A bomb at rest explodes into three pieces of equal masses. If two pieces move perpendicular to each other,…
A bomb at rest explodes into three pieces of equal masses. If two pieces move perpendicular to each other, each with a speed $v$ then the speed of the third piece is
V
$\mathrm{v} \sqrt{2}$
$\frac{\mathrm{v}}{\sqrt{2}}$
$2 \mathrm{v}$
Solution
Apply the conservation of momentum
$\begin{aligned}
& 3 m \times 0=m \vec{v}_1+m \vec{v}_2+m \vec{v}_3 \\
& 0=m v \hat{i}+m v \hat{j}+m \vec{v}_3 \\
& \vec{v}_3=-v \hat{i}-v \hat{j} \\
& \left|v_3\right|=\sqrt{(-v)^2+(-v)^2}=\sqrt{2} v
\end{aligned}$