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Unformatted text preview: Chapter 9
Impulse & Momentum Linear Momentum
n The linear momentum of a particle or an object that can be modeled as a particle of mass m moving with a velocity v is defined to be the product of the mass and velocity:
n p=mv
n The terms momentum and linear momentum will be used interchangeably in the text Linear Momentum, cont
n Linear momentum is a vector quantity
n Its direction is the same as the direction of v n n n The dimensions of momentum are ML/T The SI units of momentum are kg m / s Momentum can be expressed in component form:
n px = m vx py = m vy pz = m vz Newton and Momentum
n n Newton called the product mv the quantity of motion of the particle Newton's Second Law with constant mass (i.e. dm/dt = 0) can be used to relate the momentum of a particle to the resultant force acting on it This is a generalization of Newton's 2nd Law. Newton's Second Law
n The time rate of change of the linear momentum of a particle is equal to the net force acting on the particle
n n n This is the form in which Newton presented the Second Law It is a more general form than the one we used previously This form also allows for mass changes n Applications to systems of particles are particularly powerful Conservation of Linear Momentum
n Whenever two or more particles in an isolated system ,i.e no outside forces are present, interact the total momentum of the system remains constant
n n The momentum of the system is conserved, not necessarily the momentum of an individual particle This also tells us that the total momentum of an isolated system equals its initial momentum Conservation of Momentum, 2
n Conservation of momentum can be expressed mathematically in various ways
n n n In component form, the total momenta in each direction are independently conserved
n ptotal = p1 + p2 = constant p1i + p2i= p1f + p2f n Conservation of momentum can be applied to systems with any number of particles pix = pfx piy = pfy piz = pfz Conservation of Momentum, Archer Example
n n The archer is standing on a frictionless surface (ice) Approaches:
n n n Newton's Second Law no, we have no information about F or a Energy approach no, no information about work or energy Momentum yes Archer Example, 2
n n n n Let the system be the archer with bow (particle 1) and the arrow (particle 2) There are no external forces in the xdirection, so it is isolated in terms of momentum in the xdirection Total momentum before releasing the arrow is 0 The total momentum after releasing the arrow is p1f + p2f = 0 Archer Example, final
n The archer will move in the opposite direction of the arrow after the release
n Agrees with Newton's Third Law n Because the archer is much more massive than the arrow, his acceleration and velocity will be much smaller than those of the arrow Conservation of Momentum, Kaon Example
n The kaon decays into a positive p and a negative p particle Total momentum before decay is zero Therefore, the total momentum after the decay must equal zero
n n n p+ + p = 0 or p+ = p Momentum Conservation Demos
n n n Gas powered Tricycle Water Rocket Ballistic Cannon Ballistic Cannon
A projectile (Mp) is fired from a cannon cart (Mcc). The projectile lands in the receiver cart (Mrc). In this DEMO Mcc= Mp + Mrc There are no horizontal external forces. Thus the total momentum remains constant. Ballistic Cannon
Initially the momentum of the system is zero. Since only internal forces are involved the total system momentum stays zero. After the cannon fires and before the projectile lands in the receiver cart: MccVcc + MpVp = 0 Vcc = (Mp/Mcc)Vp When the projectile lands in the receiver cart the momentum is still zero and the cannon cart is still going to the left. 0 = MccVcc + (Mp+Mrc)Vrc Since Mcc = Mp + Mrc, Vcc =  Vrc Impulse and Momentum
n n n From Newton's Second Law, F = dp/dt Solving for dp gives dp = Fdt Integrating to find the change in momentum over some time interval
tf Dp = p f  p i = Fdt = J
n The integral is called the impulse, I, of the force F acting on an object over Dt ti ImpulseMomentum Theorem
n This equation expresses the impulsemomentum theorem:The impulse of the force F acting on a particle equals the change in the momentum of the particle
n This is Newton's Second Law (previously we assumed m is constant). More About Impulse
n n n n Impulse is a vector quantity The magnitude of the impulse is equal to the area under the forcetime curve Dimensions of impulse are M L / T Impulse is not a property of the particle, but a measure of the change in momentum of the particle Impulse, Final
n n n The impulse can also be found by using the time averaged force J = F Dt This would give the same impulse as the timevarying force does The cart's change in Momentum is
1. 2. 3. 4. 5. 6. 30kgm/s 20kgm/s 10kgm/s 10kgm/s 20kgm/s 30kgm/s
3 0k gm 35% 26% 22% 10% 6% 1% /s /s 0k gm /s /s 20 kg m /s 2 0k g 1 30 kg m /s m 10 kg m The cart's change of momentum is 1. 30 kg m/s. 2. 20 kg m/s. 3. 10 kg m/s. 4. 10 kg m/s. 5. 30 kg m/s. Impulse & System of Particles
n In some situations we might consider an object acted upon by an external force
n n e.g. a ball is pulled toward the earth There is a momentum change of the ball Conservation of Momentum Depends on the System Take Vb = 0 at t=0. FEoB(t0) = Mbvb0 (FEonB/MB)t = VB gt = VB Impulse & System
n If we consider the system of the ball and the earth and recall that the force on the ball is equal and opposite to the force of the earth on the ball.
n n The forces are internal forces No external forcetotal implies momentum is constant Conservation of Momentum Depends on the System MbVb+Mve= 0 Ve = (Mb/Me)Vb Problem Deformed Ball
A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in contact with the pavement, the lower side of the ball is temporarily flattened. Suppose that photographs show the maximum depth of the dent is 1 cm. Compute an estimate for the acceleration of the ball while it is in contact with the pavement and the time duration it is deformed. Solution Deformed Ball
To find the contact time we cam use the Impulse equation using the average force. To find the average force we use kinematics to find the average acceleration, assuming constant a gives an approximate average a. J = Fave Dt = Pf  P
To find a, first find the velocity the ball has at the instant the bottom side hits the ground assuming the ball falls 1.50 m. v 2 = v 2 + 2a(y f  yi ) f i v 2 = 0 + 2(9.80m / s 2 )(1.50m  0) f v f = 5.42m / s Solution Deformed Ball
We can use the same equation to find the acceleration, assumed constant, as the ball moves from where it first touched the ground until it stopped. v 2f = v 2 + 2a(y f  yi ) i
0 = (5.42m / s) 2 + 2a(0  .01m) a = 1470m / s 2 Since this equation used is only valid for constant acceleration this is approximately the average acceleration. The momentum of the ball goes from Pi = m(5.42 m/s) to 0 when it is being deformed. Using the impulse theorem: Problem Deformed Ball
J = Fave Dt = p f  pi Dt = p f  pi Fave = p f  pi ma 0  m(5.42m / s) @ m1.47 10 3 m / s 2 Dt @ 3.69 10 3 s
Since it bounces back to nearly the same height the time to deform and undeform is approximately twice this time. Start 3/12
n Impulse J = Fdt equals the change in momentum t
i tf tf Dp = p f  p i = Fdt = J or approximately
ti Dp = p f  p i Fave Dt Collisions Characteristics
n n n We use the term collision to represent an event during which two particles come close to each other and interact by means of forces The time interval during which the velocity changes from its initial to final values is assumed to be short The interaction force is assumed to be much greater than any external forces present
n This means the impulse approximation can be used Car Collision Example Car Crash Example
In a particular car crash test, a car of mass 1500 kg collides with a wall. The initial and final velocities of the car are vi = 15.0 m/s and vf = 2.60m/s. The collision lasts 0.150 s. Find the impulse and average force on the car. Car Crash Example
Assume force exerted by the wall on the car are greater than any other forces that cause its velocity to change. pi = (1500 kg)(15.0i m/s) = 2.25x104 i kgm/s Pf = (1500 kg)(2.60i m/s) = 0.39x104 i kgm/s J =Fave Dt = Dp = pf  pi =2.64x104 i kgm/s Fave = Dp/Dt = (2.64x104 i kgm/s)/0.150s = 1.50x105 i N (33.8x103 lbs) to the right Note that the wall force does two things. It stops the car and pushes it backwards. Collisions Example 1
n n Collisions may be the result of direct contact The impulsive forces may vary in time in complicated ways
n This force is internal to the system n Momentum is conserved Collisions Example 2
n n n The collision need not include physical contact between the objects There are still forces between the particles This type of collision can be analyzed in the same way as those that include physical contact Types of Collisions
n In an elastic collision, momentum is conserved and particles bounce off each other.
n In a perfectly elastic collision the ball bounces back at the same speed n In a perfectly inelastic collision, momentum is conserved and the objects stick together after the collision Forces in Elastic Collisions
http://www.physics.umd.edu/lecdem/services/demos/demosc7/c725.htm Show that a larger impulse is imparted by a perfectly elastic collision. Use a superball for the perfectly elastic collision and an ordinary ball for the nonperfect elastic collision. tf ti J = Fdt = p f  p i 1) p f = p i fi J = 2p i 2) p f << p f fi J @ p i Perfectly Inelastic Collision Perfectly Inelastic Collisions
n n Since the objects stick together, they share the same velocity after the collision m1v1i + m2v2i = (m1 + m2) vf note that the v's are vectors and the sign is assumed in the vector Perfectly Inelastic Collision
DEMO: collision of pellet w/stationary car BEFORE COLLISION pi =mpvp AFTER COLLISION pf = (mp+mc)vc+p pi=pf fi vp = vc+p[(mp+mc)/mp] TwoDimensional Perfectly Inelastic CollisionExample
n n n Before the collision, the car has the total momentum in the xdirection and the van has the total momentum in the ydirection After the collision, both have x and ycomponents What is the velocity after the collision and what is q? TwoDimensional Perfectly Inelastic Collisioncont'd
n n n Let 1 =car, 2 = van: m1 =1500 kg, m2 = 2500 kg, v1ix =25.0 m/s & v2iy = 20.0 m/s. TwoDimensional Perfectly Inelastic Collisioncont'd
n Momentum conservation:
n x: m1v1ix = (m1+m2)vfcosq
n n y: m2v2iy = (m1+m2)vfsinq
n 3.75x104 kgm/s = 4000vf cosq 5.00x104 kgm/s = 4000vf sinq n Divide the y equation by the x equation giving
tanq = 5.00/3.75 => q = 53.1 4 nusing the y Eq. vf = 5.00x10 /(4000 sin53.1) = 15.6 m/s
n Inelastic Collision Between a Stream of Particles and a Ball
A continuous stream of matter moving at a velocity u impacts an object of mass M that is moving with velocity v. The mass of particles that impact the mass M in a time Dt is DM. During this time the velocity increases by Dv. Applying the impulse momentum theorem taking the system to be the incoming mass DMrand the ball. r r r Fnet.ext Dt = Dp = p f  pi r r r r r Fnet.ext Dt = [(M + DM )(v + Dv]  [Mv + DMu ] r r r r r Fnet.ext Dt = MDv + DM (v  u) + DMDv r r r r r Fnet.ext Dt = MDv + DM (v  u) + DMDv r r Dv DM r r DM r Fnet.ext = M + (v  u) + Dv Dt Dt Dt Inelastic Collision Between a Stream of Particles and a Ball Taking the limit as Dt 0, which means r r dv dM r r Fnet.ext = M + (v  u) dt dt r r r dM r r r dv Fnet.ext + v rel =M where v rel = u  v the velocity of dM relative to M dt dt Notice that except for the dM term this looks like dt DM r dM r dM r r Dv = Dv << (v  u ) Dt dt dt the old version of N' s 2nd Law Rocket Launch
This is the reverse of the previous example. Here M decreases. Take M = M0  Rt, where r j R is the fuel burn rate and Fnet.ext = mg^ r r r dM dv Fnet.ext + v rel =M dt dt r dv r mg^  Rv rel = M j dt r r The quantity Rv rel is the thrust = Fthrust , r since the rocket is going up v rel = +v rel ^ j Rocket Launch
For a rocket going straight upward (+y direction) we have
Mg + Rvrel = M dv y dt Rearranging terms dv y dt = Rvrel Rvrel g= g M M 0  Rt dx ^ Integrating gives the rocket equation = ln(a + x), x = Rt ~ a+x M0 ^ v y = v rel ln ~  gt M 0  Rt (g constant) Saturn V  Apollo launch Vehicle
Given: Total mass M0=2.85x106 kg, payload mass Mp 27% of M0, fuel mass = M0  Mp Fuel burn rate R = 13.84x 103kg/s thrust Fthrust = 34.0x106 N. Saturn V  Apollo launch Vehicle
Find the: a) exhaust speed b) burn time tburn c) acceleration at liftoff & burnout d) final speed of the rocket. Saturn V Apollo Launch Vehicle
a) Exhaust speed: From Fthrust = Rvrel vrel= (34.0x106kg/s)/(13.84x 103kg/s) l= 2.46 km/s b) Burn time tburn Fuel mass M0  Mp = (1 .27) M0 = (.73) 2.85x106 kg = 2.08 x 106kg burn time to exhaust all fuel is: tb=[(Fuel Mass)/R] = 2.08 x 106kg/13.84kg/s tb = 150s Saturn V  Apollo launch Vehicle
c) Acceleration at liftoff dv y Rv rel =  9.80m / s 2 = 2.14m / s 2 dt M0 dv y Rv rel Acceleration at burnout =  9.80m / s 2 = 34.3m / s 2 dt Mp M0 ^ d) Final speed at burnout v y = v rel ln  (9.80m / s 2 )150s ~ .27M 0 v y = (2.46 10 3 m / s)(1.31) 1470m / s v y = 1.75km / s ...
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 Spring '07
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 mechanics, Mass, Momentum

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