Question

A ball of mass 0.2 kg falls from a height of 45 m. On striking the ground, it rebounds in 0.1 sec with two thirds of the velocity with which it struck the ground. Calculate

a ) change in the momentum of the ball immediately after hitting the ground
b ) the average force on the ball due to the impact.

Answer 1

mass = 0.2 kg
h = 45m

just before hitting,
u = $\sqrt{2gh}= \sqrt{2 \times 10 \times 45}= \sqrt{900}=30$ m/s
just after hitting, it rebounds with 2/3 of his initial velocity
v = - (2/3) * 30 = -20 m/s (assuming downward direction as positive)

a) change in momentum = mv - mu = 0.2*(-20) - 0.2*(30) = -10 kg m/s
So change in momentum is 10 kg m/s in upward direction.

b) time taken = 0.1 sec
average force = change in momentum / time
                     = 10 / 0.1
                     = -100 N
So average force is 100N in upward direction.

Answer 2

Answer:

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