Q87 / 90 · JEE Main 2019 (11 Jan Shift 2) · Physics

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

  • A
    $2 \sqrt{\frac{k}{p}}$
  • Correct answer
    B
    $2 \sqrt{\frac{\mathrm{p}}{k}}$
  • C
    $\sqrt{\frac{2 k}{p}}$
  • D
    $\sqrt{\frac{2 p}{\mathrm{k}}}$

Solution

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

Hence, the correct answer is option: 'B'.

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