An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, first part of mass $1\text{kg}$ moving with a velocity of $12{\text{ms}}^{-1}$ and $2\text{kg}$ second part moving with a velocity of $8{\text{ms}}^{-1}.$ If the third part moves with $4{\text{ms}}^{-1},$ its mass would be

The correct option is D $5\text{kg}$
When an explosion breaks a rock, by the law of conservation of momentum, momentum is conserved.
As initial momentum is zero
${\overrightarrow{p}}_1+{\overrightarrow{p}}_2+{\overrightarrow{p}}_3=0$

As $1\text{kg}$ and $2\text{kg}$ are moving perpendicular to each other,
Total momentum of the two pieces $1\text{kg}$ and $2\text{kg}$
$|{\overrightarrow{p}}_1+{\overrightarrow{p}}_2|=\sqrt{{12}^2+{16}^2}=20\text{kgm/s}$
The third piece has the same momentum and in the direction opposite to the resultant of these two momenta
$\therefore$ Momentum of the third piece ${\overrightarrow{p}}_3=20\text{kgm/s}$
Given velocity of third piece is $=4{\text{ms}}^{-1}$
$\therefore$
Mass of the $3^{rd}$ piece=$\frac{{\overrightarrow{p}}_3}{v}=\frac{20}{4}=5\text{kg}$
Hence,the correct answer is option (d).