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NEET 2019
Exam
NEET
Year
2019
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

NEET 2019 · Physics · Center of Mass, Momentum and Collision

A particle of mass 5 ~m at rest suddenly breaks on its own into...

A particle of mass 5 m5 \mathrm{~m} at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along mutually perpendicular direction with each speed v. The energy released during the process isA particle of mass 5 ~m at rest suddenly breaks on its own into three fragments. Two fragments of mass m each moveA particle of mass 5 ~m at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move

Options

  1. A35mv2\frac{3}{5} m v^2
  2. B53mv2\frac{5}{3} m v^2
  3. C32mv2\frac{3}{2} m v^2
  4. D43mv2\frac{4}{3} m v^2

Correct answer

D. 43mv2\frac{4}{3} m v^2

Step-by-step solution

The particle of mass 5 m5 \mathrm{~m} breaks in three fragments of mass m,mand3 mm, m and 3 \mathrm{~m} respectively. Two fragments of mass meach, move in perpendicular direction with velocity vandtheleftfragmentwillmoveinadirectionwithvelocityv and the left fragment will move in a direction with velocityv^{\prime}suchthatthetotalmomentumofthesystemmustremainconserved.<imgsrc="/api/proxy−image?url=httpsBylawofconservationofmomentum,such that the total momentum of the system must remain conserved. <img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fneet%252FZQ_C1kDEeLEjUb934LpzkY0q2S1uogU9bZfzkDTYBLc.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fneet%2FZQ_C1kDEeLEjUb934LpzkY0q2S1uogU9bZfzkDTYBLc.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fneet%2FZQ_C1kDEeLEjUb934LpzkY0q2S1uogU9bZfzkDTYBLc.original.fullsize.png?alt=media"><br> By law of conservation of momentum,5 m×0=mvi^+mvj^+3mv′⇒\mathrmv′=−v3i^−v3j^∴∣\mathrmv′∣=(−v3)2+(−v3)2=v23\begin{aligned}5 \mathrm{~m} \times0 & =\mathrm{mv} \hat{\mathrm{i}}+\mathrm{mv} \hat{\mathrm{j}}+3 \mathrm{mv}^{\prime}\\ \Rightarrow \quad \mathrmv^{\prime} & =-\frac{\mathrm{v}}{3} \hat{\mathrm{i}}-\frac{\mathrm{v}}{3} \hat{\mathrm{j}}\\ \therefore \quad\left|\mathrmv^{\prime}\right|& =\sqrt{\left(-\frac{\mathrm{v}}{3}\right)^2+\left(-\frac{\mathrm{v}}{3}\right)^2}=\frac{\mathrm{v} \sqrt{2}}{3} \end{aligned}∴\therefore Energy released E=12mv2+12mv2+12×3 m(v23)2=mv2+mv23=43mv2\begin{aligned}\mathrm{E}& =\frac{1}{2}m v²+\frac{1}{2}m v²+\frac{1}{2} \times3 \mathrm{~m}\left(\frac{\mathrm{v} \sqrt{2}}{3}\right)^2 \\ & =m v²+\frac{m v^2}{3}=\frac{4}{3}m v² \end{aligned}
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