a ball of mass 1kg dropped from height 9.8 m, strikes with ground and rebounds at height of 4.9 m, if the tym of contact between ball and ground is 0.1 s then find impulse and average force acting on ball
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m = 1 kg.
Impulse = force × time.
h = 9.8 m.
h1 = 4.9 m.
v = √(2×9.8×9.8)
= 9.8√2
v2 = √(9.8 × 9.8)
= 9.8
Change in velocity = 9.8(√2 - 1)
Force = change in momentum/time
= 1 × 9.8(√2 - 1)/0.1
= 98(√2 - 1) N
Impulse = force × time
= 9.8(√2 - 1) Ns
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