Chp 6.9 (Energy & Momentum)

Chp 6.9 (Energy & Momentum) - Physics 7A Discussion...

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Unformatted text preview: Physics 7A Discussion Notes Chapters 7, 8, and 9: Energy and Momentum Aaron Alpert October 2010 1 1.1 Work-Energy Definition of Work We know that a force causes an acceleration, but if we’re interested in the velocity, we have to know “how much” force we applied. There are two logical choices for this: the distance over which the force was applied and the time for which the force was applied. Let us use the distance over which the force was applied (later, we’ll see what happens when we use time). We define work as the force times the displacement. If you push a block for 2 m with a force of 7 N , you have done 14 N · m of work on the block. What if the force is not directed on the path of the object? For example, a block moves down a ramp inclined at θ. Gravity pulls down, but only the component along the ramp does work. That component is mg sin θ. Thus, if a 3 kg block slides 5 m down a 30◦ ramp, the work done on the block by gravity is mg sin θ · L = (3kg )(10m/s2 )(sin 30◦ )(5m) = 75 N · m. Vectorially, this is simple, W ≡ F · ∆x (1) Notice that from the dot product definition, this is equivalent to (F cos θ)(∆x), where F cos θ is the component of force in the direction of the displacement. If the force is variable over the distance, we use an integral expression for work.1 W≡ x ￿2 F dx (2) x1 1.2 Examples of Work The Fast and the Furious. Johnny Tran is driving along a road in his Honda S2000, which is 1100kg . It has a rocket which is capable of providing a horizontal thrust force. Indeed, if the force is variable and not in the direction of￿ the displacement, we need an expression to combine the vector and calculus expressions. This is W ≡ C F · dx, which is called a line integral. It requires multivariable calculus, so we won’t go into it. 1 1 Initially, the friction between his tires and the road is µ = .80, but it starts to rain and get slippery. The friction decreases steadily, and by the time he has driven 2000m along the road, µ = .20. If he was driving at a constant speed, what was the work done by the thruster on the car? By friction? By gravity? Answer: The Fast and the Furious. This problem is simple enough without a force diagram. It is not sloped, so N = mg , and Ff = µN = µmg backward. Because his speed is constant, the force exerted by the thruster must equal friction, and FT = µmg . Friction is not constant. In fact, it decreases linearly from .80 to .20 over a distance of 2000m. Thus, x µ = µ(x) = .8 − .6 2000 . What is the force done by the thruster? We integrate the force over the displacement to find the work. 2000 2000 ￿ ￿2000 ￿ ￿￿ x￿ x2 W= Frocket dx = .8 − .6 dx = .8x − .6 = 1000 J 2000 4000 0 0 0 (3) The rocket does 1000 J of work on the car. However, the force of friction is equal to the thrust, but in the opposite direction. Thus, it does negative work on the car, specifically, −1000 J . Gravity and normal act perpendicular to the displacement. Thus, they do no work on the car. 1.3 Derivation of the Work-Energy Principle Using the calculus expression of work,2 we use Newton’s Second Law to replace force by mass times acceleration. We realize that acceleration is dv/dt and that dx is v dt, and we substitute this in. x ￿2 ￿t2 dv W = ma dx = m (v dt) dt x1 t1 We “cancel” out the dt’s and get W= ￿v2 1 mv dv = m v 2 2 ￿ ￿ v2 v1 v1 We define the following quantity as kinetic energy for a particle of mass m moving at velocity v: 1 KE = T ≡ mv 2 2 We can see quite clearly the work-energy principle: The work done on a particle is equal to its change in kinetic energy. If you don’t want to do calculus, you can just use the kinematic equations for constant acceleration. We 1 (v 2 − v 2 ) can rearrange one of the equations to find a = 2 ∆x 0 . Then, if work is force times distance, work is mass times acceleration times distance, which is F ∆x = ma∆x = m 2 1 2 (v 2 − v0 ) ∆x ∆x 2 2 = 1 mv 2 − 1 mv0 . 2 2 2 Units. Since work is force times distance, work has units of Newtons times meters. This is defined to be the Joule (abbreviated J). 1 Joule = 1 J ≡ 1 Newton × 1 meter (4) Later on, we will see that units of torque are also N · m, but torque is not an energy quantity. We use Joules only for energy, never for torque. 1.4 Potential Energy General Principle. Sometimes certain forces depend only on where you are (or how far you’ve been displaced). For example, gravity depends only on how far away your are from the mass (inverse square law). These kinds of forces are called conservative forces. In contrast, you have things like air resistance (which is proportional to velocity—the faster you go, the strong the air resistance) and friction (which opposes velocity), which are called non-conservative forces. Considering that work is force times displacement, if the force is conservative, this￿does something special. We can see from calculus that if we integrate ￿ W = Fcon dx = F (x) dx, then work will be a function of the change in position only. We call this function the potential energy. In general, Fconservative = − dP E dU =− . dx dx (5) Why is the negative sign used? When a field does positive work on a particle, the speed (and KE) increases. That corresponds to a loss in PE. So, the change in PE must be equal to the negative of the work done. More generally, Fconservative = −∇U = − ∂U ∂U ∂U i− j− k ∂x ∂y ∂z (6) Gravity. When we use the simple expression that gravity is mg downward (or, more properly, −mg j), we can see that if we multiply this force by a change in the vertical position ∆y , we find that gravitational potential energy is ∆P Eg = −(−mg )∆y = mg ∆y . This is often expressed as mgh. Springs. According to Hooke’s Law, a spring exerts a force Fs = −k ∆x where k is the spring constant and ∆x is the amount the spring has been stretched from its equilibrium, or resting length. (If it is elongated, it pulls back, and if it has been compressed, it pushes ￿x back—hence the negative sign.) The work done by the spring, therefore, is 0 Fs dx = − 1 kx2 . 2 This is the negative of the change in potential energy, so, P Espring = 1 kx2 . 2 1.5 Putting It Together and Example Problems Putting It Together. Now we have the concept of work, kinetic energy, and potential energy. The work-energy principle states that the change in energy of a system is equal to 3 work done on the system. Formulaically, ∆KE + ∆P E = W (7) If there were other types of energy (chemical, thermal, etc.), we would have to include them on the left hand side of the equation. Example: Blitzkrieg. In World War II, Polish soldiers on skis attacked German tanks. (The tanks won.) Suppose a 70 kg Polish skiier is on top of a hill that is 45 m above the ground and inclined at an angle of 20◦ . The coefficient of friction between the skis and the snow is .1. If the skiier starts at 3 m/s, what is his final speed when he is annihilated by the tank at the bottom of the hill? Answer: Blitzkrieg. First, consider the potential energy. He descends 45 m, so he loses mgh = (70kg )(9.8m/s2 )(45m) = 30, 870 J of potential energy. Friction acts on the man, but we must consider it as the work done by friction on the man (because friction is nonconservative). The normal force the man experiences is mg cos θ, so the frictional force is µmg cos θ. The force of friction acts all along the surface of the slope, which has a length of H . Thus, the work done by friction is µmg cot θH = (.1)(70)(9.8)(cot 20◦ )(45) = 8481.5 J . sin θ This is negative work, i.e., work done against the direction of velocity. We can solve for the change in the skiier’s kinetic energy. ∆KE + ∆P E = W ∆KE − 30870 = −8481.5 ∆KE = 22388.5 J We must remember that he started at some initial velocity 3m/s when considering the change in KE. 12121 1 2 ∆KE = 22388.5 J = mvf − mv0 = (70kg )vf − (70kg )(3m/s)2 2 2 2 2 We solve to find that vf = 25.5 m/s. Example: Operation Bartowski. Chuck Bartowski (85 kg) and Sarah Walker (70 kg), jump off a bridge (assume his velocity at this point is zero) with a bungee cord (both of them on the same cord). The bungee cord has a resting length of 7 m and a spring constant (technically, elastic modulus) of 130 N/m. As their velocity instantaneously reaches zero, Sarah lets go and parachutes away. How high will Chuck rise above the bridge on the recoil? During the recoil, what is Chuck’s maximum speed? Answer: Operation Bartowski. For record keeping, we will say that their initial energy is zero because their initial velocity is zero, the bungee cord was unstretched, and the bridge is defined to be at height zero. Now, we have to know how far down they go at the moment Sarah lets go. Call this distance below the bridge b for bottom. The change in gravitational 4 PE is −M gb where M is the mass of both of them together, 155 kg. The spring is stretched b − 7, so it stores 1 k ∆x2 = 1 (130N/m)(b − 7m)2 . At the very bottom, the speed is zero, so 2 2 KE is zero. The total energy is constant throughout the problem, so, 1 M gy + k (∆x)2 = 0 2 1 −M gb + k (x − L)2 = 0 2￿ ￿ Mg b2 − 2 L + + L2 = 0 k Plugging in numbers, we see that b = 36 m. Then, Sarah lets go. At that moment, the total in the Chuck-bungee cord system is 1 −mc gb + k (b − L)2 = 24.7 kJ. 2 At the highest point above the bride, Chuck is at a height h, his KE is zero, his P Eg = mc gh, the P Es = 1 k (h − L)2 , and the total energy of the system is 24.7 kJ. 2 1 mc gh + k (h − L)2 = 24700 2 12 12 kh + (mc g − kL)h − kL − 24700 = 0 2 2 This is another quadratic, which yields h = 20.1 m. The only force pulling Chuck up is the bungee cord when is more than 7 meters below the bridge. After that, gravity slows him down (and the bungee cord later acts to slow him down, too). Thus, the maximum speed occurs sometime when he is between 7 and 36 meters below the bridge. When he is y below the bridge, his PE is −mc gy + 1 k (y − L)2 and his 2 total energy is 24.7 kJ. Thus, 1 KE + P E = KE − mc gy + k (y − L)2 = 24.7 J 2 Maxmimizing KE is the same as maximizing the speed. We take the derivative of KE and set it equal to zero. ￿ ￿ d KE d 1 2 = mc gy − k (y − L) + 24.7 = mc g − k (y − L) = 0 dy dy 2 c We can see directly that y = mk g + L = 13.4 m. He reaches his maximum speed when he is 13.4 m below the bridge. At this point, his kinetic energy is, 1 KE − mc g (13.4) + k (13.4 − L)2 = 24.7 J → KE = 33.2 kJ 2 ￿ From here, we can find his speed simply as v = 2KE = 27.95 m/s. m 5 Slipping Uniform Cord. A uniform cord of mass m and length L hangs on a pulley. The cord is situated such that y hangs on the right side of the pulley and L − y hangs on the left side. What will be the speed of the cord once it falls off the pulley (assuming it starts from rest)? Answer: Slipping Uniform Cord. This is a problem from Giancoli, but in the chapter on forces. It can be solved much more easily with energy. Initially on the right side of the y pulley, a fraction L of the cord sits. It has mass my . Its height, as measured from the center L of mass, is located y/2 below the pulley. (We will set the origin at the pulley.) Similarly, on the left side, a mass of cord m(y−L) sits with center of mass L−y below the pulley. The initial L 2 energy of the system is: ￿ ￿￿ ￿ ￿ my ￿ ￿ y ￿ m ( y − L) y−L E0 = mleft ghleft + mright ghright = g − +g − (8) L 2 L 2 At the moment the end of the cord slides off the pulley, the center of mass is L/2 below the pulley, and the velocity is v . The energy at this point is: ￿ ￿ 12 L 12 Ef = mgh + mv = mg − + mvf (9) 2 2 2 We set equations 8 and 9 equal to one another and solve for vf . E0 ￿ ￿￿ ￿ ￿ my ￿ ￿ y ￿ m ( y − L) y−L g − +g − L 2 L 2 2 2 −gy − gy + 2gyL − gL2 ￿ 2gy 2 2gy − L = Ef ￿ L 12 = mg − + mvf 2 2 2 2 = −gL + Lvf = vf ￿ 1.6 Power dW ˙ =W dt Power is defined as the time rate of work done on an object. P≡ It has units of J/s, commonly called the W att. By the work-energy theorem, W = ∆KE , ˙ so P = W = d KE , and we can use the time derivative of KE to express kinetic energy in dt terms of the force and velocity. ￿ ￿ dW d d dx dx P= = F dx = F dt = F =F ·v (10) dt dt dt dt dt ￿ ￿ ￿ ￿ dKE d1 1 dv dv = = mv · v = m ·v+v· = ma · v = F · v (11) dt dt 2 2 dt dt 6 Bouncy Ball. A 50g rubber ball is dropped from 2.5m. It bounces back to a height of 1.8m. A photogate timer says that this motion takes 1.400s. What was the power delivered to the ball during the bounce? If the ball has cv = 2kJ/kg · K , about how much hotter is the ball? Answer: Bouncy Ball. During free fall, ∆y = y0 + 1 at2 , but a = −g and y0 = 0 so ˙ ˙ 2 ￿ t = 2∆y . A 2.5m fall should take .7143s and a 1.8m fall should take .6061s. However, g it took 1.400s, which is an additional .08s, approximately. The difference in energy was mg (yf − y0 ) = (.05kg )(9.8m/s2 )(1.8 − 2.5m) = −.343J . This work was done over .08s, so J the (average) power during the bounce was ∆W = .343s = 4.288W. ∆t .08 The .343 Joules were converted into thermal energy. We can use the definition of specific heat from chemistry for internal energy, U = mcv ∆T . In this case, U = .343J, m = U .343 .05kg, cv = 2000J/kg · K . Thus, ∆T = mcv = (.05)(2000) = .0034◦ C. 1.7 Gravitational Potential Energy Since gravitational force is purely a function of the position (i.e., distance between centers of mass), it can be expressed as a potential energy. To find it, we just integrate the work done by gravity in changing from r1 to r2 . P EG = − ￿r2 FG dr = − ￿r2 ￿ ￿r Gm1 m2 Gm1 m2 2 Gm1 m2 Gm1 m2 − dr = − = − 2 r r r1 r2 r1 r1 r1 Suppose now that we decide to determine gravitational potential energy relative to some 1 fixed “reference” radius. Thus, r1 is a constant, and UG = c − Gmr m2 . We decide to set this Gm1 m2 fixed radius at ∞. Thus, U∞ = c − lim = c = 0. Very simply, then, r →∞ r UG = − Gm1 m2 r This allows the concept of a gravity well. As you get closer to the center of the mass’s core, your potential energy drops with 1/r. That is converted to kinetic energy, and you speed up, quite quickly. We talk about a gravity well as the amount of energy you would need to “climb out” of the gravitational field caused by a planet. If this energy came completely from kinetic energy, the speed necessary to overcome −Gm1 m2 /r is called the escape velocity. Gravity wells become very important in Einstein’s conceptualization of the curvature of spacetime. Getting a Man on the Moon. It is May 26, 1961, and you are a NASA project manager. Yesterday, the President of the United States announced that by the end of the decade, the US would put a man on the moon. You are asked by the NASA Administrator, minimally, how much fuel would it take to get a man on the moon in a 2 × 106 kg space vehicle? (Assume that the change in the mass of the fuel is negligable, the energy content of rocket fuel is 7 Figure 1: Gravity Well of the Sun. about 43M J/kg , and the following values: me = 5.9742 × 1024 kg, mm = 7.36 × 1022 kg, re = 6.378 × 106 m, rm = 1.737 × 106 m, dm→e = 3.844 × 108 m, G = 6.673 × 10−11 m3 /kg · s2 .) Answer: Getting a Man on the Moon. When the shuttle is on the surface of the earth (before launch), the gravitational potential energy is: P Ee = − Gme m Gmm m − re d − re (12) When the shuttle lands on the surface of the moon, the gravitational potential energy is: P Em = − Gme m Gmm m − d − mm rm (13) The change in PE is found by subtracting (13) from (12). We also make the assumption that d >> re and d >> rm , so we approximate d − re ≈ d − rm ≈ d. ￿ ￿ Gme m Gmm m Gme m Gmm m me mm − me mm ∆P E = − − + + = Gm + − = 1.17×1014 J d rm re d re d rm (14) 10 That energy must be supplied by the rocket fuel, which can hold 4.3 × 10 J/kg . Thus, ∆P E 1.17 × 1014 = = 2728.1kg E￿ 4.3 × 1010 (15) It’s All About Mie. The Mie Potential is a model representing the energy between two atoms separated by r. It is represented mathematically as UM ie (r) = − 8 A B +n rm r (16) For a certain atom, n = 3, m = 2, B = 10, A = 15. (A) Plot the function. (B) What would be the minimum distance that would be able to cause two atoms to separate indefinitely? (C) At this distance, what is the force between the atoms? (D) What is the equilibrium resting length, and is it stable? Figure 2: Mie Potential. Answer: It’s All About Mie. For part A, see figure 2. For part B, the energy at ∞ is zero, so U (∞) = 0 = Umin . U (r ) = − 15 10 + 3 =0 r2 r 15 10 =3 r2 r 15 10 10 = 3 →r= ≈ .667 2 r r 15 For part C, the force is −dU/dr. dU 30 30 F =− = − 3+ 4 r r ￿ dr ￿ 2 30 30 F =− + = 50.625 3 (2/3)3 (2/3)4 9 For part D, at the equilibrium position, the force is zero. 30 30 + 4 =0 r3 r 30 30 = 4 →r=1 r3 r F=− Message in a Bottle. A poor astronaut is accidentally marooned on an asteriod of radius 5000 m and mass 2.6 × 1015 kg . All he has is a small pen of mass 10 g . The pen also has a spring inside of it with spring constant 400 N/m. How much must he compress the spring in order to guarantee that the pen (containing his emergency message) is able to leave the asteroid? Answer: Message in a Bottle. The “escape energy” is equal to the energy needed to remove the pen an infite distance from the surface of the asteroid. ￿ ￿ Gm1 m2 Gma mp Gma mp Gma mp Eesc = ∆ − =− + = (17) r ∞ ra ra where mp is the mass of the pen, ma is the mass of the asteroid, and ra is the radius of the asteroid. The change in the energy of the spring is: ￿ ￿ 12 1 1 ∆ kx = k (0)2 − kx2 (18) 2 2 2p where xp is the amount the astronaut compresses the pen. From here, we just use the conservation of energy. ✿ ✘ ∆KE ✘✘✘ + ∆P E 0 ❃ ✚ =✚ W 0 (19) (20) (21) Gma mp 1 2 − kxp = 0 ra 2 ￿ 2Gma mp xp = = .029 m kra 1.8 Extra Problems Boing-y. As shown in figure 3, a block of mass mB is on a ramp. The kinetic coefficient of frition is µk . The block is connected to a spring of stiffness k on one side, and on the other, it is connected to mass mA via a string. (A) Assuming that everything starts from rest, with the spring at its resting length, and that mA >> mB , how far up the ramp will mB go? (B) What is the maximum speed of mB ? 10 Figure 3: Block on a Rough Ramp with Spring and Second Block. Answer: Boing-y. (A) At its maximum distance up the ramp, v = 0 so KEf = 0. Also, KE0 = 0. We use the conservation of energy, including the potential energy of gravity for each block, the potential energy of the spring, and the work done by friction. Assume that the block travels L up the ramp. ✿ ✘0 ∆KE ✘✘✘ + mA g ∆y 1 + mB g ∆y + k ∆(x − L0 )2 = Wf ric 2 mA g ∆y = −mA gL mB g ∆y = mB gL sin θ 1 1 k ∆(x − L0 )2 = kL2 2 2 Wf ric = −µk N L = −µk mB g cos θL 1 −mA gL + mB gL sin θ + kL2 = −µk mB g cos θL 2 1 −mA g + mB g sin θ + kL = −µk mB g cos θ 2 2 L = (mA g − mB g sin θ − mB µk g cos θ) k (22) (23) (24) (25) (26) (27) (28) (29) Note that in order for L to be positive (and thus, for our analysis to make sense), we must have mA g > mB g sin θ − mB µk g cos θ. This would usually not be a problem as θ and µk are relatively small. (B) We could maximize the velocity, but I will simply maximize the KE, which does the same thing. This time, we must include the KE in the conservation of energy equation, and 11 I will use y instead of L to denote the length along the ramp. 1 ✯ ✟0 KE (y ) − ✟✟ 0 − mA gy + mB gy sin θ + ky 2 = −µk mB g cos θy KE 2 12 KE (y ) = mA gy − mB gy sin θ − ky − µk mB g cos θy 2 dKE = mA g − mB g sin θ − ky − µk mB g cos θ = 0 dy mA g − mB g (sin θ + µk g cos θ) y= k 1 KEmax = [mA g − mB g (sin θ + µk g cos θ)]2 2k mA g − mB g (sin θ + µk g cos θ) ￿ vmax = ( mA + mB ) k (30) (31) (32) (33) (34) (35) 2 2.1 Impulse-Momentum Impulse When we started talking about work, we wanted to have a way of quantifying “how much” force (or umph, as I like to say) we applied. Work was the distance over which the force was applied, but now we turn our attention to the amount of time for which the force was applied. We shall call this quantity impulse. Simply, if you apply 13 N of force for 3 s, · you have imparted 39 N · s = 39 kgsm of momentum. Unlike work, which had to “multiply” two vectors (displacement and force), we only have one vector (force) multiplied by a scalar (time). Thus, impulse is a vector quantity! We can define momentum as J≡ ￿t2 ¯ F dt = F ∆t (36) t1 ¯ where F is the average force over the time interval. If we are given a Fx vs. t graph, we can simply find the area under the curve representing Fx (t).3 2.2 Momentum Integrating the Impulse. When we were dealing with work, we substituted ma for F, and when we integrated it, we found the work was equal to the change in a quantity 1 mv 2 . 2 It would be difficult to graph the total force F as a function of time, requiring a 3D graph for a 2D force. Usually, Fx and Fy would just be shown on separate graphs. 3 12 We perform the same substitution here, remembering that a ≡ J= ￿t2 F dt = ￿t2 ma dt = m ￿t2 dv . dt dv dt = mv(t2 ) − mv(t1 ) = m∆v dt (37) t1 t1 t1 Impulse-Momentum Theorem. We found that 1 mv 2 was a very useful quantity, and 2 we gave it a name: kinetic energy. Likewise, we give the quantity mv a name: momentum. Momentum = p ≡ mv (38) · It has units of N · s or, equivalently, kgsm . Just as we defined the Work-Energy Theorem, we define the Impulse-Momentum Theorem. J = ∆ ( m v ) = ∆p It is also quite common to describe force in terms of momentum. ￿ F= dp =p ˙ dt (39) (40) In fact, this formulation of Newton’s Second Law is sometimes referred to as Euler’s Balance Law. It is preferable to F = ma because (a) it allows us to do problems with changing mass, and (b) the law holds (for “generalized momenum”) in quantum and relativity. 2.3 Conservation of Momentum What if there are only two objects in our system and the only forces acting on these objects comes from the other object? In other words, F1 , the net force on object 1 is due entirely to object 2, and vice versa regarding F2 . (A good example of this is planetary bodies.) Newton’s Third Law tells us that these forces must be equal and opposite, so F1 = −F2 . If we calculate ￿ impulse delivered to object 1 (over some time interval from t1 to t2 ), it is the ￿t t simply J1 = t12 F1 dt = ∆p1 , and the impulse delivered to object 2 is J2 = t12 F2 dt = ∆p2 . As stated before, F1 = −F2 , so J1 = −J2 . Thus, ∆p1 + ∆p2 = 0. In a system of two particles (with no outside forces), the total momentum p1 + p2 is a conserved quantity. We can expand this to a system of N particles as follows: p = p1 + . . . + pN = N ￿ i=1 pi (41) This equation is valid only where there are no external forces. In words, momentum is conserved in a closed system. If it is an open system (i.e., there are forces that can’t be attributed to objects 1 through N), then you must include the impulse delivered by these outside forces. 13 Collisions. What happens when two objects collide? The forces between the objects are initially about zero, then they become very large for a very short amount of time, and finally the forces return to about zero. During that collision, it would be difficult for us to measure the force of impact or the time of impact, but it is easy to measure the change in momentum. Also, forces like gravity do technically act on the particle during the collision, but these other forces are so much less than the force of impact, that outside forces, in fact, impart very little impulse over the short time period. Thus, during an impact, we can ignore the outside forces and consider only the forces between the colliding particles. Thus, it acts as a “closed system” during the impact, and the momentum is conserved for collision. 2.4 Elasticity of Collision. Normally, kinetic energy is not conserved during a collision. A good amount is converted to heat, etc. However, “bouncy” materials lose very little of their KE. We call a collision perfectly elastic if there is no loss of kinetic energy.4 On the other hand, if two objects stick together after collision (imagine throwing a piece of clay on a moving cart and it just sticks), a good amount of kinetic energy is lost (possibly all of it). We call that kind of a collision perfectly inelastic, or, more rarely, perfectly plastic. 2.5 Center of Mass Definition. Now that we are working with systems of particles, it may be useful to define some sort of “average” position of the system. The “average” position of two billiard balls is right in the middle between them, but what about the “average” position of a person vs. a planet? Obviously, the planet dominates the system. Thus, we give greater importance to the massive elements of the system. We define the center of mass as follows: ¯ x≡ m1 x 1 + m 2 x 2 + . . . + m N x N ￿ mi = xi m1 + m2 + . . . + mN M i=1 N (42) Here, m1 is the mass of particle 1 and x1 is its (vector) position. Relation to Dynamics. Now, consider that each particle in the system has a net force ￿ F acting on it. These forces can be broken up into internal forces (between particles in the system) and external forces (due to particles and fields that are not part of the system. For one single particle j , ￿ Fj = Fj,int + Fj,ext = maj (43) In reality, there is no such thing as a perfectly elastic collision. We use something called the coefficient of restitution, or e, to measure the “bounciness” of an object. If e = 0.8, then the objects bounced away from each other at 80% of the (relative) speed at which they approached each other. For a perfectly elastic collision, e = 1, and for a perfectly inelastic collision, e = 0. In this course, we only have time to consider perfectly elastic and perfectly plastic collisions. 4 14 Suppose we were to add up all of the F = ma equations (such as the one above). We would get, F1,int + F1,ext + . . . + FN,int + FN,ext = ma 1 + . . . + ma N = N ￿ j =1 N ￿ j =1 Fj,int + Fj,ext = (44) ma j However, each force has an equal and opposite reaction force. Thus, the internal forces cancel each other out (as they form equal and opposite pairs). Only the external forces are left. ￿ Fext = ma1 + m2 a2 + . . . + maN (45) We might like to say that the sum of the external forces is equal to M a, where M is a mass term and a is an acceleration term. The logical choice for M is the sum of all the masses. ￿ ma 1 + m2 a 2 + . . . + ma N ¯ ¯ Fext = ma1 + m2 a2 + . . . + maN = M a → a = (46) M We notice that the term on the right is equal to the second time deriviative of the position of the center of mass. Thus, we can say that the net external force is equal to the total mass times the accleration of the center of mass. Extending this to collisions, if there is no external force, than the center of mass does not acceleratate. Thus, the velocity of the center of mass remains constant!! In terms of momentum, the system’s overall momentum (which ¯ is M v), is conserved in the absence of external forces. 2.6 Impulse-Momentum Examples Figure 4: Chancellor Birgenau Escaping the Mad Students. Whose University? In an attempt to escape from protesting UC Berkeley students (who are vandilizing his house), the 120 kg Chancellor Birgenau runs off a cliff into the 1150 kg escape vehicle driven by Vice Chancellor Breslauer. Assuming that the Chancellor runs straight off the cliff (no jumping), falling 8 meters down and landing 4 meters from the edge of the cliff horizontally, and that the car is initially going 5 m/s in the direction of the Chancellor, what is the velocity of the car after the Chancellor lands in it? 15 Answer: Whose University? The first order of business is finding the Chancellor’s velocity when he lands in the car. We only care about the horizontal component of his landing velocity. If he falls 8 meters, we can find the time of flight by −8 = y0 t + 1 y t2 = −gt2 , ˙ ¨ 2 ￿ and t = 2 ∗ 8/g = 1.278 s. Thus, his x velocity is 4 = xt = 1.278x, and x = 3.13 m/s. ˙ ˙ ˙ Now, we can simply use the conservation of momentum. The Chancellor’s initial speed shall be v1 , the car’s initial speed w1 , the final speed w2 . p1 m B v1 + m c w 1 (120)(3.13) + (1150)(5) w2 = = = = p2 ( mB + mc ) w 2 (1270)w2 4.82 m/s Figure 5: Billard Balls Colliding. Billard Balls. Billard balls are approximately 0.016 kg , but in this imperfect set, the cue ball is 0.014 kg . Initially, the cue ball is coming toward the nine ball at a speed of 12.4 m/s at an angle of 42◦ relative to the wall of the table. The cue ball glances the nine ball, which moves at 8.2 m/s. What is the final velocity of the cue ball (magnitude and direction)? Assume the collision is perfectly elastic. Answer: Billard Balls. The initial velocity of the cue ball is (v1x , v1y ) = (12.4 cos 42◦ , 12.4 sin 42◦ ) = (9.22, 8.30). The initial velocity of the nine ball is (0￿ The fi, 0). 2 2 nal velocity of the cue ball can be (v2x , v2y ) = v (cos φ, sin φ), with final speed v = v2x + v2y . The final velocity of the nine ball is (w2x , w2y ) = 8.2(cos θ, sin θ), with final speed v = ￿ p 1x = p 2x m c v 1x + m 9 w 1x = m c v 2x + m 9 w 2x (.014)(9.22) = (.014)v2 cos φ + (.016)(8.2) cos θ 2 2 v2x + v2y . There are three unknowns: v, φ, and θ. We get our three equations from: conservation of x momentum, conservation of y momentum, and conservation of energy. We start with the x momentum. 16 Now the y momentum. p 1y = p 2y m c v 1y + m 9 w 1y = m c v 2y + m 9 w 2y (.014)(8.30) = (.014)v2 sin φ + (.016)(8.2) sin θ Now the conservation of energy (because the collision is perfectly elastic). Note that the factor of 1/2 has been cancelled out. 2 mc v1 KE1 = KE2 2 2 2 + m 9 w 1 = mc v2 + m 9 w 2 2 (.014)(12.4)2 = (.014)v2 + (.016)(8.2)2 We solve this last line to find v2 = 8.77m/s. (.014)(9.22) = (.014)(8.77) cos φ + (.016)(8.2) cos θ (.014)(8.30) = (.014)(8.77) sin φ + (.016)(8.2) sin θ We now have two equations with two unknowns. Unfortunately, they are not very easy to solve by hand. We can plug them into a CAS (I used We get two solution sets, and we have to ignore the one that doesn’t make sense. That leaves us with θ = 86.9◦ , φ = −6.9◦ . Thus, the cue ball ends up going at 8.77 m/s at a direction of 86.9◦ above the horizontal. Sumo Battle. Arguing about who played the better Mr. Miyagi/Mr. Han, Pat Morita and Jackie Chan decide to have a sumo wrestling fight. Pat, weighing 90 kg, runs toward Jackie at 4 m/s. Jackie, weighing 80 kg, runs toward Pat at 6 m/s. When they collide (assuming they grab on to each other), where at what speed are they travelling, and in what direction? Use a zero-momentum frame of reference. Answer: Sumo Battle. The initial momentum (arbitrarily calling Pat’s velocity positive) is p1 = mP v1 + mJ w1 = (90)(4) + (80)(−6) = −120 N · s. We, however, want to use a reference frame travelling at velocity u such that p1 = 0. To find the momentum in the new reference frame, we simply subtract off u to get the velocities relative to this new frame of reference, and we set it equal to zero. p 1 = m P ( v 1 − u) + m J ( w 1 − u) 0 = (90)(4 − u) + (80)(−6 − u) −(90)(4) + (80)(6) = (−90 − 80)u 17 Solving, we find u = −.706. This means the reference frame is going the same direction in which Jackie Chan is running. After the collision, the momentum will be p2 = (mJ + mP )V2 . However, since p1 = 0 and there are no outside forces, p2 = 0, which implies V2 = 0. However, this is in the zero momentum frame. In the absolute frame, v2 = V2 + u = 0 − .706 = −.706. Thus, they are travelling at .706 m/s in the direction Jackie Chan was originally going. These notes ©2010 by Aaron Alpert. 18 ...
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