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JEE Main 20137 April
Exam
JEE Main
Year
2013
Date
7 April
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2013 · 7 April · Physics · Center of Mass, Momentum and Collision

This question has Statement - I and Statement - II of the four...

This question has Statement−IStatement - I and Statement−IIStatement - II of the four choices given after the Statements, choose the one that best describes the two Statements.

Statement - I:ApointparticleofmassA point particle of massmmovingwithspeedmoving with speedνcollideswithstationarypointparticleofmasscollides with stationary point particle of massM.Ifthemaximumenergylosspossibleisgivenas<br/>. If the maximum energy loss possible is given as<br />f(12mν2)\left(\dfrac{1}{2}m{ν}^{2}\right) then f=(mM+m)f = \left(\dfrac{m}{M + m}\right).

Statement - II: Maximum energy loss occurs when the particles get stuck together as a result of the collision.

Options

  1. AStatement-II is true, Statement-IIII is false.
  2. BStatement-II is false, Statement-IIII is true.
  3. CStatement-II is true, Statement-IIII is true, Statement-IIII is a correct explanation of Statement-II.
  4. DStatement-II is true, Statement-IIII is true, Statement-IIII is not a correct explanation of Statement-II.

Correct answer

B. Statement-II is false, Statement-IIII is true.

Step-by-step solution

Maximum energy loss will take place in perfectly inelastic collision, and is given by, ∆K=P22m−P22(m+M)∆ K = \dfrac{{P}^{2}}{2m} - \dfrac{{P}^{2}}{2\left(m + M\right)}
⇒∆K=P22m[1−m(m+M)]⇒ ∆ K = \dfrac{{P}^{2}}{2m}\left[1 - \dfrac{m}{\left(m + M\right)}\right]
⇒∆K=P22m[M(m+M)]⇒ ∆ K = \dfrac{{P}^{2}}{2m}\left[\dfrac{M}{\left(m + M\right)}\right]
⇒∆K=12mv2[M(m+M)]⇒ ∆ K = \dfrac{1}{2}m{v}^{2}\left[\dfrac{M}{\left(m + M\right)}\right]
On comparing with the given form of equation, we have, \left(f = \dfrac{M}{m + M}\right)<br/><br/>HenceStatement−<br /><br />Hence Statement-Iisfalse,Statement−is false, Statement-II is true. ( As it is perfect Inelastic collision)

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