An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of$12m{s}^{-1}$and the second part of mass 2 kg moves with $8m{s}^{-1}$speed. If the third part flies off with$4m{s}^{-1}$speed, then its mass is

The situation is as shown in the figure.

According to law of conservation of linear momentum

$$\begin{gathered} {\overline{p}}_1+{\overline{p}}_2+{\overline{p}}_3=0 \\ \therefore {\overline{p}}_3=-\left({\overline{p}}_1+{\overline{p}}_2\right) \end{gathered}$$

$Here, {\overline{p}}_1=\left(1kg\right)\left(12m{s}^{-1}\right)\widehat{i}=12\widehat{i}kgm{s}^{-1}$

${\overline{p}}_2=\left(2kg\right)\left(8m{s}^{-1}\right)\widehat{j}=16\widehat{j}kgm{s}^{-1};$

$\therefore {\overline{p}}_3=-\left(12\widehat{i}+16\widehat{j}\right)kgm{s}^{-1} ;$

$The magnitude of {\overline{p}}_3 is \left|{\overline{p}}_3\right|=\sqrt{{\left(12\right)}^2+{\left(16\right)}^2}=20kgm{s}^{-1} ;$

$\therefore m_3=\frac{p_3}{v_3}=\frac{20kgm{s}^{-1}}{4m{s}^{-1}}=5 kg$