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

Correct answer: D. starts moving in an elliptical orbit around the planet Conserving momentum (m)/(2)(v)/(2) + mv = (m + (m)/(2)) v _ f v _ f = (2mV)/(4 × (3m)/(2)) = (5V)/(6) v _ f < v _ orb (= v) thus the combined mass will go on to an elliptical path

JEE Main 20209 JanuaryMorning Shift

Exam: JEE Main
Year: 2020
Date: 9 January
Shift: Morning Shift
Subject: Physics
Chapter: Gravitation

JEE Main 2020 · 9 January · Morning Shift · Physics · Gravitation

A body A of mass m is moving in a circular orbit of radius R about...

A body A of mass

m

is moving in a circular orbit of radius

R

about a planet. Another body B of mass

\dfrac{m}{2}

collides with A with a velocity which is half \left(\dfrac{\overset{\to}{v}}{2}\right)

the instantaneous velocity

\overset{\to}{v} of A. The collision is completely inelastic. Then, the combined body:

Options

Acontinues to move in a circular orbit
BEscapes from the Planet's Gravitational field
CFalls vertically downwards towards the planet
Dstarts moving in an elliptical orbit around the planet

Correct answer

D. starts moving in an elliptical orbit around the planet

Step-by-step solution

Conserving momentum

\dfrac{m}{2}\dfrac{v}{2} + mv = \left(m + \dfrac{m}{2}\right){v}_{f}

{v}_{f} = \dfrac{2mV}{4 \times \dfrac{3m}{2}} = \dfrac{5V}{6}

{v}_{f} < {v}_{orb}\left(= v\right)

thus the combined mass will go on to an elliptical path

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