A particle of mass $m$ , originally at rest, is subjected to a force whose direction is constant but whose magnitude varies with the time according to the relation
$F = F_{0} [1 - {(\frac{t - T}{T})}^{2}]$
where $F_{0}$ and $T$ are constant. The force acts only for the time interval $2 T$ . Find the speed $v$ of the particle after a time $2 T$ .

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$\frac{3 F_{0} T}{2 m}$

$\frac{4 F_{0} T}{4 m}$

$\frac{4 F_{0} T}{3 m}$

$\frac{8 F_{0} T}{2 m}$

answer is B.

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Detailed Solution

We have acceleration of the particle,

$a = \frac{F}{m} = \frac{F_{0}}{m} [1 - {(\frac{t - T}{T})}^{2}]$

or $\frac{d v}{d t} = \frac{F_{0}}{m} [1 - {(\frac{t - T}{T})}^{2}]$

or $v = \int_{0}^{2 T} \frac{F_{0}}{m} [1 - {(\frac{t - T}{T})}^{2}] d t$

$= \frac{F_{0}}{m} {[t - \frac{{(\frac{t - T}{T})}^{3} T}{3}]}_{0}^{2 T}$

$= \frac{F_{0}}{m} [\{2 T - \frac{{(\frac{2 T - T}{T})}^{3} T}{3}\} - \{0 - \frac{{(\frac{0 - T}{T})}^{3} T}{3}\}]$

$= \frac{4 F_{0} T}{3 m}$ .

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