Question

A bomb explodes into three parts in a horizontal plane. Two masses 1 $kg$ and 2 $kg$goes off at right angles to each other. The first part of mass $1$ $kg$ moves with a speed of $12$ $m/s$ and the second part of mass $2$ $kg$ moves with speed of $8$ $m/s$. If the third part flies with a speed of $4$ $m/s$, then its mass is

• A. $3$ $kg$
• B. $5$ $kg$
• C. $7$ $kg$
• D. $17$ $kg$

Answer 1

The correct option is B $5$ $kg$

Given that two parts moves at right angles with each other.
Let first part of $1$ $kg$ move along $x$-axis with $12m/s$, so its velocity vector is $\left(12\hat{i}\right)m/s$.
Let second part of $2kg$ move along $y$-axis with velocity $8m/s$, so its velocity vector is $\left(8\hat{j}\right)m/s$.
Let $v$ be the velocity of third part of mass $m$ having velocity $4$ $m/s$.
Now according to law of conservation of momentum,
Momentum after explosion = momentum before explosion
$\overrightarrow{P_1}+\overrightarrow{P_2}+{\overrightarrow{P_3}}^{}=01\times \left(12\hat{i}\right)+2\times \left(8\hat{j}\right)+\overrightarrow{P_3}=0\overrightarrow{P_3}=-(12\hat{i}+16\widehat{\mathrm{j)}}\left(\overrightarrow{P_3}\right|=\sqrt{{\left(\mathrm{-12}\right)}^2+{\left(\mathrm{-16}\right)}^2}=20kg\mathrm{m/s}$Momentum = mass $\times$velocity
$20=m\times 4\therefore m=5kg$