JEE PYQ: Oscillations Question 39
Question 39 - 2019 (11 Jan 2019 Shift 2)
A particle of mass $m$ is moving in a straight line with momentum $p$. Starting at time $t = 0$, a force $F = kt$ acts in the same direction on the moving particle during time interval $T$ so that its momentum changes from $p$ to $3p$. Here $k$ is a constant. The value of $T$ is:
(1) $2\sqrt{\frac{k}{p}}$
(2) $2\sqrt{\frac{p}{k}}$
(3) $\sqrt{\frac{2k}{p}}$
(4) $\sqrt{\frac{2p}{k}}$
Show Answer
Answer: (2)
Solution
$\int_p^{3p} dp = \int_0^T kt,dt$. $2p = \frac{kT^2}{2}$. $T = 2\sqrt{\frac{p}{k}}$.
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