Question

A bullet of mass 0.01$kg$ and traveling at a speed of 500$m/sec$ strikes a block of which suspended by a string of length 5$m$ . The centre of gravity of the block is found to vertical distance of 0.1$m$ . What is the speed of the bullet after it emerges from the block?

• A. 359$m/s$
• B. 220$m/s$
• C. 204$m/s$
• D. 284$m/s$

Answer 1

The correct option is A 359$m/s$

By energy conservation of the suspended block we can say

initial $KE=$ final potential energy

$\frac{1}{2}m{v}^2=mgH$

$\frac{1}{2}m{v}^2=gH$

$v=\sqrt{2gH}$

$v=\sqrt{2\times 10\times 0.1}=1.41m/s$

now$,$ by momentum conservation

$P_{bullet}=P_{block}+P_{bullet}$

$0.01\times 500=1\times 1.41+0.01\times v_f$

$5-1.41=0.01\times v_f$

$0.01v_f=3.59$

$v_f=359m/s$

Hence,

option $\left(A\right)$ is correct answer.