Q. A stationary bomb explodes into three pieces. One piece of $2\ kg$ mass moves with a velocity of $8\ ms^{- 1}$ at right angles to the other piece of mass $1\ kg$ moving with a velocity of $12\ ms^{- 1}$. If the mass of the third piece is $0.5\ kg$, then its velocity is
A) $10\ ms^{- 1}$
B) $20\ ms^{- 1}$
C) $30\ ms^{- 1}$
D) $40\ ms^{- 1}$
Explanation
(d) Applying the law of conservation of linear momentum, we get
$0.5 \times v = \sqrt{(2 \times 8)^{2} + (1 \times 12)^{2}} = \sqrt{256 + 144} = \sqrt{400}$
$0.5\ v = 20 \Rightarrow v = \frac{20}{0.5} = 40\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.