Question

A bullet of mass 10gms moving with the velocity of 400m/s gets embedded in a freely suspended wooden block of mass 900gms. What is the velocity acquired by the block?

Answer 1

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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Answer 2

Answer:

Explanation:

bullet :

mass= 10 g = 10/1000=0.01kg

velocity = 400 m/s

wodden block:

mass=900g+0.01 kg ( bullet gets emedded)

900/1000+0.01 kg = 0.91kg

velocity = v

we know that

p of bullet = recoil of gun

0.01×400=0.91×v

4=0.91v

v= 4/0.91

v = 4.39 m/s