A proton of mass m collides elastically with | vedclass

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Question

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

Vedclass · Accepted Answer

$m$

Vedclass · Answer

Apply principle of conservation of momentum along $x-direction,$

$mu=mv_1$ $\cos {45^ \circ } + M{v_2}\cos {45^ \circ }$

$mu = \frac{1}{{\sqrt 2 }}\left( {m{v_1} + M{v_2}} ight)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i ight)$

Along $y-direction,$

$0 = m{v_1}\sin {45^ \circ } - M{v_2}\sin {45^ \circ }$

$0 = \left( {m{v_1} - M{v_2}} ight)\frac{1}{{\sqrt 2 }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} ight)$

$m,{u_1} = u\,\,\,\,\,\,M,{u_2} = 0$

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} ight) \hfill \ 
\end{gathered} $

Solving eqn$ (i) ,(ii) \& (iii),$ we get mass of unknown patricle, $M=m$

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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