A body A of mass m is moving in a circular orbit of radius R abou… - Vidyadip

MCQ

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

✓
starts moving in an elliptical orbit around the planet
B
continues to move in a circular orbit
C
Falls vertically downwards towards the planet
D
Escapes from the Planet's Gravitational field.

✓

Answer

Correct option: A.

starts moving in an elliptical orbit around the planet

a
Initially, the body of mass $\mathrm{m}$ is moving in a circular orbit of radius $\mathrm{R} .$ So it must be moving with orbital speed.

$\mathrm{v}_{0}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$

After collision, let the combined mass moves with speed $v_{1 }$

${\mathrm{mv}_{0}+\frac{\mathrm{m}}{2} \frac{\mathrm{v}_{0}}{2}=\left(\frac{3 \mathrm{m}}{2}\right) \mathrm{v}_{1}}$

${\mathrm{v}_{1}=\frac{5 \mathrm{v}_{0}}{6}}$

since after collision, the speed is not equal to orbital speed at that point. So motion cannot be circular. since velocity will remain tangential, so it cannot fall vertically towards the planet. Their speed after collision is less than escape speed $\sqrt{2} v_{0}, \quad$

so they cannot escape gravitational field. So their motion will be elliptical around the planet.

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