JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 18)
A body A of mass m is moving in a circular orbit
of radius R about a planet. Another body B of
mass
$${m \over 2}$$
collides with A with a velocity which is half $$\left( {{{\overrightarrow v } \over 2}} \right)$$ the instantaneous velocity$${\overrightarrow v }$$
of A.
The collision is completely inelastic. Then, the
combined body :
starts moving in an elliptical orbit around
the planet.
Falls vertically downwards towards the
planet
Escapes from the Planet's Gravitational field.
continues to move in a circular orbit
Explanation
Orbital speed for of A is v = $$\sqrt {{{GM} \over R}} $$
After collision, let the combined mass moves with speed v1
$$ \therefore $$ mv + $${m \over 2}{v \over 2}$$ = $$\left( {{{3m} \over 2}} \right){v_1}$$
$$ \Rightarrow $$ v1 = $${{5v} \over 6}$$
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 $$\sqrt 2 v$$, so they cannot escape gravitational field. So their motion will be elliptical around the planet.
After collision, let the combined mass moves with speed v1
$$ \therefore $$ mv + $${m \over 2}{v \over 2}$$ = $$\left( {{{3m} \over 2}} \right){v_1}$$
$$ \Rightarrow $$ v1 = $${{5v} \over 6}$$
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 $$\sqrt 2 v$$, so they cannot escape gravitational field. So their motion will be elliptical around the planet.
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