Part complete during a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. if the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball?

Question

Physics

Stephen Decker

23 July, 17:22

Part complete during a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. if the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball?

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Answers (1)

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Preston House · Answer 1

The average force applied to the ball = 106.7 N

Explanation:

Force is given by

f = ΔP/t

ΔP = change in momentum = m Vf - m Vi

m = mass = 0.2 kg

Vf = final velocity = 12 m/s

Vi=initial velocity = - 20 m/s (negative because it is going towards the wall which is treated as negative axis)

t = time = 60 ms = 0.06 s

now ΔP = 0.2 [ 12 - (-20) ]

ΔP=0.2 (32) = 6.4 kg m/s

now force F = ΔP/t

F = 6.4/0.06

F=106.7 N