Question

A bullet of mass 20 g and moving with 600 m/s 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 m/s
• B. 150 m/s
• C. 400 m/s
• D. 300 m/s

Answer 1

Answer: (A)

200 m/s

Before collision,

Momentum of bullet $=\frac{20}{1000}\times 600=12kgm/s$

Let $V_2,V_3$ be velocities of bullet and block after the collision.

Using conservation of momentum,

$\Rightarrow 12=\frac{20}{1000}\times V_2+4V_3$

Given height reached by the block $=0.2m$

Using conservation of energy for the block:

Change in K.E $=$ Work done

$\Rightarrow \frac{1}{2}\times 4\times {V_3}^2=4\times 10\times 0.2\Rightarrow V_3=2m/s$

On solving for $V_2$:

$V_2=\frac{12-8}{\left(\frac{20}{1000}\right)}=200m/s$

Hence, option A is correct.