JEE MAIN - Physics (2020 - 5th September Evening Slot - No. 17)
A spaceship in space sweeps stationary
interplanetary dust. As a result, its mass
increases at a rate $${{dM\left( t \right)} \over {dt}}$$ = bv2(t), where v(t) is its instantaneous velocity. The instantaneous acceleration of the satellite is :
increases at a rate $${{dM\left( t \right)} \over {dt}}$$ = bv2(t), where v(t) is its instantaneous velocity. The instantaneous acceleration of the satellite is :
-bv3(t)
$$ - {{2b{v^3}} \over {M\left( t \right)}}$$
$$ - {{b{v^3}} \over {M\left( t \right)}}$$
$$ - {{b{v^3}} \over {2M\left( t \right)}}$$
Explanation
Given $${{dM\left( t \right)} \over {dt}}$$ = bv2(t)
In free space no external force so there in only thrust force on rocket.
Fthrust = v$${{dm} \over {dt}}$$
Force on satellite = $$ - \overrightarrow v {{dm\left( t \right)} \over {dt}}$$
M(t)a = – v (bv2)
$$ \Rightarrow $$ a = $$ - {{b{v^3}} \over {M\left( t \right)}}$$
In free space no external force so there in only thrust force on rocket.
Fthrust = v$${{dm} \over {dt}}$$
Force on satellite = $$ - \overrightarrow v {{dm\left( t \right)} \over {dt}}$$
M(t)a = – v (bv2)
$$ \Rightarrow $$ a = $$ - {{b{v^3}} \over {M\left( t \right)}}$$
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