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Even the lightest bowling ball (a 6-pounder) is 1000  heavier than a ping pong ball (~2.7 oz).

A bowling ball and ping-pong ball are rolling towards you with the same momentum . Which ball is moving toward you with the greater speed? A ) the bowling ball B ) the ping pong ball C ) same speed for both. Even the lightest bowling ball (a 6-pounder) is

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Even the lightest bowling ball (a 6-pounder) is 1000  heavier than a ping pong ball (~2.7 oz).

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  1. A bowling ball and ping-pong ball are rolling towards you with the same momentum. Which ball is moving toward you with the greater speed? A) the bowling ball B) the ping pong ball C) same speed for both Even the lightest bowling ball (a 6-pounder) is 1000 heavier than a ping pong ball (~2.7 oz). The ping ball would have to move ~1000 times faster!

  2. A bowling ball and ping-pong ball are rolling towards you with the same momentum. If you exert the same force in stopping each, which takes a longer time to bring to rest? A) the bowling ball B) the ping pong ball C) same time for both stopping each, which takes a longer time to bring to rest? for which is the stopping distance greater?

  3. If a cue ball moving with velocity v strikes a stationary v v = 0 billiard ball head-on and comes to an abrupt halt, v v = 0 its target ball moves off with the same velocity v. During their brief contact Force of red Force of billiard ball = cue ball on on cue ball billiard ball stopping cue ball launching target ball Notice: mv + 0 = 0 + mv

  4. During their brief contact FA pushes B = -FB pushes A over the same time! FABt = -FBAt Here I use: vB to represent B’s initial velocity v0 and v'B to represent B’s final velocity. The total momentum remains unchanged! We say: Momentum is conserved.

  5. v A fast moving car traveling with a speed v rear-ends an identical model (and total mass) car idling in neutral at the intersection. They lock bumpers on impact and move forward at • A) 0 (both stop). • B) v/4 • v/2 • v

  6. A heavy truck and light car both traveling at the speed limit v, have a head-on collision. If they lock bumpers on impact they skid together to the A) right B) left Under what conditions would they stop dead?

  7. A heavy truck and light car have a head-on collision bringing them to a sudden stop. Which vehicle experienced the greater force of impact? the greater impulse? the greater change in momentum? the greater acceleration? A) the truck B) the car C) both the same

  8. A100 kg astronaut at rest catches a 50 kg meteor moving toward him at 9 m/sec. If the astronaut manages to hold onto the meteor after catching it, what speed does he pick up? A) 3 m/sec B) 4.5 m/sec C) 9 m/sec D) 15 m/sec E) 18 m/sec F) some other speed v' =450 kg·m/s 150kg (100 kg0)+(50 kg9m/s) = (150kg)v'

  9. For these two vehicles to be stopped dead in their tracks by a collision at this intersection A) They must have equal mass B) They must have equal speed C) both A and B D) is IMPOSSIBLE

  10. Car A has a mass of 900 kg and is travelling east at a speed of 10 m/sec. Car B has a mass of 600 kg and is travelling north at a speed of 20 m/sec. The two cars collide, and lock bumpers. Neglecting friction which arrow best represents the direction the combined wreck travels? A B C A 600 kg 20 m/sec 900 kg 10 m/sec B

  11. Car A has a mass of 900 kg and is travelling east at a speed of 10 m/sec. Car B has a mass of 500 kg and is travelling north at a speed of 25 m/sec. The two cars collide and stick together. Neglecting friction Which of the arrows best represents the direction the combined wreck travels? A B C A 600 kg 20 m/sec 900 kg 10 m/sec B The answer is (A). Momentum is conserved, so the total momentum before the collision must equal the total momentum after. But momentum is a vector, so we have to add the vector arrows (by sliding them and placing them head to tail). The momentum of car A is mass  velocity = 9000 kg m/s in the direction of east. The momentum of car B is mass  velocity = 12000 kg m/s north. Draw these vectors, making sure to make the lengths proportional to the momenta.

  12. A projectile with initial speed v0scatters off a target (as shown) with final speed vf. mvf mv0 The direction its target is sent recoiling is best represented by B C A T D E G F

  13. A projectile with initial speed v0scatters off a target (as shown) with final speed vf. mvf mv0 The sum of the final momentum (the scattered projectile and the recoiling target) must be the same as the initial momentum of the projectile! F

  14. SOME ANSWERS Question 1 B) the ping pong ball Even the lightest bowling ball (a 6-pounder) is 1000 heavier than a ping pong ball (~2.7 oz). The ping ball would have to move ~1000 times faster! Question 2 C) same time for both v V So the bowling ball is not moving very fast, while the ping pong ball must be moving at a pretty high speed. But we’re told both have thesame momentum! To stop either one means to remove its momentum completely.All it has (mv) to 0. So both must undergo the exact same loss in momentum. The stopping time can be figured out from the momentum change needed: t must be the same for each! same for each B) the ping pong ball Question 3 The bowling ball reaches you will a small speed, v, which you slow to zero. During those t seconds, it travels with an average speed v/2, moving a distance (v/2)t before stopping. In the same amount of time, the ping pong ball travels much farther: (V/2)t. Or we can note that stopping distance can be directly computed using: bigger v needs bigger d same for each

  15. F SOME MORE ANSWERS and since they lock bumpers and move together v’1=v’2 Question 4 • v/2 v2=0 and since m1 = m2 we don’t need to distinguish them by different labels. Question 5 A) right Question 6 • Since forces are equal and opposite, both experience the same force. • Since both experience the same force in the same time, they both receive the same impulse. • Since they both have the same impulse, they both must undergo the same • change in momentum. • Since they both experience the same force, the less massive car has a greater acceleration, since a = F/m. Question 7 (100 kg0)+(50 kg9m/s) = (150kg)v' A) 3 m/sec v' =450 kg·m/s 150kg Momentum is conserved, so the total momentum before the collision must equal the total momentum after. But momentum is a vector - we have to add the vector arrows (by sliding them so they meet head to tail). The momentum of car A is mass velocity = 9000 kg m/s EAST. The momentum of car B is mass velocity = 12000 kg m/s NORTH. Just try drawing these vectors, making sure their lengths are proportional to the momenta. Question 8 D) is IMPOSSIBLE A Question 9 The question becomes: Question 10 mvf plus WHAT? = F mv0

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