Question

A particle of mass M originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation
$F = F_0 \left[1 - \left(\frac{t}{T}\right)^2\right]$
Where $F_0$ and T are constants. The force acts only for the time interval 2T. The velocity v of the particle after time 2T is:

• A. $F_0T/3M$
• B. $4F_0T/3M$
• C. $F_0T/2M$
• D. $2F_0T/M$

Hint:

The impulse-momentum theorem ($J = \int F dt = \Delta p$) is a powerful tool for problems involving time-varying forces. Pay close attention to the limits of integration defined by the time interval.

Answer 1

The Correct Option is B

According to Newton’s second law, \[ F = \frac{dp}{dt} \] Hence, impulse equals change in momentum: \[ \int F\,dt = \Delta p \] The particle starts from rest, so initial momentum is zero. Final momentum is \(Mv\). \[ Mv = \int_{0}^{2T} F_0\left[1 - \left(\frac{t}{T}\right)^2\right] dt \] \[ Mv = F_0 \int_{0}^{2T} \left(1 - \frac{t^2}{T^2}\right) dt \] Integrating, \[ \int \left(1 - \frac{t^2}{T^2}\right) dt = t - \frac{t^3}{3T^2} \] Apply limits: \[ Mv = F_0 \left[ t - \frac{t^3}{3T^2} \right]_{0}^{2T} \] \[ Mv = F_0 \left[ 2T - \frac{(2T)^3}{3T^2} \right] \] \[ Mv = F_0 \left[ 2T - \frac{8T}{3} \right] \] \[ Mv = F_0 \left( \frac{6T - 8T}{3} \right) = \frac{4F_0T}{3} \] Therefore, \[ v = \frac{4F_0T}{3M} \]