Question

A bullet of mass 20g moving with a velocity of 180m/s hits a wooden block of mass 1 kg at rest and gets embedded in it calculate the velocity with which both the bullet and the block move​

Answer 1

Answer:

Explanation:

Initial velocity of the bullet(u₁) = 180 m/s.

Initial velocity of Block(u₂) = 0 m/s.

Mass of the Bullet(m) = 20 g.

Mass of the Wooden block(M) = 1 kg. =1000 g.

Total Initial momentum of the system = mu₁ + Mu₂

= 20 × 180 + 1000 × 0

= 3600 gm/s.

Let the final velocity of the bullet along with block be v.

Then, Final Momentum of the system = (M + m)v

= 1020 × v

Since, Final momentum = Initial momentum. [By law of conservation of momentum.]

∴ 1020v = 3600

∴ v = 2.55 m/s.

Hence, the velocity of bullet sticked in Wooden block is 2.55 m/s.

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