Q. 74.5( 4 Votes )

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 :

Initial velocity = 0m/s = u

Mass = 50g = 0.05kg


Final velocity = v


Using 3rd equation of motion


v2 = u2 + 2gh


v2 = 02 + 2(9.8)(0.1)


v2 = 1.96


v = 1.4m/s


For 50cm rebound


Using 3rd equation of motion


v2 = u2 – 2gh


02 = u2 – 2(9.8)(0.5)


u = 3.13m/s


Impulse = change in momentum


= mv–(–mu)


= 0.05kg × 1.4m/s + 0.05kg × 3.13m/s


= 0.2265Ns


Force =


=


= 2.265N


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