JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Conservation of Momentum

A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate

\frac{d M ( t )}{d t} = b v^{2} (t)

, where

v (t)

its instantaneous velocity. The instantaneous acceleration of the satellite is:

Answer: (B)

Solution

No external force acts on the spaceship-plus-dust system as the stationary dust is swept up, so the total momentum is conserved:

\frac{d}{d t} (M v) = 0

. Expanding,

M \frac{d v}{d t} + v \frac{d M}{d t} = 0

, so the acceleration is

\frac{d v}{d t} = - \frac{v}{M} \frac{d M}{d t}

. With the given rate

\frac{d M}{d t} = b v^{2}

, the instantaneous acceleration is

- \frac{b v ^{3}}{M ( t )}

, the minus sign showing that collecting dust slows the ship.

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option