Question

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 1

Initial velocity of the ball = u=0m/s
Let the final velocity = vm/s
g=9.8m/s²; h=10cm=0.1m
Using v²=u²+2gh we get,
v²=0+2×9.8×0.1=1.96
∴, v=√1.96=1.4m/s
In the second time,
Final velocity=0m/s
h=50cm=0.5m
Using, v²=u²-2gh [as the ball goes against gravitational acceleration]
0=u²-2×9.8×0.5
or, u=√9.8=3.13m/s
Now, Impulse= change in linear momentum
                       =mv-m(-u)
                       =0.05×(1.4+3.13)
                       =0.05×4.53
                       =0.2265Ns
∴, Force=Impulse/Time
               =0.2265/0.1
               =2.265N

Answer 2

Answer:

2.265 Newton is the answer