Question

A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate $\frac{dM\left(t\right)}{dt}=b{v}^2\left(t\right)$, where $v\left(t\right)$ is its instantaneous velocity. The instantaneous acceleration of the satellite is

• A. $-b{v}^3\left(t\right)$
• B. $-\frac{b{v}^3}{M\left(t\right)}$
• C. $-\frac{2b{v}^3}{M\left(t\right)}$
• D. $-\frac{b{v}^3}{2M\left(t\right)}$

Answer 1

The correct option is B $-\frac{b{v}^3}{M\left(t\right)}$

From the Newton's second law,

$F=\frac{dp}{dt}=\frac{d\left(mv\right)}{dt}=v\left(\frac{dm}{dt}\right)...\left(i\right)$

We have given,
$\frac{dM\left(t\right)}{dt}=b{v}^2\left(t\right)...\left(ii\right)$

Thrust on the satellite,

$F=-v\left(\frac{dm}{dt}\right)=-v\left(b{v}^2\right)=-b{v}^3$ [Using (i) and (ii)]

$\Rightarrow F=M\left(t\right)a=-b{v}^3\Rightarrow a=\frac{-b{v}^3}{M\left(t\right)}$