A missile of mass $M$ moving with velocity $v = 200 m / s$ explodes in midair breaking into two parts of mass $M / 4$ and $3 M / 4$ . 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 $400 m / s$ , what is the magnitude and direction of the initial velocity of the other piece.

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Solution

Parallel and perpendicular momentum is conserved

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

Parallel momentum

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

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Parallel and perpendicular momentum is conserved m4400sin60=3m4usinθu=200√33sinθ Parallel momentum m(200)=400(m4)cos60+3m4ucosθ200=400(m4)12+34m(200√33sinθcosθ)4=1+√3cotθ3=√3cotθcotθ=√3θ=30°u=200√33sin30°=400√3m/s