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4-1.Newton's Laws of Motion
medium
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
A
$2\sqrt {\frac{k}{p}} $
B
$2\sqrt {\frac{p}{k}} $
C
$\sqrt {\frac{{2k}}{p}}$
D
$\sqrt {\frac{{2p}}{k}} $
(JEE MAIN-2019)
Solution
$\begin{array}{l}
\frac{{dp}}{{dt}} = F = kt\\
\int_p^{3p} {dP} = \int_0^T {kt\,dt} \\
2p = \frac{{K{T^2}}}{2}\,\,;\,\,\,T = 2\sqrt {\frac{p}{k}}
\end{array}$
Standard 11
Physics
