JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 _______.
- A \(v_B: v_A\)
- B \(m_B: m_A\)
- C \(m_B v_B: m_A v_A\)
- D \(1: 1\)
Answer & Solution
Correct Answer
(A) \(v_B: v_A\)
Step-by-step Solution
Detailed explanation
Initial momentum is zero. \(\text { Hence }\left|P_A\right|=\left|P_B\right|\) \(\Rightarrow m_A v_B=m_B v_B\) \(\frac{(\mathrm{KE})_A}{(\mathrm{KE})_B}=\frac{\frac{1}{2} m_A v_A^2}{\frac{1}{2} m_B v_B^2}=\frac{v_A}{v_B}\)…
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