1. Momentum is defined as the product of an object's mass and velocity. It is a conserved quantity such that the total momentum of an isolated system remains constant.
2. During collisions, conservation of momentum states that the total momentum of colliding objects before the collision equals the total momentum after. If no external forces are applied, momentum is conserved.
3. Collisions can be elastic, where both momentum and kinetic energy are conserved, or inelastic where kinetic energy is not conserved but momentum still is. The analysis of collisions uses conservation laws to solve for unknown velocities.
5. Example 9-1: Force of a tennis serve. For a top player, a tennis ball may leave the racket on the serve with a speed of 55 m/s (about 120 mi/h). If the ball has a mass of 0.060 kg and is in contact with the racket for about 4 ms (4 x 10 -3 s), estimate the average force on the ball. Would this force be large enough to lift a 60-kg person?
6. Example 9-2: Washing a car: momentum change and force. Water leaves a hose at a rate of 1.5 kg/s with a speed of 20 m/s and is aimed at the side of a car, which stops it. (That is, we ignore any splashing back.) What is the force exerted by the water on the car?
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8. Conservation of momentum can also be derived from Newton’s laws. A collision takes a short enough time that we can ignore external forces. Since the internal forces are equal and opposite, the total momentum is constant.
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10. Example 9-3: Railroad cars collide: momentum conserved. A 10,000-kg railroad car, A, traveling at a speed of 24.0 m/s strikes an identical car, B, at rest. If the cars lock together as a result of the collision, what is their common speed immediately after the collision?
11. Momentum conservation works for a rocket as long as we consider the rocket and its fuel to be one system, and account for the mass loss of the rocket.
12. Example 9-4: Rifle recoil. Calculate the recoil velocity of a 5.0-kg rifle that shoots a 0.020-kg bullet at a speed of 620 m/s.
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14. During a collision, objects are deformed due to the large forces involved. The force exerted on the objects will jumps from zero to a very large force in a short time then will go to zero again
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16. Since the time of the collision is very short, we need not worry about the exact time dependence of the force, and can use the average force.
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18. The impulse tells us that we can get the same change in momentum with a large force acting for a short time, or a small force acting for a longer time. This is why you should bend your knees when you land; why airbags work; and why landing on a pillow hurts less than landing on concrete.
19. Conservation of Energy and Momentum in Collisions Momentum is conserved in all collisions. Collisions in which kinetic energy is conserved as well are called elastic collisions, and those in which it is not are called inelastic.
20. Elastic Collisions Here we have two objects colliding elastically. We know the masses and the initial speeds. Since both momentum and kinetic energy are conserved, we can write two equations. This allows us to solve for the two unknown final speeds.
21. Example 9-7: Equal masses. Billiard ball A of mass m moving with speed v A collides head-on with ball B of equal mass. What are the speeds of the two balls after the collision, assuming it is elastic? Assume (a) both balls are moving initially ( v A and v B ), (b) ball B is initially at rest ( v B = 0).
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23. Example 9-9: A nuclear collision. A proton (p) of mass 1.01 u (unified atomic mass units) traveling with a speed of 3.60 x 10 4 m/s has an elastic head-on collision with a helium (He) nucleus ( m He = 4.00 u) initially at rest. What are the velocities of the proton and helium nucleus after the collision? Assume the collision takes place in nearly empty space.
24. Inelastic Collisions With inelastic collisions, some of the initial kinetic energy is lost to thermal or potential energy. Kinetic energy may also be gained during explosions, as there is the addition of chemical or nuclear energy. A completely inelastic collision is one in which the objects stick together afterward, so there is only one final velocity.
25. Example 9-10: Railroad cars again. A 10,000-kg railroad car, A, traveling at a speed of 24.0 m/s strikes an identical car, B, at rest. If the cars lock together as a result of the collision, how much of the initial kinetic energy is transformed to thermal or other forms of energy? Before collision After collision
26. Example 9-11: Ballistic pendulum. The ballistic pendulum is a device used to measure the speed of a projectile, such as a bullet. The projectile, of mass m , is fired into a large block of mass M , which is suspended like a pendulum. As a result of the collision, the pendulum and projectile together swing up to a maximum height h . Determine the relationship between the initial horizontal speed of the projectile, v , and the maximum height h .
27. Collisions in Two or Three Dimensions Conservation of energy and momentum can also be used to analyze collisions in two or three dimensions, but unless the situation is very simple, the math quickly becomes unwieldy. Here, a moving object collides with an object initially at rest. Knowing the masses and initial velocities is not enough; we need to know the angles as well in order to find the final velocities.
28. Example 9-12: Billiard ball collision in 2-D. Billiard ball A moving with speed v A = 3.0 m/s in the + x direction strikes an equal-mass ball B initially at rest. The two balls are observed to move off at 45° to the x axis, ball A above the x axis and ball B below. That is, θ A ’ = 45° and θ B ’ = -45 °. What are the speeds of the two balls after the collision?
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30. Example 9-13: Proton-proton collision. A proton traveling with speed 8.2 x 10 5 m/s collides elastically with a stationary proton in a hydrogen target. One of the protons is observed to be scattered at a 60° angle. At what angle will the second proton be observed, and what will be the velocities of the two protons after the collision?
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33. ConcepTest 9.2a Momentum and KE I A system of particles is known to have a total kinetic energy of zero. What can you say about the total momentum of the system? 1) momentum of the system is positive 2) momentum of the system is positive 3) momentum of the system is zero 4) you cannot say anything about the momentum of the system
34. ConcepTest 9.2a Momentum and KE I A system of particles is known to have a total kinetic energy of zero. What can you say about the total momentum of the system? 1) momentum of the system is positive 2) momentum of the system is positive 3) momentum of the system is zero 4) you cannot say anything about the momentum of the system Since the total kinetic energy is zero, this means that all of the particles are at rest ( v = 0). Therefore, since nothing is moving, the total momentum of the system must also be zero.
35. ConcepTest 9.2b Momentum and KE II A system of particles is known to have a total momentum of zero. Does it necessarily follow that the total kinetic energy of the system is also zero? 1) yes 2) no
36. ConcepTest 9.2b Momentum and KE II A system of particles is known to have a total momentum of zero. Does it necessarily follow that the total kinetic energy of the system is also zero? 1) yes 2) no Momentum is a vector, so the fact that p tot = 0 does not mean that the particles are at rest! They could be moving such that their momenta cancel out when you add up all of the vectors. In that case, since they are moving, the particles would have non-zero KE.
37. ConcepTest 9.2c Momentum and KE III Two objects are known to have the same momentum. Do these two objects necessarily have the same kinetic energy? 1) yes 2) no
38. ConcepTest 9.2c Momentum and KE III Two objects are known to have the same momentum. Do these two objects necessarily have the same kinetic energy? 1) yes 2) no If object #1 has mass m and speed v and object #2 has mass 1/2 m and speed 2 v , they will both have the same momentum. However, since KE = 1/2 mv 2 , we see that object #2 has twice the kinetic energy of object #1, due to the fact that the velocity is squared.
39. ConcepTest 9.3a Momentum and Force A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 130-g pebble. How does the rate of change of the boulder’s momentum compare to the rate of change of the pebble’s momentum? 1) greater than 2) less than 3) equal to
40. ConcepTest 9.3a Momentum and Force A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 130-g pebble. How does the rate of change of the boulder’s momentum compare to the rate of change of the pebble’s momentum? 1) greater than 2) less than 3) equal to The rate of change of momentum is, in fact, the force. Remember that F = p / t . Since the force exerted on the boulder and the pebble is the same, then the rate of change of momentum is the same.
41. ConcepTest 9.3b Velocity and Force 1) greater than 2) less than 3) equal to A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 130-g pebble. How does the rate of change of the boulder’s velocity compare to the rate of change of the pebble’s velocity?
42. ConcepTest 9.3b Velocity and Force 1) greater than 2) less than 3) equal to The rate of change of velocity is the acceleration. Remember that a = v / t . The acceleration is related to the force by Newton’s 2 nd Law ( F = ma ), so the acceleration of the boulder is less than that of the pebble (for the same applied force) because the boulder is much more massive. A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 130-g pebble. How does the rate of change of the boulder’s velocity compare to the rate of change of the pebble’s velocity?
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47. ConcepTest 9.5b Two Boxes II In the previous question, which box has the larger velocity after the force acts? 1) the heavier one 2) the lighter one 3) both the same
48. ConcepTest 9.5b Two Boxes II In the previous question, which box has the larger velocity after the force acts? 1) the heavier one 2) the lighter one 3) both the same The force is related to the acceleration by Newton’s 2 nd Law ( F = ma ). The lighter box therefore has the greater acceleration and will reach a higher speed after the 1-second time interval. Follow-up: Which box has gone a larger distance after the force acts? Follow-up: Which box has gained more KE after the force acts?
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51. ConcepTest 9.7 Impulse A small beanbag and a bouncy rubber ball are dropped from the same height above the floor. They both have the same mass. Which one will impart the greater impulse to the floor when it hits? 1) the beanbag 2) the rubber ball 3) both the same
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53. ConcepTest 9.8 Singing in the Rain A person stands under an umbrella during a rainstorm. Later the rain turns to hail, although the number of “drops” hitting the umbrella per time and their speed remains the same. Which case requires more force to hold the umbrella? 1) when it is hailing 2) when it is raining 3) same in both cases
54. ConcepTest 9.8 Singing in the Rain A person stands under an umbrella during a rainstorm. Later the rain turns to hail, although the number of “drops” hitting the umbrella per time and their speed remains the same. Which case requires more force to hold the umbrella? 1) when it is hailing 2) when it is raining 3) same in both cases When the raindrops hit the umbrella, they tend to splatter and run off, whereas the hailstones hit the umbrella and bounce back upward. Thus, the change in momentum (impulse) is greater for the hail. Since p = F t , more force is required in the hailstorm. This is similar to the situation with the bouncy rubber ball in the previous question.
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63. ConcepTest 9.11 Golf Anyone? You tee up a golf ball and drive it down the fairway. Assume that the collision of the golf club and ball is elastic. When the ball leaves the tee, how does its speed compare to the speed of the golf club? 1) greater than 2) less than 3) equal to
64. ConcepTest 9.11 Golf Anyone? You tee up a golf ball and drive it down the fairway. Assume that the collision of the golf club and ball is elastic. When the ball leaves the tee, how does its speed compare to the speed of the golf club? 1) greater than 2) less than 3) equal to If the speed of approach (for the golf club and ball) is v , then the speed of recession must also be v . Since the golf club is hardly affected by the collision and it continues with speed v , then the ball must fly off with a speed of 2 v .
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77. ConcepTest 9.15 Gun Control When a bullet is fired from a gun, the bullet and the gun have equal and opposite momenta. If this is true, then why is the bullet deadly? (whereas it is safe to hold the gun while it is fired) 1) it is much sharper than the gun 2) it is smaller and can penetrate your body 3) it has more kinetic energy than the gun 4) it goes a longer distance and gains speed 5) it has more momentum than the gun
78. ConcepTest 9.15 Gun Control When a bullet is fired from a gun, the bullet and the gun have equal and opposite momenta. If this is true, then why is the bullet deadly? (whereas it is safe to hold the gun while it is fired) 1) it is much sharper than the gun 2) it is smaller and can penetrate your body 3) it has more kinetic energy than the gun 4) it goes a longer distance and gains speed 5) it has more momentum than the gun While it is true that the magnitudes of the momenta of the gun and the bullet are equal, the bullet is less massive and so it has a much higher velocity. Since KE is related to v 2 , the bullet has considerably more KE and therefore can do more damage on impact.
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85. ConcepTest 9.18 Baseball Bat Where is center of mass of a baseball bat located? 1) at the midpoint 2) closer to the thick end 3) closer to the thin end (near handle) 4) it depends on how heavy the bat is
86. ConcepTest 9.18 Baseball Bat Where is center of mass of a baseball bat located? 1) at the midpoint 2) closer to the thick end 3) closer to the thin end (near handle) 4) it depends on how heavy the bat is Since most of the mass of the bat is at the thick end, this is where the center of mass is located. Only if the bat were like a uniform rod would its center of mass be in the middle.
Editor's Notes
Figure 9-12. Caption: Two equal-mass objects (a) approach each other with equal speeds, (b) collide, and then (c) bounce off with equal speeds in the opposite directions if the collision is elastic, or (d) bounce back much less or not at all if the collision is inelastic.
Figure 9-13. Caption: Two small objects of masses m A and m B , (a) before the collision and (b) after the collision.
Figure 9-14. Caption: In this multi-flash photo of a head-on collision between two balls of equal mass, the white cue ball is accelerated from rest by the cue stick and then strikes the red ball, initially at rest. The white ball stops in its tracks and the (equal mass) red ball moves off with the same speed as the white ball had before the collision. See Example 9–7. Solution: a. Use both conservation of momentum and conservation of energy; the balls exchange velocities. b. Ball A stops, and ball B moves on with ball A’s original velocity.
Solution: a. Both momentum and kinetic energy are conserved. The rest is algebra. b. In this case, v A ′ doesn’t change much, and v B ′ = 2 v A . c. In this case, v B ’ remains zero, and mass A reverses its direction at the same speed.
Figure 9-15. Caption: Example 9–9: (a) before collision, (b) after collision. Solution: We don’t need to know what u is; all we need are the relative masses of the proton and the helium nucleus. Momentum and kinetic energy are conserved; solving the equations gives the helium nucleus’s velocity as 1.45 x 10 4 m/s and the proton’s as -2.15 x 10 4 m/s (backwards).
Figure 9-5. Solution: Momentum is conserved; the initial kinetic energy is 2.88 x 10 6 J and the final kinetic energy is 1.44 x 10 6 J, so 1.44 x 10 6 J of kinetic energy have been transformed to other forms.
Figure 9-16. Solution: This has two parts. First, there is the inelastic collision between the bullet and the block; we need to find the speed of the block. Then, the bullet+block combination rises to some maximum height; here we can use conservation of mechanical energy to find the height, which depends on the speed.
Figure 9-18. Caption: Object A, the projectile, collides with object B, the target. After the collision, they move off with momenta p A ’ and p B ’ at angles θ A ’ and θ B ’. The objects are shown here as particles, as we would visualize them in atomic or nuclear physics. But they could also be macroscopic pool balls.
Figure 9-19. Solution: Apply conservation of momentum; the masses are the same, as are the outgoing angles. The final speeds are equal; both are 2.1 m/s.
Figure 9-18. Solution: We have three unknowns – the two outgoing velocities and the angle of the second proton – and three equations – momentum conservation in x and y, and conservation of kinetic energy. Solving gives the speed of the first proton to be 4.1 x 10 5 m/s, of the second to be 7.1 x 10 5 m/s, and the angle of the second to be -30 °.