Solveeit Logo

Question

Physics Question on Newtons Laws of Motion

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

0.6 m/s

Explanation

Solution

Given,
Weight of bullet (mB)=10gm\left(m_{B}\right)=10 \,gm
=10×103kg=10 \times 10^{-3} \,kg
Weight of block (M)=10kg(M)=10 \,kg
Initial velocity of bullet (vi)=1000m/s\left(v_{i}\right)=1000\, m / s
Final velocity of bullet (vf)=400m/s\left(v_{f}\right)=400\, m / s
By the law of conservation of momentum,
mBVi=mBVf+10Vm_{B} V_{i}=m_{B} V_{f}+10 \,V
[where, v=v= velocity of block]
(10×103)(1000)=(10×103)(400)+10v\left(10 \times 10^{-3}\right)(1000) =\left(10 \times 10^{-3}\right)(400)+10 v
10=4+10V10 =4+10 \,V
10V=10410 V =10-4
V=610=0.6m/sV =\frac{6}{10}=0.6\, m / s