Question

A 6 kg bowling ball rolling at 5 m/s strikes a stationary 4 kg bowling ball

A 6 kg bowling ball rolling at 5 m/s strikes a stationary 4 kg bowling ball. If ball #1 is moving forward at 2 m/s after the collision, what is the speed and direction of ball #2? What is the impulse of the system? If the collision last for .5 seconds, how much force is exerted?
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Answer #1
Initial momentum = 6*5 = 30 kg m/s

Final momentum is therefore 30
30 = 6*2 + 4*v
4 v = 18
v = 4.5 m/s forward
answered by: natilie
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Answer #2
There is no change of momentum of the system so no impulse for the system.
However there is a change of momentum or impulse for Ball 2 of 18 kg m/s

Force = change of momentum or impulse / time
=18/.5 = 36 Newtons
answered by: jolene
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