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 \[ F=F_{0}\left[1-\left(\frac{t-T}{T}\right)^{2}\right] \] Where $F_{0}$ and $T$ are constants. The force acts only for the time interval 2T. The velocity $v$ of the particle after time 2 T is :

(C) $4 \mathrm{F}_{0} \mathrm{~T} / 3 \mathrm{M}$

(A) $\mathrm{F}_{0} \mathrm{~T} / 3 \mathrm{M}$

(B) $2 \mathrm{F}_{0} \mathrm{~T} / \mathrm{M}$

(D) $\mathrm{F}_{0} \mathrm{~T} / 2 \mathrm{M}$

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Conservation of Momentum

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} [1 - (\frac{t - T}{T})^{2}]

Where

F_{0}

and

T

are constants. The force acts only for the time interval 2T. The velocity

v

of the particle after time 2 T is :

Answer: (C)

The impulse delivered over the interval equals the change in momentum, and the particle starts from rest:

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

Putting

u = \frac{t - T}{T}

, so

d t = T d u

and

u

runs from

- 1

1

M v = F_{0} T \int_{- 1}^{1} (1 - u^{2}) d u = F_{0} T [u - \frac{u ^{3}}{3}]_{- 1}^{1} = \frac{4 F _{0} T}{3}

The velocity of the particle after time

2 T

\frac{4 F _{0} T}{3 M}

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Single correct option