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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2024
  • Hard
  • MCQ

A stationary particle breaks into two parts of masses \(m_A\) and \(m_B\) which move with velocities \(v_A\) and \(V_B\) respectively. The ratio of their kinetic energies \(\left(\mathrm{K}_B: \mathrm{K}_{\mathrm{A}}\right)\) is _______.

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Verified Solution

Answer & Solution

Correct Answer

(A) \(v_B: v_A\)

Step-by-step Solution

Detailed explanation

Initial momentum is zero.

\(\text { Hence }\left|P_A\right|=\left|P_B\right|\)

\(\Rightarrow m_A v_B=m_B v_B\)

\(\frac{(\mathrm{KE})_A}{(\mathrm{KE})_B}=\frac{\frac{1}{2} m_A v_A^2}{\frac{1}{2} m_B v_B^2}=\frac{v_A}{v_B}\)

\(\frac{(\mathrm{KE})_B}{(\mathrm{KE})_A}=\frac{v_B}{v_A}\)

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