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JEE Main 201911 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Evening Shift
Subject
Physics
Chapter
Laws of Motion

JEE Main 2019 · 11 January · Evening Shift · Physics · Laws of Motion

A particle of mass m is moving in a straight line with momentum p....

A particle of mass m\mathrm{m} is moving in a straight line with momentum p. Starting at time t=0,t=0, a force F=kF=k t acts in the same direction on the moving particle during time interval T so that its momentum changes from p\mathrm{p} to 3p3 \mathrm{p}. Here kk is a constant. The value of T\mathrm{T} is

Options

  1. A2kp2 \sqrt{\frac{k}{p}}
  2. B2pk2 \sqrt{\frac{\mathrm{p}}{k}}
  3. C2kp\sqrt{\frac{2 k}{p}}
  4. D2pk\sqrt{\frac{2 p}{\mathrm{k}}}

Correct answer

B. 2pk2 \sqrt{\frac{\mathrm{p}}{k}}

Step-by-step solution

From Newton's second law dpdt=F=kt\frac{\mathrm{dp}}{\mathrm{dt}}=\mathrm{F}=\mathrm{kt} Integrating both sides we get, ∫p3pdp=∫0Tktdt⇒[p]p3p=k[t22]0T\int_{\mathrm{p}}^{3 \mathrm{p}} \mathrm{dp}=\int_{0}^{\mathrm{T}} \mathrm{ktdt} \Rightarrow[\mathrm{p}]_{\mathrm{p}}^{3 \mathrm{p}}=\mathrm{k}\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{\mathrm{T}} ⇒2p=kT22⇒T=2pk\Rightarrow 2 \mathrm{p}=\frac{\mathrm{kT}^{2}}{2} \Rightarrow \mathrm{T}=2 \sqrt{\frac{\mathrm{p}}{\mathrm{k}}}
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