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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2018
  • Hard
  • MCQ

A proton of mass \(m\) collides elastically with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of \(90^o\) with respect to each other. The mass of unknown particle is

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Answer & Solution

Correct Answer

(D) \(m\)

Step-by-step Solution

Detailed explanation

Apply principle of conservation of momentum along \(x-direction,\)

\(mu=mv_1\) \(\cos {45^ \circ } + M{v_2}\cos {45^ \circ }\)

\(mu = \frac{1}{{\sqrt 2 }}\left( {m{v_1} + M{v_2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\)

Along \(y-direction,\)

\(0 = m{v_1}\sin {45^ \circ } - M{v_2}\sin {45^ \circ }\)

\(0 = \left( {m{v_1} - M{v_2}} \right)\frac{1}{{\sqrt 2 }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)\)

\(m,{u_1} = u\,\,\,\,\,\,M,{u_2} = 0\)

proton        Unknown mass     After collision

         Beforce collision

     Coefficient of restution  \(e = 1\)

\(\begin{gathered}
   = {v_2} - {v_1}\,\cos 90 \hfill \\
  \,\,\,\,\,u\,\cos \,45 \hfill \\ 
\end{gathered} \)

                                    (\(\because \) coillsion is elastic)

\(\begin{gathered}
   \Rightarrow \,\frac{{{v_2}}}{{\frac{u}{{\sqrt 2 }}}} = 1 \hfill \\
   \Rightarrow \,u = \sqrt 2 {v_{2\,}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right) \hfill \\ 
\end{gathered} \)

Solving eqn\( (i) ,(ii) \& (iii),\) we get mass of unknown patricle, \(M=m\)

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