Q 11-04-179JEE MainJEE Main 2020 (5 Sep, Shift 2)Medium
A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate $\dfrac{dM(t)}{dt} = bv^2(t)$, where $v(t)$ is its instantaneous velocity. The instantaneous acceleration of the spaceship is:
Answer: (B) $-\dfrac{bv^3}{M(t)}$
No external force acts on the spaceship + dust system, so its momentum $Mv$ stays constant:
$$\frac{d(Mv)}{dt} = M\frac{dv}{dt} + v\frac{dM}{dt} = 0$$
$$a = \frac{dv}{dt} = -\frac{v}{M}\cdot bv^2 = -\frac{bv^3}{M(t)}$$
Solution by Sreeraj P, M.Sc Physics