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 = F₀[1 - (t-T/T)²] Where F0 and T are constants. The force acts only for the time interval 2T. The velocity v of the particle after time 2T is: (a) 2F0T/M (b) F0T/2M (c) 4F0T/3M (d) F0T/3M

Answer: (c). Impulse = ∫₀2T F dt = Mv. ∫₀2T F₀[1 - ((t-T)²)/(T²)]dt = F₀[2T - 1/T²·2T³/3] = F₀(2T - 2T/3) = (4F₀T)/(3) v = (4F₀T)/(3M), so the answer is (c).

A particle of mass M originally at rest is subjected to a force whose… | JEE Main 2021 (27 Jul, Shift 2)

Q15JEE Main 2021 · 27 JUL · SHIFT 2Medium

Impulse & Momentum Conservation

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 F0 and T are constants. The force acts only for the time interval 2T. The velocity v of the particle after time 2T is:

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Answer: (c)

Impulse $= \int_0^{2T} F\,dt = Mv$ .
$\int_0^{2T} F_0\left[1 - \frac{(t-T)^2}{T^2}\right]dt = F_0\left[2T - \frac{1}{T^2}\cdot\frac{2T^3}{3}\right]$
$= F_0\left(2T - \frac{2T}{3}\right) = \frac{4F_0T}{3}$
$v = \frac{4F_0T}{3M}$ , so the answer is (c).

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Impulse & Momentum Conservation question from JEE Main 2021 (27 Jul, Shift 2)

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