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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2019
  • Hard
  • MCQ

Two particles, of masses \(M\) and \(2M\), moving as shown, with speeds of \(10\, m/s\) and \(5\, m/s\), collide elastically at the origin. After the collision, they move along the indicated directions with speeds \(v_1\) and \(v_2\) respectively. The value of \(v_1\) and \(v_2\) are nearly

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Answer & Solution

Correct Answer

(B) \(6.5\, m/s\) and \(6.3\, m/s\)

Step-by-step Solution

Detailed explanation

\(\begin{array}{l}
2MV,\cos {30^ \circ } + M{v_2}\cos {45^ \circ }\\
 = 10\,M\,\cos {30^ \circ } + 10\cos {45^ \circ }\\
 \Rightarrow {v_1}\sqrt 3  + \frac{{{v_2}}}{{\sqrt 2 }} = 5\sqrt 3  + 5\sqrt 2 \,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\\
2MV,\,\sin {30^ \circ } - M{V_2}\sin {45^ \circ }\\
 =  - 10\,M\sin {30^ \circ } + 10\,M\,\sin {45^ \circ }
\end{array}\)

\(\begin{array}{l}
V - \frac{{{V_2}}}{{\sqrt 2 }} =  - 5 + 5\sqrt 2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)\\
{V_1} = \frac{{5\left( {\sqrt 3  - 1} \right) + 10\sqrt 2 }}{{\sqrt 3  + 1}} = \frac{{17.5}}{{2.7}} = 6.5\,m/s\\
{V_2} = 6.3\,m/s
\end{array}\)

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