A particle of mass M is originally at rest. It is subjected to a force whose direction is constant but the magnitude varies with the time according to the following relation :
$F = F_o[1-(\frac{t-T}{R})^2]$
where $F_o$ and T are constant. The force acts only for the time interval 2T. What will be the velocity v of the particle after time 2T ?
Answer & explanation
Correct answer: option 3
$F = F_o [1 - (\frac{t - T}{R})^2]$
As $F = Ma = M\frac{dv}{dt}$,
$M\frac{dv}{dt} = F_o [1-(\frac{t-T}{R})^2]$
$\frac{dv}{dt} = \frac{F_o}{M} [1-(\frac{t-T}{R})^2]$
$v = \frac{F_o}{M}\int_o^{F_o} [1 - (\frac{t-T}{R})^2]dt $
$\Rightarrow v = \frac{F_o}{M}\int_o^{F_o} [t - \frac{T}{3}(\frac{t}{T}-1)^3]_0^{2T} dt $
$\Rightarrow v = \frac{4F_o}{3M}$