Question

A bomb at rest explodes into three fragments of equal masses. Two fragments fly off at right angles to each other with velocities $9m{s}^{-1}$ and $12m{s}^{-1}$ respectively. Calculate the speed of the third fragment.

Answer 1

conservation of momentum
$\begin{array}{c}3m\times 0=m{\overrightarrow{V}}_1+m{\overrightarrow{V}}_2+m{\overrightarrow{V}}_3\\ 0=qm\hat{i}+12m\hat{j}+m{\overrightarrow{V}}_3\\ {\overrightarrow{V}}_3=-9\hat{i}-12\hat{j}\\ |{\overrightarrow{V}}_3|=\sqrt{9^2+{12}^2}=\sqrt{225}=15m/s\end{array}$