Question

A bullet of mass $20 \,g$ and moving with $600\, ms^{-1}$ collides with a block of mass $4 \,kg$ hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height $0.2 \,m$ after collision

• A. $200 \, ms^{-1}$
• B. $150 \, ms^{-1}$
• C. $400 \, ms^{-1}$
• D. $300 \, ms^{-1}$

Answer 1

Answer: (A)

According to conservation of linear momentum
$m_1v_1 =m_1 v+m_2v_2$
where $v_1$, is velocity of bullet before collision, $v$ is velocity of bullet after the collision and $v_2$ is the velocity of block.
= $0.02 \times 600 = 0.02v + 4v_2$
Here $v_2 =\sqrt{2gh}=\sqrt{2 \times 10 \times 0.2 }=2ms^{-1}$
=$0.02 \times 600 = 0.02 v + 4 \times 2$
$\Rightarrow 0.02v=12-8$
$\Rightarrow v=\frac{4}{0.02}=200 ms^{-1}$