Q 11-04-158JEE MainJEE Main 2021 (27 Jul, Shift 2)Medium
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 $F = F_0\left[1 - \left(\dfrac{t - T}{T}\right)^2\right]$ 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:
Answer: (C) $\dfrac{4F_0T}{3M}$
Impulse $= \displaystyle\int_0^{2T}F\,dt$. Put $u = \dfrac{t - T}{T}$, $dt = T\,du$, $u$ from $-1$ to $1$:
$$J = F_0T\int_{-1}^{1}(1 - u^2)\,du = F_0T\left(2 - \frac23\right) = \frac{4F_0T}{3}$$
$$v = \frac JM = \frac{4F_0T}{3M}$$
Solution by Sreeraj P, M.Sc Physics