JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
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 \(\sqrt {\frac{{2p}}{k}} \)
Answer & Solution
Correct Answer
(B) \(2\sqrt {\frac{p}{k}} \)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l} \frac{{dp}}{{dt}} = F = kt\\ \int_p^{3p} {dP} = \int_0^T {kt\,dt} \\ 2p = \frac{{K{T^2}}}{2}\,\,;\,\,\,T = 2\sqrt {\frac{p}{k}} \end{array}\)
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