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  • 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 °.

11 momentum Presentation Transcript

  • 1. Linear Momentum Topic 7
  • 2. Lecture Outline
    • Momentum
    • Conservation Of Momentum
    • Collision and Impulse
  • 3. Momentum
    • Product of mass and velocity
    • Momentum is a vector symbolized by the symbol p, and is defined as
    • [Unit: kg m/s]
    • Direction of momentum is the same as direction of velocity
    • Faster moving object have higher momentum than slower moving object even when they has a same mass
    • Two objects with different mass moving at same speed, object with higher mass will have higher momentum
  • 4.
    • To change the momentum of an object you need force, whether to increase, decrease of change the direction of the momentum
    • Originally Newton’s Second Law is stated in term of momentum
    • The rate of change of momentum of object is equal to the net force applied to it
  • 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?
  • 7. Conservation of Momentum
    • In certain circumstances, momentum is a conserved quantities
    • Two objects in collision with each other, the total momentum of these objects are conserved
    • During a collision, measurements show that the total momentum does not change:
  • 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.
  • 9.
    • This is not only true for only two colliding objects, but also true for more than two objects, as long as the total external force is zero
    • Law of conservation of momentum
    • when the net external force on a system of objects is zero, the total momentum of the system remains constant.
    • Equivalently
    • The total momentum of an isolated system of objects remains constant
  • 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.
  • 13. Collisions and Impulse
    • Collision happen when two or more objects are in contact with each other
      • Tennis racket or baseball bat striking a ball
      • Billiard ball colliding
      • Hammer hitting a nail
    • The interaction force between these objects usually stronger than any interaction between the system and their environment. – other forces can be ignore
  • 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
  • 15.
    • Since , we can write
    • the definition of impulse:
    • = momentum change
    • Symbol J [unit: kg m/s]
  • 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.
  • 17.
    • Example: Bend your knees when landing
    • a) calculate the impulse experience when a 70 kg person lands on firm ground after jumping from height 3.0m
    • b) estimate the average force exerted on the person’s feet by the ground if the landing is stiff-legged assuming the body move 1.0 cm during impact
    • c) with bend legs, when the legs are bent about 50 cm
  • 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).
  • 22.
    • Example 9-8: Unequal masses, target at rest.
    • A very common practical situation is for a moving object ( m A ) to strike a second object ( m B , the “target”) at rest ( v B = 0). Assume the objects have unequal masses, and that the collision is elastic and occurs along a line (head-on).
    • Derive equations for v B ’ and v A ’ in terms of the initial velocity v A of mass m A and the masses m A and m B .
    • Determine the final velocities if the moving object is much more massive than the target ( m A >> m B ).
    • Determine the final velocities if the moving object is much less massive than the target ( m A << m B ).
  • 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?
  • 29.
    • If the collision is elastic, we have learn that kinetic energy is conserved.
  • 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?
  • 31. ConcepTest 9.1 Rolling in the Rain
    • An open cart rolls along a frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (assume that the rain falls vertically into the box)
    1) speeds up 2) maintains constant speed 3) slows down 4) stops immediately
  • 32. ConcepTest 9.1 Rolling in the Rain
    • An open cart rolls along a frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (assume that the rain falls vertically into the box)
    Since the rain falls in vertically, it adds no momentum to the box, thus the box’s momentum is conserved. However, since the mass of the box slowly increases with the added rain, its velocity has to decrease . 1) speeds up 2) maintains constant speed 3) slows down 4) stops immediately Follow-up: What happens to the cart when it stops raining?
  • 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?
  • 43. ConcepTest 9.4 Collision Course
    • A small car and a large truck collide head-on and stick together. Which one has the larger momentum change?
    1) the car 2) the truck 3) they both have the same momentum change 4) can’t tell without knowing the final velocities
  • 44. ConcepTest 9.4 Collision Course
    • A small car and a large truck collide head-on and stick together. Which one has the larger momentum change?
    Since the total momentum of the system is conserved, that means that  p = 0 for the car and truck combined . Therefore,  p ca r must be equal and opposite to that of the truck ( –  p truck ) in order for the total momentum change to be zero. Note that this conclusion also follows from Newton’s 3 rd Law. 1) the car 2) the truck 3) they both have the same momentum change 4) can’t tell without knowing the final velocities Follow-up: Which one feels the larger acceleration?
  • 45. ConcepTest 9.5a Two Boxes I
    • Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each one for exactly 1 second . Which box has more momentum after the force acts ?
    1) the heavier one 2) the lighter one 3) both the same F F light heavy
  • 46. ConcepTest 9.5a Two Boxes I
    • Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each one for exactly 1 second . Which box has more momentum after the force acts ?
    We know: so impulse  p = F av  t. In this case F and  t are the same for both boxes ! Both boxes will have the same final momentum . 1) the heavier one 2) the lighter one 3) both the same F F light heavy av  t  p F 
  • 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?
  • 49. ConcepTest 9.6 Watch Out!
    • You drive around a curve in a narrow one-way street at 30 mph when you see an identical car heading straight toward you at 30 mph . You have two options: hit the car head-on or swerve into a massive concrete wall (also head-on). What should you do?
    1) hit the other car 2) hit the wall 3) makes no difference 4) call your physics prof!! 5) get insurance!
  • 50. ConcepTest 9.6 Watch Out!
    • You drive around a curve in a narrow one-way street at 30 mph when you see an identical car heading straight toward you at 30 mph . You have two options: hit the car head-on or swerve into a massive concrete wall (also head-on). What should you do?
    In both cases your momentum will decrease to zero in the collision. Given that the time  t of the collision is the same, then the force exerted on YOU will be the same!! If a truck is approaching at 30 mph, then you’d be better off hitting the wall in that case. On the other hand, if it’s only a mosquito , well, you’d be better off running him down ... 1) hit the other car 2) hit the wall 3) makes no difference 4) call your physics prof!! 5) get insurance!
  • 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
  • 52. 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
    • Both objects reach the same speed at the floor. However, while the beanbag comes to rest on the floor, the ball bounces back up with nearly the same speed as it hit. Thus, the change in momentum for the ball is greater, because of the rebound . The impulse delivered by the ball is twice that of the beanbag.
      • For the beanbag:  p = p f – p i = 0 – (– mv ) = mv
      • For the rubber ball:  p = p f – p i = mv – (– mv ) = 2 mv
    Follow-up: Which one imparts the larger force to the floor?
  • 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.
  • 55. ConcepTest 9.9a Going Bowling I
    • A bowling ball and a ping-pong ball are rolling toward you with the same momentum . If you exert the same force to stop each one, which takes a longer time to bring to rest?
    1) the bowling ball 2) same time for both 3) the ping-pong ball 4) impossible to say p p
  • 56. ConcepTest 9.9a Going Bowling I
    • A bowling ball and a ping-pong ball are rolling toward you with the same momentum . If you exert the same force to stop each one, which takes a longer time to bring to rest?
    We know: Here, F and  p are the same for both balls! It will take the same amount of time to stop them. so  p = F av  t 1) the bowling ball 2) same time for both 3) the ping-pong ball 4) impossible to say p p av  t  p F 
  • 57. ConcepTest 9.9b Going Bowling II
    • A bowling ball and a ping-pong ball are rolling toward you with the same momentum . If you exert the same force to stop each one, for which is the stopping distance greater?
    1) the bowling ball 2) same distance for both 3) the ping-pong ball 4) impossible to say p p
  • 58. ConcepTest 9.9b Going Bowling II
    • A bowling ball and a ping-pong ball are rolling toward you with the same momentum . If you exert the same force to stop each one, for which is the stopping distance greater?
    Use the work-energy theorem: W =  KE . The ball with less mass has the greater speed (why?) , and thus the greater KE (why again?) . In order to remove that KE, work must be done, where W = Fd . Since the force is the same in both cases, the distance needed to stop the less massive ball must be bigger . 1) the bowling ball 2) same distance for both 3) the ping-pong ball 4) impossible to say p p
  • 59. ConcepTest 9.10a Elastic Collisions I
    • Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on. In which case does the golf ball have the greater speed after the collision?
    1) situation 1 2) situation 2 3) both the same v 2 v 1 at rest at rest
  • 60. ConcepTest 9.10a Elastic Collisions I
    • Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on. In which case does the golf ball have the greater speed after the collision?
    Remember that the magnitude of the relative velocity has to be equal before and after the collision! In case 1 the bowling ball will almost remain at rest, and the golf ball will bounce back with speed close to v . In case 2 the bowling ball will keep going with speed close to v , hence the golf ball will rebound with speed close to 2 v . 1) situation 1 2) situation 2 3) both the same v 1 v 2 2 v
  • 61. ConcepTest 9.10b Elastic Collisions II
    • Carefully place a small rubber ball (mass m ) on top of a much bigger basketball (mass M ) and drop these from some height h . What is the velocity of the smaller ball after the basketball hits the ground, reverses direction and then collides with small rubber ball?
    1) zero 2) v 3) 2v 4) 3v 5) 4v
  • 62. ConcepTest 9.10b Elastic Collisions II
    • Carefully place a small rubber ball (mass m ) on top of a much bigger basketball (mass M ) and drop these from some height h . What is the velocity of the smaller ball after the basketball hits the ground, reverses direction and then collides with small rubber ball?
    • Remember that relative velocity has to be equal before and after collision! Before the collision, the basketball bounces up with v and the rubber ball is coming down with v , so their relative velocity is –2v . After the collision, it therefore has to be +2v !!
    1) zero 2) v 3) 2 v 4) 3 v 5) 4 v Follow-up: With initial drop height h , how high does the small rubber ball bounce up? v v v v 3v v (a) (b) (c) m M
  • 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 .
  • 65. ConcepTest 9.12a Inelastic Collisions I
    • A box slides with initial velocity 10 m/s on a frictionless surface and collides inelastically with an identical box. The boxes stick together after the collision. What is the final velocity?
    1) 10 m/s 2) 20 m/s 3) 0 m/s 4) 15 m/s 5) 5 m/s v f v i M M M M
  • 66. ConcepTest 9.12a Inelastic Collisions I
    • A box slides with initial velocity 10 m/s on a frictionless surface and collides inelastically with an identical box. The boxes stick together after the collision. What is the final velocity?
    1) 10 m/s 2) 20 m/s 3) 0 m/s 4) 15 m/s 5) 5 m/s The initial momentum is: M v i = ( 10 ) M v f v i M M M M The final momentum is: M tot v f = (2 M ) v f = (2 M ) (5) The final momentum must be the same!!
  • 67. ConcepTest 9.12b Inelastic Collisions II
    • On a frictionless surface, a sliding box collides and sticks to a second identical box which is initially at rest. What is the final KE of the system in terms of the initial KE?
    v i v f 1) KE f = KE i 2) KE f = KE i / 4 3) KE f = KE i /  2 4) KE f = KE i / 2 5) KE f =  2 KE i
  • 68. ConcepTest 9.12b Inelastic Collisions II
    • On a frictionless surface, a sliding box collides and sticks to a second identical box which is initially at rest. What is the final KE of the system in terms of the initial KE?
    Momentum : mv i + 0 = (2 m ) v f So we see that: v f = 1/2 v i Now, look at kinetic energy : First, KE i = 1/2 mv i 2 So: KE f = 1/2 m f v f 2 = 1/2 (2 m ) (1/2 v i ) 2 = 1/2 ( 1/2 mv i 2 ) = 1/2 KE i v i v f 1) KE f = KE i 2) KE f = KE i / 4 3) KE f = KE i /  2 4) KE f = KE i / 2 5) KE f =  2 KE i
  • 69. ConcepTest 9.13a Nuclear Fission I
    • A uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater momentum?
    1) the heavy one 2) the light one 3) both have the same momentum 4) impossible to say 1 2
  • 70. ConcepTest 9.13a Nuclear Fission I
    • A uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater momentum?
    1) the heavy one 2) the light one 3) both have the same momentum 4) impossible to say The initial momentum of the uranium was zero, so the final total momentum of the two fragments must also be zero. Thus the individual momenta are equal in magnitude and opposite in direction. 1 2
  • 71. ConcepTest 9.13b Nuclear Fission II
    • A uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater speed?
    1) the heavy one 2) the light one 3) both have the same speed 4) impossible to say 1 2
  • 72. ConcepTest 9.13b Nuclear Fission II
    • A uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater speed?
    We have already seen that the individual momenta are equal and opposite. In order to keep the magnitude of momentum mv the same, the heavy fragment has the lower speed and the light fragment has the greater speed . 1) the heavy one 2) the light one 3) both have the same speed 4) impossible to say 1 2
  • 73.
    • Amy (150 lbs) and Gwen (50 lbs) are standing on slippery ice and push off each other. If Amy slides at 6 m/s, what speed does Gwen have?
    ConcepTest 9.14a Recoil Speed I (1) 2 m/s (2) 6 m/s (3) 9 m/s (4) 12 m/s (5) 18 m/s 150 lbs 50 lbs
  • 74.
    • Amy (150 lbs) and Gwen (50 lbs) are standing on slippery ice and push off each other. If Amy slides at 6 m/s, what speed does Gwen have?
    ConcepTest 9.14a Recoil Speed I The initial momentum is zero , so the momenta of Amy and Gwen must be equal and opposite . Since p = mv , then if Amy has 3 times more mass , we see that Gwen must have 3 times more speed . (1) 2 m/s (2) 6 m/s (3) 9 m/s (4) 12 m/s (5) 18 m/s 150 lbs 50 lbs
  • 75. ConcepTest 9.14b Recoil Speed II
    • A cannon sits on a stationary railroad flatcar with a total mass of 1000 kg . When a 10-kg cannon ball is fired to the left at a speed of 50 m/s , what is the recoil speed of the flatcar?
    1) 0 m/s 2) 0.5 m/s to the right 3) 1 m/s to the right 4) 20 m/s to the right 5) 50 m/s to the right
  • 76. ConcepTest 9.14b Recoil Speed II
    • A cannon sits on a stationary railroad flatcar with a total mass of 1000 kg . When a 10-kg cannon ball is fired to the left at a speed of 50 m/s , what is the recoil speed of the flatcar?
    Since the initial momentum of the system was zero, the final total momentum must also be zero. Thus, the final momenta of the cannon ball and the flatcar must be equal and opposite. p cannonball = (10 kg)(50 m/s) = 500 kg-m/s p flatcar = 500 kg-m/s = (1000 kg) (0.5 m/s) 1) 0 m/s 2) 0.5 m/s to the right 3) 1 m/s to the right 4) 20 m/s to the right 5) 50 m/s to the right
  • 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.
  • 79. ConcepTest 9.16a Crash Cars I
    • If all three collisions below are totally inelastic , which one(s) will bring the car on the left to a complete halt?
    1) I 2) II 3) I and II 4) II and III 5) all three
  • 80. ConcepTest 9.16a Crash Cars I
    • If all three collisions below are totally inelastic , which one(s) will bring the car on the left to a complete halt?
    In case I , the solid wall clearly stops the car. In cases II and III , since p tot = 0 before the collision , then p tot must also be zero after the collision , which means that the car comes to a halt in all three cases. 1) I 2) II 3) I and II 4) II and III 5) all three
  • 81. ConcepTest 9.16b Crash Cars II
    • If all three collisions below are totally inelastic , which one(s) will cause the most damage (in terms of lost energy)?
    1) I 2) II 3) III 4) II and III 5) all three
  • 82. ConcepTest 9.16b Crash Cars II
    • If all three collisions below are totally inelastic , which one(s) will cause the most damage (in terms of lost energy)?
    The car on the left loses the same KE in all 3 cases, but in case III , the car on the right loses the most KE because KE = 1/2 m v 2 and the car in case III has the largest velocity . 1) I 2) II 3) III 4) II and III 5) all three
  • 83. ConcepTest 9.17 Shut the Door!
    • You are lying in bed and you want to shut your bedroom door. You have a superball and a blob of clay (both with the same mass) sitting next to you. Which one would be more effective to throw at your door to close it?
    1) the superball 2) the blob of clay 3) it doesn’t matter -- they will be equally effective 4) you are just too lazy to throw anything
  • 84. ConcepTest 9.17 Shut the Door!
    • You are lying in bed and you want to shut your bedroom door. You have a superball and a blob of clay (both with the same mass) sitting next to you. Which one would be more effective to throw at your door to close it?
    The superball bounces off the door with almost no loss of speed, so its  p (and that of the door) is 2 mv . The clay sticks to the door and continues to move along with it, so its  p is less than that of the superball , and therefore it imparts less  p to the door. 1) the superball 2) the blob of clay 3) it doesn’t matter -- they will be equally effective 4) you are just too lazy to throw anything
  • 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.