A bullet of mass 10g moving with a velocity of 400m/s gets embedded in a freely suspended wooden block of mass 900g. What is the velocity acquired by the block?
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By using law of conservation of momentum
v = mu / (M + m)
= (10 g × 10^-3 kg/g × 400 m/s) / [(10 + 900)g × 10^-3 kg/g]
= 4 m/s
Velocity acquired by the block is 4 m/s
v = mu / (M + m)
= (10 g × 10^-3 kg/g × 400 m/s) / [(10 + 900)g × 10^-3 kg/g]
= 4 m/s
Velocity acquired by the block is 4 m/s
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