Question

A missile of mass $M$ moving with velocity $v=200m/s$ explodes in midair breaking into two parts of mass $M/4$ and $3M/4$. If the smaller piece flies off at an angle of ${60}^o$ with respect to the original direction of motion with an initial speed of $400m/s$, what is the magnitude and direction of the initial velocity of the other piece.

Answer 1

Parallel and perpendicular momentum is conserved

$\frac{m}{4}400\sin 60=\frac{3m}{4}u\sin \theta u=\frac{200\sqrt{3}}{3\sin \theta }$

Parallel momentum

$m\left(200\right)=400\left(\frac{m}{4}\right)\cos 60+\frac{3m}{4}u\cos \theta 200=400\left(\frac{m}{4}\right)\frac{1}{2}+\frac{3}{4}m\left(\frac{200\sqrt{3}}{3\sin \theta }\cos \theta \right)4=1+\sqrt{3}\cot \theta 3=\sqrt{3}\cot \theta \cot \theta =\sqrt{3}\theta =30°u=\frac{200\sqrt{3}}{3\sin 30°}=\frac{400}{\sqrt{3}}m/s$