- 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
Answer & Solution
(D) \(m\)
Step-by-step Solution
Detailed explanation
\(mu=mv_1\) \(\cos {45^ \circ } + M{v_2}\cos {45^ \circ }\)
\(mu = \frac{1}{{\sqrt 2 }}\left( {m{v_1} + M{v_2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\)
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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