Q. 74.5( 4 Votes )
A rubber ball of mass 50 g falls from a height of 10 cm and rebounds to a height of 50 cm. Determine the change in linear momentum and average force between the ball and the ground taking time of contact as 0.1 s.
Answer :
Initial velocity = 0m/s = u
Mass = 50g = 0.05kg
Final velocity = v
Using 3rd equation of motion
v2 = u2 + 2gh
v2 = 02 + 2(9.8)(0.1)
v2 = 1.96
v = 1.4m/s
For 50cm rebound
Using 3rd equation of motion
v2 = u2 – 2gh
02 = u2 – 2(9.8)(0.5)
u = 3.13m/s
Impulse = change in momentum
= mv–(–mu)
= 0.05kg × 1.4m/s + 0.05kg × 3.13m/s
= 0.2265Ns
Force = 
= 
= 2.265N
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