1.

A ball of mass 1 kg dropped from 9.8 m height, strikes the ground and rebounds to a height of 4.9 m. If thetime of contact between ball and ground is 0.1 s, then find impulse and average force acting on ball.

Answer»

Velocity when it hits the groundu = sqrt(2gH)= sqrt(2 * 9.8 * 9.8)= 9.8 sqrt(2) m/s

Velocity just after hitting the groundv = sqrt(2gH)= sqrt(2 * 9.8 * 4.9)= 9.8 m/s

Impulse = change in momentum= mass × change in velocity= m * (v - u)= 1kg * (9.8 - (-9.8*sqrt(2)) m/s= 9.8 * (1 + sqrt(2)) kgm/s= 9.8 * 1.414 kgm/s= 13.8572 kgm/s

Average Force = Impulse / Time taken= 13.8572 / 0.1= 1.38 N

wrong ans



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