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:

(A) starts moving in an elliptical orbit around the planet.

(B) continues to move in a circular orbit.

(C) Escapes from the Planet's Gravitational field.

(D) Falls vertically downwards towards the planet.

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Conservation of Momentum

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

(\frac{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.✓ Correct
(B)continues to move in a circular orbit.
(C)Escapes from the Planet's Gravitational field.
(D)Falls vertically downwards towards the planet.

Answer: (A)

Solution

Momentum conservation for the sticking collision gives the new velocity:

m v + \frac{m}{2} \cdot \frac{v}{2} = \frac{3 m}{2} v_{f}

, so

v_{f} = \frac{5 v}{6}

. The speed needed to stay on the circle of radius

R

is the original

v

, and since

\frac{5 v}{6} < v

the combined body is now too slow to hold that circle. It therefore falls inward along an ellipse, with the point of collision as the farthest point of the new orbit.

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Single correct option

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JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Conservation of Momentum

Solution

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