Question

A rubber ball of mass 50g falls from a height of 1m and rebpunds to a height of 50cm. Calculate the impulse and the average force between the ball and the ground, if the time during which they are in contact was 0.1 sec.

Answer 1

Velocity of rubber ball just before hitting the ground is
u = √(2gH) = √(2 × 10 m/s² × 1 m) = √20 m/s

Velocity of rubber ball just after hitting the ground is
v = -√(2gH) = -√(2 × 10 m/s² × 0.5 m) = -√10 m/s

Impulse = Mass × Change in velocity
= [50 g × 10^-3 kg/g] × [(√20 - (-√10)) m/s]
= [0.05 kg] × [√10(√2 + 1) m/s]
= [0.05 × 2.414√10] kg m/s
= 0.38 kg m/s

Force = Impulse / Time
= [0.38 kg m/s] / 0.1 s
= 3.8 N

[Note: Answer will be slight different if we take g = 9.8 m/s²]