Find out the velocity of the bullet:

A bullet of weight W₁  = 5 N is fired into a body B₂ of weight W₂  = 45 N suspended by a string of length L = 1 metres. Because of impact, body B₂ swings through an ∠ θ = 60^o. If the impact is fully plastic, find out the velocity of the bullet.

Solution

 Refer Figure. Let initial velocity of the bullet by V₁.

 Initial velocity of B₂ = V₂ = 0

After impact, we have, by principle of conservation of momentum

                                   (m₁ + m₂) V_C = m₁ V₁            --------- (I)

By using principle of conservation of energy for the second event when body B₂ swings through a maximum ∠ θ = 60^o,

((W₁ + W₂) /2 g)(V_c) 2 = (W₁ + W₂) × L × (1 - cosθ)

∴ V ₂  = 2 g L (1/2)   = 9.8 × 1

∴ V_c  = 3.13 m / sec .