Consider a ball falls from a height of 2 m and rebounds to a height of 0.5 m.If the mass of the ball is 60g, then find the impules and the average force between the ball and the ground.The time for which the ball and the ground are in contact was 0.2s
Consider a ball falls from a height of 2 m and rebounds to a height of 0.5 m.If the mass of the ball is 60g, then find the impules and the average force between the ball and the ground.The time for which the ball and the ground are in contact was 0.2s
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 = 2m
The velocity of the ball with which it hits the ground can be found using,
v^ 2 = u 2 + 2as
=> v^ 2 = 0 + 2×9.8×2
=> v = 6.26 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]
=> u =3.13 m/s (upward)
So,Magnitude of the change in velocity = 6.26-3.13=3.13 m/s
=> Δv = 3.13m/s
∴ Impulse=change in linear momentum,dp = mΔv = 100 x 3.13=313 kg m/s
So, Force because of acceleration achieved during the strike is, F = dp/dt = 313/0.2 = 1565 N
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 = 2m
The velocity of the ball with which it hits the ground can be found using,
v^ 2 = u 2 + 2as
=> v^ 2 = 0 + 2×9.8×2
=> v = 6.26 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]
=> u =3.13 m/s (upward)
So,Magnitude of the change in velocity = 6.26-3.13=3.13 m/s
=> Δv = 3.13m/s
∴ Impulse=change in linear momentum,dp = mΔv = 100 x 3.13=313 kg m/s
So, Force because of acceleration achieved during the strike is, F = dp/dt = 313/0.2 = 1565 N
