Q. A stationary bomb explodes into two parts of masses in the ratio of 1:3 .If the heavier mass moves with a velocity $4ms^{- 1}$ , 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 heavier mass
Explanation
(a) The ratio of masses=1:3
Therefore ,$m_{1} = xkg,m_{2} = 3x\ kg$
Applying law of conservation of momentum
$m_{1}v_{1} + m_{2}v_{2} = 0$
$\Rightarrow \ \ \ \ \ \ \ \ x \times v_{1} + 3x \times 4 = 0$
Or $v_{1} = - 12ms^{- 1}$
Therefore, velocity of lighter mass is opposite to that of heavier mass.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.