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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. |
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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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