A bullet of mass 10 kg moving with a velocity of 400 m per second gets embedded in a freely suspended wooden block of mass 900 grams. what is the velocity acquired by the block?

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Solution

By using law of conservation of momentum
m₁u₁ = (m₁ + m₂)v

v = m₁u₁ / (m₁ + m₂)
= (10 × 10^-3 kg × 400 m/s) / [(10 + 900) × 10^-3 kg]
= 4.395 m/s

Velocity acquired by the wooden block is 4.395 m/s

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A bullet of mass 10 kg moving with a velocity of 400 m per second gets embedded in a freely suspended wooden block of mass 900 grams. what is the velocity acquired by the block?

By using law of conservation of momentum m₁u₁ = (m₁ + m₂)v v = m₁u₁ / (m₁ + m₂) = (10 × 10^-3 kg × 400 m/s) / [(10 + 900) × 10^-3 kg] = 4.395 m/s Velocity acquired by the wooden block is 4.395 m/s In previous case, if the image is not clear, please download the image

By using law of conservation of momentum
m₁u₁ = (m₁ + m₂)v

v = m₁u₁ / (m₁ + m₂)
= (10 × 10^-3 kg × 400 m/s) / [(10 + 900) × 10^-3 kg]
= 4.395 m/s

Velocity acquired by the wooden block is 4.395 m/s

In previous case, if the image is not clear, please download the image