t. A bail of mass 0.2 kg falls from a height of 45 m. On striking the

1.	t. A bail of mass 0.2 kg falls from a height of 45 m. On striking the ground, it rebounds in 0.1sedwith two third 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» m = 0.2 kg h = 45m v^2 = u^2 +2as v^2 = 2gs v^2 = 21045 v^2=900m/s v= 30m/s rebound velocity =2/3v 20m/s since it is opp direction speed =-20m/s change in momentum = m(v-u) =0.2(30+20) =10Nsec average force = rate of change in momentum =10/0.1 100N

t. A bail of mass 0.2 kg falls from a height of 45 m. On striking the ground, it rebounds in 0.1sedwith two third 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»

m = 0.2 kg

h = 45m

v^2 = u^2 +2as

v^2 = 2gs

v^2 = 2*10*45

v^2=900m/s

v= 30m/s

rebound velocity =2/3v

20m/s

since it is opp direction speed =-20m/s

change in momentum = m(v-u)

=0.2(30+20)

=10Nsec

average force = rate of change in momentum

=10/0.1

100N

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