the elapsed time is 2.0 s, find vw and vr: vw = x w,f − x w,i Δt 2.0 m − 6.0 m = = −2.0 m/s 2.0 s relative to the dock, and x − xr,i vr = r,f Δt 2.50 m − 0.50 m = = 1.0 m/s 2.0 s also relative to the dock. Substitute numerical values and evaluate Ktot: K tot = (60 kg )(− 2.0 m/s)2 2 + 1 (120 kg )(1.0 m/s ) 2 1 2 = 0.18 kJ Evaluate K with the raft tied to the pier: 2 K tot = 1 m w v w = 2 1 2 (60 kg )(3.0 m/s)2 = 0.27 kJ (c) All the kinetic energy derives from the chemical energy of the woman and, assuming she stops via static friction, the kinetic energy is transformed into her internal energy. (d) After the shot leaves the woman’s hand, the raft-woman system constitutes an inertial reference frame. In that frame, the shot has the same initial velocity as did the shot that had a range of 6.0 m in the reference frame of the land. Thus, in the raft-woman frame, the shot also has a range of 6.0 m and lands at the front of the raft. A 1.0-kg steel ball and a 2.0-m cord of negligible mass make up a 103 •• simple pendulum that can pivot without friction about the point O, as in Figure 853. This pendulum is released from rest in a horizontal position and when the ball is at its lowest point it strikes a 1.0-kg block sitting at rest on a shelf. Assume that the collision is perfectly elastic and take the coefficient of kinetic friction between the block and shelf to be 0.10. (a) What is the velocity of the block just after impact? (b) How far does the block slide before coming to rest (assuming the shelf is long enough)? Picture the Problem Let the zero of gravitational potential energy be at the elevation of the 1.0-kg block. We can use conservation of energy to find the speed of the bob just before its perfectly elastic collision with the block and 818 Chapter 8 conservation of momentum to find the speed of the block immediately after the collision. We’ll apply Newton’s second law to find the acceleration of the sliding block and use a constant-acceleration equation to find how far it slides before coming to rest. ΔK + ΔU = 0 (a) Use conservation of energy to find the speed of the bob just before its collision with the block: or Because Ki = Uf = 0: 1 2 Kf − Ki + U f − U i = 0 2 mball vball + mball gΔh = 0 and vball = 2 gΔh ( ) Substitute numerical values and evaluate vball: vball = 2 9.81m/s 2 (2.0 m ) = 6.26 m/s Because the collision is perfectly elastic and the ball and block have the same mass: vblock = vball = 6.3 m/s (b) Using a constant-acceleration equation, relate the displacement of the block to its acceleration and initial speed: vf2 = vi2 + 2ablock Δx Solving for Δx yields: Apply r r F = ma to the sliding ∑ block: or, because vf = 0, 0 = vi2 + 2a block Δx 2 − vblock − vi2 Δx = = 2ablock 2ablock ∑F x and ∑F y Using the definition of fk (=μkFn) eliminate fk and Fn between the two equations and solve for ablock: Substitute for ablock to obtain: = − f k = mablock = Fn − mblock g = 0 ablock = − μ k g Δx = 2 v2 − vblock = block − 2μ k g 2μ k g Conservation of Linear Momentum 819 Substitute numerical values and evaluate Δx: Δx = (6.26 m/s)2 = 2(0.10) (9.81m/s 2 ) 20 m 104 •• Figure 8-54 shows a World War I cannon mounted on a railcar so that it will project a shell at an angle of 30º. With the car initially at rest, the cannon fires a 200-kg projectile at 125 m/s. (All values are for the frame of reference of the track.) Now consider a system composed of a cannon, shell, and railcar, all on the frictionless track. (a) Will the total vector momentum of that system be the same just before and just after the shell is fired? Explain your answer. (b) If the mass of the railcar plus cannon is 5000 kg, what will be the recoil velocity of the car along the track after the firing? (c) The shell is observed to rise to a maximum height of 180 m as it moves through its trajectory. At this point, its speed is 80.0 m/s. On the basis of this information, calculate the amount of thermal energy produced by air friction on the shell on its way from firing to this maximum height. Picture the Problem We can use conservation of momentum in the horizontal direction to find the recoil velocity of the car along the track after the firing. Because the shell will neither rise as high nor be moving as fast at the top of its trajectory as it would be in the absence of air friction, we can apply the workenergy theorem to find the amount of thermal energy produced by the air friction. (a) No. The vertical reaction force of the rails is an external force and so the momentum of the system will not be conserved. (b) Use conservation of momentum in the horizontal (x) direction to obtain: Δp x = 0 or mv cos 30° − Mvrecoil = 0 Solving for vrecoil yields: vrecoil = Substitute numerical values and evaluate vrecoil : vrecoil = mv cos 30° M (200 kg )(125 m/s)cos30° 5000 kg = 4.3 m/s (c) Using the work-energy theorem, relate the thermal energy produced by air friction to the change in the energy of the system: Wext = Wf = ΔEsys = ΔU + ΔK 820 Chapter 8 Substitute for ΔU and ΔK to obtain: Wext = mgyf − mgyi + 1 mvf2 − 1 mvi2 2 2 ( = mg ( yf − yi ) + 1 m vf2 − vi2 2 ) Substitute numerical values and evaluate Wext: ( ) [ Wext = (200 kg ) 9.81 m/s 2 (180 m ) + 1 (200 kg ) (80.0 m/s ) − (125 m/s ) 2 2 2 = − 569 kJ 105 ••• [SSM] One popular, if dangerous, classroom demonstration involves holding a baseball an inch or so directly above a basketball, holding the basketball a few feet above a hard floor, and dropping the two balls simultaneously. The two balls will collide just after the basketball bounces from the floor; the baseball will then rocket off into the ceiling tiles with a hard ″thud″ while the basketball will stop in midair. (The author of this problem once broke a light doing this.) (a) Assuming that the collision of the basketball with the floor is elastic, what is the relation between the velocities of the balls just before they collide? (b) Assuming the collision between the two balls is elastic, use the result of Part (a) and the conservation of momentum and energy to show that, if the basketball is three times as heavy as the baseball, the final velocity of the basketball will be zero. (This is approximately the true mass ratio, which is why the demonstration is so dramatic.) (c) If the speed of the baseball is v just before the collision, what is its speed just after the collision? Picture the Problem Let the numeral 1 refer to the basketball and the numeral 2 to the baseball. The left-hand side of the diagram shows the balls after the basketball’s elastic collision with the floor and just before they collide. The right-hand side of the diagram shows the balls just after their collision. We can apply conservation of momentum and the definition of an elastic collision to obtain equations relating the initial and final velocities of the colliding balls that we can solve for v1f and v2f. r v 2f m2 r v 2i r v1i m1 m2 r v1f m1 (a) Because both balls are in free-fall, and both are in the air for the same amount of time, they have the same velocity just before the basketball rebounds. After the basketball rebounds elastically, its velocity will have the same magnitude, but the opposite direction than just before it hit the ground. The velocity of the basketball will be equal in magnitude but opposite in direction to the velocity of the baseball. Conservation of Linear Momentum 821 (b) Apply conservation of linear momentum to the collision of the balls to obtain: Use conservation of mechanical energy to set the speed of recession equal to the negative of the speed of approach: m1v1f + m2 v2f = m1v1i + m2 v2i (1) v2f − v1f = −(v1i − v2i ) or v1f − v2f = v2i − v1i (2) Multiply equation (2) by m2 and add it to equation (1) to obtain: (m1 + m2 )v1f = (m1 − m2 )v1i + 2m2v2i Solve for v1f to obtain: m1 − m2 2m2 v1i + v2 i m1 + m2 m1 + m2 or, because v2i = −v1i, 2m2 m − m2 v1f = 1 v1i − v1i m1 + m2 m1 + m2 v1f = = For m1 = 3m2 and v1i = v: v1f = m1 − 3m2 v1i m1 + m2 3m2 − 3m2 v= 0 3m2 + m2 (c) Multiply equation (2) by m1 and subtract it from equation (1) to obtain: (m1 + m2 )v2f = (m2 − m1 )v2i + 2m1v1i Solve for v2f to obtain: m − m1 2m1 v1i + 2 v 2i m1 + m2 m1 + m2 or, because v2i = −v1i, m − m1 2m1 v2 f = v1i − 2 v1i m1 + m2 m1 + m2 v2 f = = For m1 = 3m2 and v1i = v: v2 f = 3m1 − m2 v1i m1 + m2 3(3m2 ) − m2 v = 2v 3m2 + m2 106 ••• (a) Referring to Problem 105, if we held a third ball above the baseball and basketball, and wanted both the basketball and baseball to stop in mid-air, what should the ratio of the mass of the top ball to the mass of the baseball be? 822 Chapter 8 (b) If the speed of the top ball is v just before the collision, what is its speed just after the collision? Picture the Problem In Problem 105 only two balls are dropped. They collide head on, each moving at speed v, and the collision is elastic. In this problem, as it did in Problem 105, the solution involves using the conservation of momentum equation m1v1f + m2 v2 f = m1v1i + m2 v2i and the elastic collision equation v1f − v2 f = v2i − v1i where the numeral 1 refers to the baseball, and the numeral 2 to the top ball. The diagram shows the balls just before and just after their collision. From Problem 105 we know that v1i = 2v and v2i = −v. (a) Express the final speed v1f of the baseball as a function of its initial speed v1i and the initial speed of the top ball v2i (see Problem 64): Substitute for v1i and , v2i to obtain: Divide the numerator and denominator of each term by m2 to introduce the mass ratio of the upper ball to the lower ball: Set the final speed of the baseball v1f equal to zero and let x represent the mass ratio m1/m2 to obtain: Solving for x yields: (b) Apply the second of the two equations in Problem 64 to the collision between the top ball and the baseball: r v 2f m2 r v 2i m2 r v1i r v1f = 0 m1 v1f = 2m2 m1 − m2 v2 i v1i + m1 + m2 m1 + m2 v1f = m1 − m2 (2v ) + 2m2 (− v ) m1 + m2 m1 + m2 m1 −1 m2 (2v ) + m 2 (− v ) v1f = m1 1 +1 +1 m2 m2 0= x −1 (2v ) + 2 (− v ) x +1 x +1 x= m1 = m2 v2 f = 1 2 2m1 m − m1 v 2i v1i + 2 m1 + m2 m1 + m2 m1 Conservation of Linear Momentum 823 Substitute v1i = 2v and v2i = −v to obtain: v2 f = 2m1 (2v ) + m2 − m1 (− v ) m1 + m2 m1 + m2 In part (a) we showed that m2 = 2m1. Substitute and simplify to obtain: v3f = 2(2m1 ) (2v ) − 2m1 − m1 v m1 + 2m1 m1 + 2m1 = 7 3 v 107 ••• [SSM] In the ″slingshot effect,″ the transfer of energy in an elastic collision is used to boost the energy of a space probe so that it can escape from the solar system. All speeds are relative to an inertial frame in which the center of the sun remains at rest. Figure 8-55 shows a space probe moving at 10.4 km/s toward Saturn, which is moving at 9.6 km/s toward the probe. Because of the gravitational attraction between Saturn and the probe, the probe swings around Saturn and heads back in the opposite direction with speed vf. (a) Assuming this collision to be a one-dimensional elastic collision with the mass of Saturn much greater than that of the probe, find vf. (b) By what factor is the kinetic energy of the probe increased? Where does this energy come from? Picture the Problem Let the direction the probe is moving after its elastic collision with Saturn be the positive direction. The probe gains kinetic energy at the expense of the kinetic energy of Saturn. We’ll relate the speed of approach relative to the center of mass to urec and then to v. The +x direction is also the direction of the motion of Saturn. (a) Relate the speed of recession to the speed of recession relative to the center of mass: v = u rec + vcm Find the speed of approach: uapp = −9.6 km/s − 10.4 km/s (1) = −20.0 km/s Relate the relative speed of approach to the relative speed of recession for an elastic collision: u rec = −uapp = 20.0 km/s Because Saturn is so much more massive than the space probe: vcm = vSaturn = 9.6 km/s Substitute numerical values in equation (1) and evaluate v: v = 20 km/s + 9.6 km/s = 30 km/s 824 Chapter 8 2 (b) Express the ratio of the final kinetic energy to the initial kinetic energy and simplify: 2 K f 1 Mvrec ⎛ vrec ⎞ 2 =1 =⎜ 2 ⎜v ⎟ ⎟ Ki ⎝ i⎠ 2 Mvi Substitute numerical values and evaluate Kf/Ki: K f ⎛ 29.6 km/s ⎞ ⎟ = 8.1 =⎜ K i ⎜ 10.4 km/s ⎟ ⎝ ⎠ 2 The energy comes from an immeasurably small slowing of Saturn. 108 •• A 13-kg block is at rest on a level floor. A 400-g glob of putty is thrown at the block so that the putty travels horizontally, hits the block, and sticks to it. The block and putty slide 15 cm along the floor. If the coefficient of kinetic friction is 0.40, what is the initial speed of the putty? Picture the Problem Let the system include the block, the putty, and Earth. Then Fext,net = 0 and momentum is conserved in this perfectly inelastic collision. We’ll use conservation of momentum to relate the after-collision velocity of the block plus blob and conservation of energy to find their after-collision velocity. Noting that, because this is a perfectly elastic collision, the final velocity of the block plus blob is the velocity of the center of mass, use conservation of momentum to relate the velocity of the center of mass to the velocity of the glob before the collision: pi = pf or Use conservation of energy to find the initial energy of the block plus glob: ΔK + ΔU + Wf = 0 Because ΔU = Kf = 0, 2 − 1 Mvcm + f k Δx = 0 2 Because fk = μkMg: 2 − 1 Mvcm + μ k MgΔx = 0 2 Solve for vcm to obtain: vcm = 2 μ k gΔx Substitute numerical values and evaluate vcm : vcm = 2(0.40) 9.81m/s 2 (0.15 m ) ⎛M ⎞ ⎟v mgl vgl = Mvcm ⇒ vgl = ⎜ (1) ⎜ m ⎟ cm gl ⎠ ⎝ where M = mgl + mbl . ( = 1.08 m/s ) Conservation of Linear Momentum 825 Substitute numerical values in equation (1) and evaluate vgl : ⎛ 13 kg + 0.400 kg ⎞ ⎟ (1.08 m/s ) vgl = ⎜ ⎜ ⎟ 0.400 kg ⎝ ⎠ = 36 m/s 109 ••• [SSM] Your accident reconstruction team has been hired by the local police to analyze the following accident. A careless driver rear-ended a car that was halted at a stop sign. Just before impact, the driver slammed on his brakes, locking the wheels. The driver of the struck car had his foot solidly on the brake pedal, locking his brakes. The mass of the struck car was 900 kg, and that of the initially moving vehicle was 1200 kg. On collision, the bumpers of the two cars meshed. Police determine from the skid marks that after the collision the two cars moved 0.76 m together. Tests revealed that the coefficient of kinetic friction between the tires and pavement was 0.92. The driver of the moving car claims that he was traveling at less than 15 km/h as he approached the intersection. Is he telling the truth? Picture the Problem Let the direction the moving car was traveling before the collision be the +x direction. Let the numeral 1 denote this car and the numeral 2 the car that is stopped at the stop sign and the system include both cars and Earth. We can use conservation of momentum to relate the speed of the initially-moving car to the speed of the meshed cars immediately after their perfectly inelastic collision and conservation of energy to find the initial speed of the meshed cars. Using conservation of momentum, relate the before-collision speed to the after-collision speed of the meshed cars: Solving for v1 and simplifying yields: Using conservation of energy, relate the initial kinetic energy of the meshed cars to the work done by friction in bringing them to a stop: pi = pf or m1v1 = (m1 + m2 )V v1 = ⎛ m⎞ m1 + m2 V = ⎜1 + 2 ⎟ V ⎜ m⎟ m1 1⎠ ⎝ (1) ΔK + ΔEthermal = 0 or, because Kf = 0 and ΔEthermal = fΔs, − K i + f k Δs = 0 Substitute for Ki and, using fk = μkFn = μkMg, eliminate fk to obtain: − 1 MV 2 + μ k MgΔx = 0 2 Solving for V yields: V = 2 μ k gΔx 826 Chapter 8 ⎛ m⎞ v1 = ⎜1 + 2 ⎟ 2 μ k gΔx ⎜ m1 ⎟ ⎝ ⎠ Substitute for V in equation (1) to obtain: Substitute numerical values and evaluate v1: ⎛ 900 kg ⎞ 2 v1 = ⎜1 + ⎜ 1200 kg ⎟ 2(0.92) 9.81 m/s (0.76 m ) = 6.48 m/s = 23 km/h ⎟ ⎝ ⎠ ( ) The driver was not telling the truth. He was traveling at 23 km/h. 110 •• A pendulum consists of a compact 0.40-kg bob attached to a string of length 1.6 m. A block of mass m rests on a horizontal frictionless surface. The pendulum is released from rest at an angle of 53º with the vertical. The bob collides elastically with the block at the lowest point in its arc. Following the collision, the maximum angle of the pendulum with the vertical is 5.73º. Determine the mass m. Picture the Problem Let the zero of gravitational potential energy be at the lowest point of the bob’s swing and note that the bob can swing either forward or backward after the collision. We’ll use both conservation of momentum and conservation of energy to relate the velocities of the bob and the block before and after their collision. Choose the +x direction to be in the direction of the motion of the block. 2 2 pm pm ⇒m = Km = 2m 2K m Express the kinetic energy of the block in terms of its after-collision momentum: Use conservation of energy to relate Km to the change in the potential energy of the bob: ΔK + ΔU = 0 or, because Ki = 0, Km + Uf − Ui = 0 Solve for Km, substitute for Uf and Ui and simplify to obtain: K m = −U f + U i (1) = mbob g [L(1 − cos θ i ) − L(1 − cos θ f )] = mbob gL[cos θ f − cos θ i ] Substitute numerical values and evaluate Km: ( ) K m = (0.40 kg ) 9.81 m/s 2 (1.6 m )[cos5.73° − cos53°] = 2.47 J Conservation of Linear Momentum 827 Use conservation of energy to find the velocity of the bob just before its collision with the block: ΔK + ΔU = 0 or, because Ki = Uf = 0, Kf − Ui = 0 mbob v 2 − mbob gL(1 − cos θ i ) = 0 Substitute for Kf and Ui to obtain: 1 2 Solving for v yields: v = 2 gL(1 − cos θ i ) Substitute numerical values and evaluate v: v = 2 9.81 m/s 2 (1.6 m )(1 − cos53°) Use conservation of energy to find the velocity of the bob just after its collision with the block: ΔK + ΔU = 0 or, because Kf = Ui = 0, − Ki + U f = 0 Substitute for Ki and Uf to obtain: − 1 mbob v'2 + mbob gL(1 − cos θ f ) = 0 2 Solve for v′: v' = 2 gL(1 − cos θ f ) Substitute numerical values and evaluate v′: v' = 2 9.81 m/s 2 (1.6 m )(1 − cos5.73°) Use conservation of momentum to relate pm after the collision to the momentum of the bob just before and just after the collision: pi = p f or mbob v = ± mbob v'+ p m Solve for and evaluate pm: p m = mbob v ± mbob v' ( ) = 3.536 m/s ( ) = 0.396 m/s = (0.40 kg )(3.536 m/s ± 0.396 m/s ) = 1.414 kg ⋅ m/s ± 0.158 kg ⋅ m/s Find the larger value for pm: p m = 1.414 kg ⋅ m/s + 0.158 kg ⋅ m/s = 1.573 kg ⋅ m/s Find the smaller value for pm: p m = 1.414 kg ⋅ m/s − 0.158 kg ⋅ m/s = 1.256 kg ⋅ m/s 828 Chapter 8 Substitute numerical values in equation (1) to determine the two values for m: m= or m= (1.573 kg ⋅ m/s) 2 2(2.47 J ) = 0.50 kg (1.256 kg ⋅ m/s) 2 2(2.47 J ) = 0.32 kg 111 ••• [SSM] A 1.00-kg block and a second block of mass M are both initially at rest on a frictionless inclined plane (Figure 8-56) Mass M rests against a spring that has a force constant of 11.0 kN/m. The distance along the plane between the two blocks is 4.00 m. The 1.00-kg block is released, making an elastic collision with the unknown block. The 1.00-kg block then rebounds a distance of 2.56 m back up the inclined plane. The block of mass M comes momentarily comes to rest 4.00 cm from its initial position. Find M. Picture the Problem Choose the zero of gravitational potential energy at the location of the spring’s maximum compression. Let the system include the spring, the blocks, and Earth. Then the net external force is zero as is work done against friction. We can use conservation of energy to relate the energy transformations taking place during the evolution of this system. Apply conservation of energy to the system: ΔK + ΔU g + ΔU s = 0 Because ΔK = 0: ΔU g + ΔU s = 0 Express the change in the gravitational potential energy: ΔU g = − mgΔh − Mgx sin θ Express the change in the potential energy of the spring: ΔU s = 1 kx 2 2 Substitute to obtain: − mgΔh − Mgx sin θ + 1 kx 2 = 0 2 Solving for M and simplifying yields: kx 2 − mgΔh kx 2mΔh M= =− gx sin 30° g x Relate Δh to the initial and rebound positions of the block whose mass is m: Δh = (4.00 m − 2.56 m )sin 30° 1 2 = 0.72 m Conservation of Linear Momentum 829 Substitute numerical values and evaluate M: M= (11.0 × 10 ) N/m (0.0400 m ) 2(1.00 kg ) (0.72 m ) − = 8.9 kg 0.0400 m 9.81 m/s 2 3 112 ••• A neutron of mass m makes an elastic head-on collision with a stationary nucleus of mass M. (a) Show that the kinetic energy of the nucleus after the collision is given by Knucleus = [4mM/(m + M)2]Kn, where Kn is the initial kinetic energy of the neutron. (b) Show that the fractional change in the kinetic energy of the neutron is given by ΔK n 4(m M ) . =− Kn (1 + [m M ])2 (c) Show that this expression gives plausible results both if m << M and m = M. What is the best stationary nucleus for the neutron to collide head-on with if the objective is to produce a maximum loss in the kinetic energy of the neutron? Picture the Problem In this elastic head-on collision, the kinetic energy of recoiling nucleus is the difference between the initial and final kinetic energies of the neutron. We can derive the indicated results by using both conservation of energy and conservation of momentum and writing the kinetic energies in terms of the momenta of the particles before and after the collision. (a) Use conservation of energy to relate the kinetic energies of the particles before and after the collision: 2 2 2 pni pnf pnucleus = + 2m 2m 2M (1) Apply conservation of momentum to obtain a second relationship between the initial and final momenta: pni = pnf + pnucleus (2) Eliminate pnf in equation (1) using equation (2): pnucleus pnucleus pni + − =0 2M 2m m (3) Use equation (3) to write 2 pni 2m in terms of pnucleus: 2 (M + m ) p2 pni = K n = nucleus 2 2m 8M m (4) ⎡ 4Mm ⎤ K nucleus = K n ⎢ 2⎥ ⎣ (M + m ) ⎦ (5) Use equation (4) to express 2 K nucleus = pnucleus 2M in terms of Kn: 2 830 Chapter 8 (b) Relate the change in the kinetic energy of the neutron to the aftercollision kinetic energy of the nucleus: ΔK n = − K nucleus Using equation (5), express the fraction of the energy lost in the collision: ΔK n 4Mm =− Kn (M + m )2 (c) If m << M: ΔK n → 0 as expected. Kn I f m = M: ΔK n 4 =− = − 1 as expected. Kn (1 + 1)2 =− 4(m M ) (1 + (m M ))2 113 ••• The mass of a carbon nucleus is approximately 12 times the mass of a neutron. (a) Use the results of Problem 112 to show that after N head-on collisions of a neutron with carbon nuclei at rest, the kinetic energy of the neutron is approximately 0.716N K0, where K0 is its initial kinetic energy. (b) Neutrons emitted during the fission of a uranium nucleus have kinetic energies of about 2.0 MeV. For such a neutron to cause the fission of another uranium nucleus in a reactor, its kinetic energy must be reduced to about 0.020 eV. How many head-on collisions are needed to reduce the kinetic energy of a neutron from 2.0 MeV to 0.020 eV, assuming elastic head-on collisions with stationary carbon nuclei? Picture the Problem Problem 112 (b) provides an expression for the fractional loss of kinetic energy per collision. (a) Using the result of Problem 112 (b), express the fractional loss of energy per collision: K nf K ni − ΔK n (M − m ) = = K ni E0 (M + m )2 Evaluate this fraction to obtain: K nf (12m − m ) = = 0.716 E0 (12m + m )2 Express the kinetic energy of one neutron after N collisions: K nf = 0.716 N E0 (b) Substitute for Knf and E0 to obtain: 0.716 N = 10 −8 2 2 Conservation of Linear Momentum 831 Take the logarithm of both sides of the equation and solve for N: N= −8 ≈ 55 log 0.716 114 ••• On average, a neutron actually loses only 63 percent of its energy in an elastic collision with a hydrogen atom (not 100 percent) and 11 percent of its energy during an elastic collision with a carbon atom (not 28 percent). (These numbers are an average over all types of collisions, not just head-on ones. Thus the results are lower than the ones determined from analyses like that in Problem 113 because most collisions are not head-on.) Calculate the actual number of collisions, on average, needed to reduce the energy of a neutron from 2.0 MeV to 0.020 eV if the neutron collides with stationary (a) hydrogen atoms and (b) carbon atoms. Picture the Problem We can relate the number of collisions needed to reduce the energy of a neutron from 2 MeV to 0.02 eV to the fractional energy loss per collision and solve the resulting exponential equation for N. (a) Using the result of Problem 113 (b), express the fractional loss of energy per collision: K nf K ni − ΔK n K ni − 0.63K ni = = K ni E0 K ni Express the kinetic energy of one neutron after N collisions: K nf = 0.37 N K 0 Substitute for Knf and K0 to obtain: 0.37 N = 10 −8 Take the logarithm of both sides of the equation and solve for N: N= (b) Proceed as in (a) to obtain: K nf K ni − ΔK n K ni − 0.11K ni = = K ni E0 K ni = 0.37 −8 ≈ 19 log 0.37 = 0.89 Express the kinetic energy of one neutron after N collisions: K nf = 0.89 N K 0 Substitute for Knf and K0 to obtain: 0.89 N = 10 −8 Take the logarithm of both sides of the equation and solve for N: N= −8 ≈ 158 log 0.89 832 Chapter 8 115 ••• [SSM] Two astronauts at rest face each other in space. One, with mass m1, throws a ball of mass mb to the other, whose mass is m2. She catches the ball and throws it back to the first astronaut. Following each throw the ball has a speed of v relative to the thrower. After each has made one throw and one catch, (a) How fast are the astronauts moving? (b) How much has the two-astronaut system’s kinetic energy changed and where did this energy come from? Picture the Problem Let the direction that astronaut 1 first throws the ball be the positive direction and let vb be the initial speed of the ball in the laboratory frame. Note that each collision is perfectly inelastic. We can apply conservation of momentum and the definition of the speed of the ball relative to the thrower to each of the perfectly inelastic collisions to express the final speeds of each astronaut after one throw and one catch. (a) Use conservation of linear momentum to relate the speeds of astronaut 1 and the ball after the first throw: m1v1 + mb v b = 0 (1) Relate the speed of the ball in the laboratory frame to its speed relative to astronaut 1: v = v b − v1 (2) Eliminate vb between equations (1) and (2) and solve for v1: v1 = − Substitute equation (3) in equation (2) and solve for vb: vb = Apply conservation of linear momentum to express the speed of astronaut 2 and the ball after the first catch: mb v m1 + mb m1 v m1 + mb 0 = mb v b = (m2 + mb )v 2 Solving for v2 yields: v2 = mb vb m2 + mb Express v2 in terms of v by substituting equation (4) in equation (6): v2 = mb m1 v m2 + mb m1 + mb ⎡ ⎤ mb m1 =⎢ v (m2 + mb )(m1 + mb )⎥ ⎣ ⎦ (3) (4) (5) (6) (7) Conservation of Linear Momentum 833 (m2 + mb )v2 = mb vbf + m2v2f (8) Relate the speed of the ball in the laboratory frame to its speed relative to astronaut 2: v = v 2f − v bf (9) Eliminate vbf between equations (8) and (9) and solve for v2f: ⎛ mb ⎞ ⎛ m1 ⎞ v2 f = ⎜ ⎜ m + m ⎟ ⎜1 + m + m ⎟v (10) ⎟⎜ ⎟ b ⎠⎝ 1 b⎠ ⎝2 Substitute equation (10) in equation (9) and solve for vbf: ⎡ mb ⎤⎡ m1 ⎤ − 1⎥ ⎢1 + v bf = ⎢ ⎥ v (11) ⎣ m2 + mb ⎦ ⎣ m1 + mb ⎦ Use conservation of momentum to express the speed of astronaut 2 and the ball after she throws the ball: (m1 + mb )v1f Apply conservation of momentum to express the speed of astronaut 1 and the ball after she catches the ball: Using equations (3) and (11), eliminate vbf and v1 in equation (12) and solve for v1f: v1f = − = mb vbf + m1v1 m2 mb (2m1 + mb ) (m1 + mb )2 (m2 + mb ) (12) v ΔK = K f − K i or, because Ki = 0, ΔK = K f = K 1f + K 2f (b) The change in the kinetic energy of the system is: 2 2 = 1 m1v1f + 1 m2 v 2f 2 2 Substitute for v1f and v2f to obtain: ⎛ m2 mb (2m1 + mb ) ⎞ 2 1 ⎛ mb ⎟ ⎜ ΔK = 1 m1 ⎜ − 2 ⎜ (m + m )2 (m + m ) ⎟ v + 2 m2 ⎜ m + m b ⎝2 1 b 2 b⎠ ⎝ 2 ⎞ ⎟ ⎟ ⎠ 2 2 ⎛ m1 ⎞ 2 ⎜1 + ⎜ m +m ⎟ v ⎟ 1 b⎠ ⎝ Simplify to obtain: ΔK = 1 2 2 2 m2 mb (2m1 + mb ) ⎛ m1m2 ⎞ 2 ⎜1 + ⎟v (m2 + mb )2 (m1 + mb )2 ⎜ (m1 + mb )2 ⎟ ⎝ ⎠ This additional energy came from chemical energy in the astronaut’s bodies. 834 Chapter 8 116 ••• A stream of elastic glass beads, each with a mass of 0.50 g, comes out of a horizontal tube at a rate of 100 per second (see Figure 8-57). The beads fall a distance of 0.50 m to a balance pan and bounce back to their original height. How much mass must be placed in the other pan of the balance to keep the pointer at zero? Picture the Problem Take the zero of gravitational potential energy to be at the elevation of the pan and let the system include the balance, the beads, and Earth. We can use conservation of energy to find the vertical component of the velocity of the beads as they hit the pan and then calculate the net downward force on the pan from Newton’s second law. Let the +y direction be upward. Use conservation of energy to relate the y component of the bead’s velocity as it hits the pan to its height of fall: ΔK + ΔU = 0 or, because Ki = Uf = 0, 2 1 2 gh 2 mv y − mgh = 0 ⇒ v y = Substitute numerical values and evaluate vy: v y = 2 9.81 m/s 2 (0.50 m ) = 3.13 m/s Express the change in momentum in the y direction per bead: Use Newton’s second law to express the net force in the y direction exerted on the pan by the beads: ( ) Δp y = p yf − p yi = mv y − (− mv y ) = 2mv y Fnet, y = − N Δp y Δt Δp y Letting M represent the mass to be placed on the other pan, equate its weight to the net force exerted by the beads, substitute for Δpy, and solve for M: and Substitute numerical values and evaluate M: M = (100 / s ) − Mg = − N M= Δt N ⎛ 2mv y ⎞ ⎜ ⎟ Δt ⎜ g ⎟ ⎝ ⎠ = 32 g [2(0.00050 kg )(3.13 m/s)] 9.81m/s 2 Conservation of Linear Momentum 835 117 ••• A dumbbell, consisting of two balls of mass m connected by a massless 1.00-m-long rod, rests on a frictionless floor against a frictionless wall with one ball directly above the other. The center-to-center distance between the balls is equal to 1.00. The dumbbell then begins to slide down the wall as in Figure 8-58. Find the speed of the bottom ball at the moment when it equals the speed of the top ball. Picture the Problem Assume that the connecting rod goes halfway through both balls, i.e., the centers of mass of the balls are separated by 1.00 m. Let the system include the dumbbell, the wall and floor, and Earth. Let the zero of gravitational potential be at the center of mass of the lower ball and use conservation of energy to relate the speeds of the balls to the potential energy of the system. By symmetry, the speeds will be equal when the angle with the vertical is 45°. Use conservation of energy to express the relationship between the initial and final energies of the system: Ei = E f Express the initial energy of the system: Ei = mgL where L is the length of the rod. Express the energy of the system when the angle with the vertical is 45°: Ef = mgL sin 45° + 1 (2m ) v 2 2 Substitute to obtain: ⎛1⎞ 2 gL = gL⎜ ⎟+v ⎝ 2⎠ Solving for v yields: 1⎞ ⎛ v = gL⎜1 − ⎟ 2⎠ ⎝ Substitute numerical values and evaluate v: v= (9.81m/s )(1.00 m )⎛1 − ⎜ = 1.70 m/s 2 ⎝ 1⎞ ⎟ 2⎠ 836 Chapter 8 ...
View Full Document

This note was uploaded on 11/10/2011 for the course PHYS 1301W taught by Professor Marshak during the Fall '08 term at Minnesota.

Ask a homework question - tutors are online