Question

A 10gm bullet moving directly upward at 1000 m/s strikes and passes through the center of mass of a 10 kg block initially at rest .The bullet emerges from the block moving directly upward at 400 m/s. What will be velocity of the block just after the bullet comes out of it ?

• A. 0.6 m/s
• B. 1 m/s
• C. 0.4 m/s
• D. 1.4 m/s

Answer 1

Answer: (A)

Given,
Weight of bullet $\left(m_{B}\right)=10 \,gm$
$=10 \times 10^{-3} \,kg$
Weight of block $(M)=10 \,kg$
Initial velocity of bullet $\left(v_{i}\right)=1000\, m / s$
Final velocity of bullet $\left(v_{f}\right)=400\, m / s$
By the law of conservation of momentum,
$m_{B} V_{i}=m_{B} V_{f}+10 \,V$
[where, $v=$ velocity of block]
$\left(10 \times 10^{-3}\right)(1000) =\left(10 \times 10^{-3}\right)(400)+10 v $
$10 =4+10 \,V$
$10 V =10-4 $
$V =\frac{6}{10}=0.6\, m / s$