Find out the velocity of the bullet:
A bullet of weight W_{1} = 5 N is fired into a body B_{2} of weight W_{2} = 45 N suspended by a string of length L = 1 metres. Because of impact, body B_{2} 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 V_{1}.
Initial velocity of B_{2 }= V_{2} = 0
After impact, we have, by principle of conservation of momentum
(m_{1} + m_{2}) V_{C} = m_{1} V_{1} --------- (I)
By using principle of conservation of energy for the second event when body B_{2} swings through a maximum ∠ θ = 60^{o},
((W_{1} + W_{2}) /2 g)(V_{c}) 2 = (W_{1} + W_{2}) × L × (1 - cosθ)
∴ V_{ 2} = 2 g L (1/2) = 9.8 × 1
∴ V_{c} = 3.13 m / sec .