Question

A bullet of mass $20\text{g}$ and moving with $600\text{m/s}$ collides with a block of mass $4\text{kg}$ hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height $0.2\text{m}$ after collision?
(Take $g=10{\text{m/s}}^2$)

• A. $200\text{m/s}$
• B. $150\text{m/s}$
• C. $400\text{m/s}$
• D. $300\text{m/s}$

Answer 1

The correct option is A $200\text{m/s}$

Conservation of linear momentum holds here, According to conservation of linear momentum,
$m_1{v}_1=m_{1}v+m_2{v}_2$
where $v$ is the velocity of bullet after the collision and $v_2$ is the velocity of block.
$\therefore 0.02\times 600=0.02v+4v_2$
From the conservation of mechanical energy,
$m_{2}gh=\frac{1}{2}{m}_2{v}_2^2$
$\Rightarrow v_2=\sqrt{2gh}$
$=\sqrt{2\times 10\times 0.2}$
$=2\text{m/s}$
$\therefore 0.02\times 600=0.02v+4v_2$
$\Rightarrow 0.02v=12-8$
$\therefore v=\frac{4}{0.02}$
$v=200\text{m/s}$