Question

A bullet of mass $0.01kg$ is travelling at a speed of $500m/s$ strikes a block of mass $2kg$ which is suspended by a string of length $5m$. The centre of gravity of the block is found to rise a vertical distance of $0.2m$. What is the speed of the bullet after it emerges from the block?

Answer 1

By law of conservation of energy,

$\begin{array}{c}\frac{1}{2}{m}_2{v_2}^2=m_{2}gh\\ v_2=\sqrt{2gh}=\sqrt{2\times 9.8\times 0.01}=1.4m{s}^{-1}\end{array}$

If $u_1$ is the initial velocity of bullet,

Applying law of conservation of momentum,

$\begin{array}{c}{m}_1{u}_1=m_1{v}_1+m_2{v}_2\\ v_1=\frac{0.01\times 500-2\times 1.4}{0.01}=220m{s}^{-1}\end{array}$