Q. A bullet of mass 0.01 kg and travelling at a speed of 500$ms^{- 1}$ strikes a block of mass 2 kg, which is suspended by a string of length 5m. The centre of gravity of the block is found to rise a vertical distance of 0.1 m. What is the speed of the bullet after it emerges from the block?
A) $580\ ms^{- 1}$
B) 220$\ ms^{- 1}$
C) 1.4$\ ms^{- 1}$
D) 7.8$\ ms^{- 1}$
Explanation
(b) The speed acquired by block, on account of collision of bullet with it, be $v_{0}\ ms^{- 1}.$ Since the block rise by 0.1 m, hence
$0.1 = \frac{v_{0}^{2}}{2g}$
$\Rightarrow \ \ v_{0}^{2} = 2 \times g \times 0.1$ or $v_{0} = \sqrt{2}\ ms^{- 1}$
Now as per conservation of momentum law for collision between bullet and block,
$mu = mu + Mv_{0}$
$\Rightarrow v = u - \frac{M}{m}v_{0} = 500 - \frac{2kg}{0.01kg} \times \sqrt{2\ }ms^{- 1}$
$= \left( 500 - 200\sqrt{2} \right)ms^{- 1}$
$= 220ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.