Q. 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 after collision?
A) $200\ ms^{- 1}$
B) $150\ \ {ms}^{- 1}$
C) $400\ {ms}^{- 1}$
D) $300\ {ms}^{- 1}$
Explanation
(a) 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 ,
$\therefore 0.02 \times 600 = 0.02v + 4v_{2}$
Here, $v_{2} = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.2} = 2ms^{- 1}$
$\therefore 0.02 \times 600 = 0.02v + 4x_{2}$
Or $0.02v = 12 - 8$
Or $v = \frac{4}{0.02} = 200\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.