Question

A stationary bomb explodes into two parts of masses in the ratio of $ 1:3 $ . If the heavier mass moves with a velocity $ 4\,m/s, $ what is the velocity of lighter part?

• A. $ 12\,ms^{-1}$ opposite to heavier mass
• B. $ 12\,ms^{-1}$ in the direction of heavier mass
• C. $6 \,ms^{-1}$ opposite to heavier mass
• D. $6 \,ms^{-1}$ in the direction of heavier mass

Answer 1

Answer: (A)

The ratio of masses $=1: 3$
Therefore, $m_{1}=x kg,\, m_{2}=3 x\, kg$
Applying law of conservation of momentum
$m_{1} v_{1}+m_{2} v_{2}=0$
$\Rightarrow x \times v_{1}+3 x \times 4=0$
$\Rightarrow v_{1}=-12\, m / s$
Therefore, velocity of lighter mass is opposite to that of heavier mass.