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

Initial velocity of ball at A = 0.
Final velocity of ball at B = v
                                    s = 1 m
As v² - u² = 2as
v² - 0² = 2 x 9.8 x 1 or v = ?19.6 m/s
Let u' be the velocity of rebound of the ball.
s' = 50 cm = 0.5 m, g = - 9.8 m/s2
Final velocity at C = v' = 0
As v² - u² = 2as
Therefore 0² - (u')² = 2 x (-9.8) x 0.5
or    u' = ?9.8 m/s
Now, Impulse =  Change in momentum
                      = mv - (-mu') = mv + mu' = m(v+ u)
                      = 50/1000 (?19.6 + ?9.8)
                      = 1/20 (4.427 + 3.130) = 0.378 Ns
Average force = Impulse/Time = 0.378/0.1 = 3.78 N