A ball with an initial momentum of 6.00 kg m/s bounces off a walland travels in

Question

A ball with an initial momentum of 6.00 kg m/s bounces off a walland travels in the opposite direction with a momentum of 4.00 kgm/s. what is the magnitude of the impulse acting on the ball?

If the ball in previous problem interacts with the wall for a timeinterval of 0.22 s, what is the average force exerted on thewall?

A 42.0 kg skateboarder traveling at 1.50 m/s hits a wall andbounces off of it.If the magnitude of the impulse is 150.0 kg m/s,calculate the final velocity of the skateboarder.

Please help me with these complicated physics problem...Please showyour work so I can understand how to do this in the future.

Explanation / Answer

The impulse = change in momentum = 4.00kg -m/s - (-6.00kg-m/s) =10.0 kg-m/s (assuming away from the wall is positive)...Impulse =10.0N-s Now if the ball was in contact for 0.22s, then F= impulse/t =10.0N-s/0.22s = 45.5N Now the impulse = 150.0kg-ms/ = m*vf - m*vi = 42.0*v - 42.0*(-1.50)= 42v + 63... 42v = 150 - 63 = 87...=> v = 87/42 = 2.07m/s

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