Laws of Motion

JEE Main level, single correct, medium

45 of 210

Question 45 of 210

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

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Answer: (B) 2pk2 \sqrt{\frac{\mathrm{p}}{k}}

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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