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 ?

Updated On: Jan 18, 2023
  • 0.6 m/s
  • 1 m/s
  • 0.4 m/s
  • 1.4 m/s
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The Correct Option is A

Solution and Explanation

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 $
$V =\frac{6}{10}=0.6\, m / s$
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Concepts Used:

Newtons Laws of Motion

Newton’s First Law of Motion:

Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.

Newton’s Second Law of Motion:

Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass.

Mathematically, we express the second law of motion as follows:

Newton’s Third Law of Motion:

Newton’s 3rd law states that there is an equal and opposite reaction for every action.