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Physics · WORK, ENERGY AND POWER
हिन्दीA particle of mass \( m_1 \) collides head-on elastically with a stationary particle of mass \( m_2 \). If after the collision, the incident particle moves backwards with a speed equal to one-third of its initial speed, the ratio of the masses \( m_1 / m_2 \) is:
Correct answer · Option 3
\( 1:2 \)
Explanation
For 1D elastic collision, \( v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i} \). Here, \( v_{1f} = -\frac{v_{1i}}{3} \implies \frac{m_1 - m_2}{m_1 + m_2} = -\frac{1}{3} \implies 3m_1 - 3m_2 = -m_1 - m_2 \implies 4m_1 = 2m_2 \implies \frac{m_1}{m_2} = \frac{1}{2} \).
