Question

A bullet of mass 0.01 kg and travelling at a speed of 500 m/s strike horizontally to a block of mass of 2 kg which is suspended by a string of length 5 m. The centre of gravity of the block is found to rise a vertical distance of 0.1 m. If the speed of the bullet after it emerge from the block is 55n m/s, find n.

Answer 1

4

Answer 2

1. The velocity of the block immediately after the collision is found using energy conservation ($KE = PE$), which gives $v_B' = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.1} = 1.4$ m/s.
2. Conservation of linear momentum is applied to the collision: $m_b v_b = m_b v_b' + M_B v_B'$.
3. Substituting known values ($0.01 \times 500 = 0.01 v_b' + 2 \times 1.4$) gives $5 = 0.01 v_b' + 2.8$.
4. Solving for $v_b'$ yields $0.01 v_b' = 2.2$, so $v_b' = 220$ m/s.
5. Given $v_b' = 55n$, we have $220 = 55n$, which gives $n = 4$.