(C) The force acting on the satellite is given by Newton's second law for a variable mass system: $F = \frac{d}{dt}(Mv)$.Since the satellite is in for

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(C) The force acting on the satellite is given by Newton's second law for a variable mass system: $F = \frac{d}{dt}(Mv)$.Since the satellite is in force-free space,the net external force $F = 0$.Expanding the derivative,we get: $F = M \frac{dv}{dt} + v \frac{dM}{dt} = 0$.Given the rate of mass accumulation is $\frac{dM}{dt} = \alpha v$,we substitute this into the equation:$M \frac{dv}{dt} + v(\alpha v) = 0$.Rearranging to solve for acceleration $a = \frac{dv}{dt}$:$M \frac{dv}{dt} = -\alpha v^2$.Therefore,the deceleration (or acceleration) is $a = -\frac{\alpha v^2}{M}$.

Class 11 Physics Newton's Laws of Motion and Friction Variable Mass System and Rocket Problem

Medium

$A$ satellite in force-free space sweeps stationary interplanetary dust at a rate $\frac{dM}{dt} = \alpha v$,where $M$ is the mass,$v$ is the velocity of the satellite,and $\alpha$ is a constant. What is the deceleration of the satellite?

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A
$ - 2\alpha v^2/M$
B
$ - \alpha v^2/2M$
C
$ - \alpha v^2/M$
D
$ - \alpha v^2$

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Newton's Laws of Motion and Friction

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Variable Mass System and Rocket Problem

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$A$ satellite in force-free space sweeps stationary interplanetary dust at a rate $\frac{dM}{dt} = \alpha v$,where $M$ is the mass,$v$ is the velocity of the satellite,and $\alpha$ is a constant. What is the deceleration of the satellite?

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$A$ rocket moves straight upward with zero initial velocity and with an acceleration of $20 \,m/s^2$. It runs out of fuel and stops accelerating at the end of $5^{th} \,sec$. It reaches a maximum height and falls back to the earth. The speed when it hits the ground is (Take $g = 10 \,m/s^2$):

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