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

While falling down, initial velocity = 0, final velocity with which the body strikes the ground = v m/s, acceleration due to gravity = 9.8 m/s ^{ 2 }, height = 0.1m

The velocity of the ball with which it hits the ground can be found using,

v ^{ 2 } = u ^{ 2 } + 2as

=> v ^{ 2 } = 0 + 2×9.8×0.1

=> v = 1.96 m/s (downward)

While it moves up, initial velocity = u, final velocity = 0, acceleration due to gravity = -9.8 m/s ^{ 2 }, height = 0.5 m

The velocity of the ball with which it leaves the ground can be found using,

v ^{ 2 } = u ^{ 2 } + 2as

=> 0 = u ^{ 2 } - 2×9.8×0.5 [this time the velocity and acceleration are oppositely directed so we have the negative sign]

=> v = 9.8 m/s (upward)

So, Change in velocity = 9.8 m/s (upward) – 1.96 m/s (downward)

=> Δv = 7.84 m/s

∴ change in linear momentum,dp = mΔv = 0.392 kg m/s

So, Force because of acceleration achieved during the strike is, F = dp/dt = 0392/0.1 = 3.92 N^{ }

**
**