એક એકરેખીક અથડામણમાં v_0 જેટલી પ્રારંભિક ઝડપ | vedclass

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Question

Before collision                     after collision

$\begin{gathered}
  \frac{1}{2}mv

Vedclass · Accepted Answer

$\sqrt 2 \;{v_0}$

Vedclass · Answer

Before collision                     after collision

$\begin{gathered}
  \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 = \frac{3}{2}\left( {\frac{1}{2}mv_0^2} ight) \hfill \
   \Rightarrow \,v_1^2 + v_2^2 = \frac{3}{2}v_0^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( i ight) \hfill \ 
\end{gathered} $

From momentum conservation 

$m{v_0} = m\left( {{v_1} + {v_2}} ight)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( {ii} ight)$

Squarring both sides,

$\begin{gathered}
  {\left( {{v_1} + {v_2}} ight)^2} = v_0^2 \hfill \
   \Rightarrow \,v_1^2 + v_2^2 + 2{v_1}{v_2} = v_0^2 \hfill \
  \,\,\,\,\,\,2{v_1}{v_2} = \frac{{v_0^2}}{2} \hfill \
  {\left( {{v_1} - {v_2}} ight)^2} = v_1^2 + v_2^2 - 2{v_1}{v_2} = \frac{3}{2}v_0^2 + \frac{{v_0^2}}{2} \hfill \ 
\end{gathered} $

Solving we get relative velocity between the two particles

${v_1} - {v_2} = \sqrt 2 {v_0}$

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