A body A of mass m is moving in a circular orbit of radius R about a planet. Another body B of mass m/2 collides with A with a velocity which is half (v/2) the instantaneous velocity v of A. The collision is completely inelastic. Then, the combined body: (a) starts moving in an elliptical orbit around the planet. (b) Falls vertically downwards towards the planet (c) Escapes from the Planet's Gravitational field. (d) continues to move in a circular orbit

Answer: (a). Momentum conservation (B moves along v): m v + m/2/2 = 3m/2v'. v' = (5mv/4)/(3m/2) = 5/6v. Orbital speed at radius R is v (independent of mass), and v' < v < √(2) v (escape speed). Speed differs from circular speed but is below escape speed, so the orbit becomes elliptical. So the answer is (a) starts moving in an elliptical orbit.

A body A of mass m is moving in a circular orbit of radius R about a… | JEE Main 2020 (9 Jan, Shift 1)

Q23JEE Main 2020 · 9 JAN · SHIFT 1Difficult

Satellite Energy & Weightlessness

A body A of mass m is moving in a circular orbit of radius R about a planet. Another body B of mass $\frac{m}{2}$ collides with A with a velocity which is half $\left(\frac{\vec{v}}{2}\right)$ the instantaneous velocity $\vec{v}$ of A. The collision is completely inelastic. Then, the combined body:

Show answer and solution

Answer: (a)

Momentum conservation (B moves along $\vec{v}$ ): $m v + \frac{m}{2}\cdot\frac{v}{2} = \frac{3m}{2}v'$ .
$v' = \frac{5mv/4}{3m/2} = \frac{5}{6}v$ .
Orbital speed at radius $R$ is $v$ (independent of mass), and $v' < v < \sqrt{2}\,v$ (escape speed).
Speed differs from circular speed but is below escape speed, so the orbit becomes elliptical.
So the answer is (a) starts moving in an elliptical orbit.

Report a mistake

Satellite Energy & Weightlessness question from JEE Main 2020 (9 Jan, Shift 1)

More questions on Satellite Energy & Weightlessness