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JEE Mains · Physics · STD 11 - Newtons laws of motion
  • JEE Mains 2019
  • Medium
  • MCQ

A body of mass \(m_1\) moving with an unknown velocity of \(v_1 \hat i\) undergoes a collinear collision with a body of mass \(m_2\) moving with a velocity \(v_2 \hat i\) . After collision \(m_1\) and \(m_2\) move with velocities of \(v_3 \hat i\) and \(v_4 \hat i\) respectively. If \(m_2 = 0.5\, m_1\) and \(v_3 = 0.5\, v_1\) then \(v_1\) is:

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Answer & Solution

Correct Answer

(C) \({v_4} - {v_2}\)

Step-by-step Solution

Detailed explanation

Applying linear momentum conservation

\(\begin{array}{l}
{m_1}{v_1}\hat i + {m_2}{v_2}\hat i = {m_1}{v_3}\hat i + {m_2}{v_4}\hat i\\
{m_1}{v_1} + 0.5\,{m_1}{v_2} = {m_1}\left( {0.5{v_1}} \right) + 0.5\,{m_1}\,{v_4}\\
0.5\,{m_1}{v_1} = 0.5\,{m_1}\left( {{v_4} - {v_2}} \right)\\
{v_1} = {v_4} - {v_2}
\end{array}\)

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