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Q.

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

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a

$- b v^{3} (t)$

b

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

c

$- \frac{2 b v^{3}}{M (t)}$

d

$- \frac{b v^{3}}{2 M (t)}$

answer is B.

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Detailed Solution

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

$F = \frac{d p}{d t} = M \frac{d v}{d t} + \frac{d M}{d t}$

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

$F = M \frac{d v}{d t} + b v^{3} (t)$

Force acting on satellite is zero $F = 0$

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

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A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increase at a rate dM(t)dt=bv2(t) , where v(t) is its instantaneous velocity. The instantaneous acceleration of the satellite is