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Physics · WORK, ENERGY AND POWER
हिन्दीA body of mass \( m_1 \) moving with speed \( v_1 \) collides completely inelastically with a stationary body of mass \( m_2 \). The fractional loss of kinetic energy of the system \( \frac{\Delta K}{K_i} \) is given by:
Correct answer · Option 1
\( \frac{m_2}{m_1 + m_2} \)
Explanation
Loss \( \Delta K = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} v_1^2 \). Since initial kinetic energy \( K_i = \frac{1}{2}m_1 v_1^2 \), the fractional loss is \( \frac{\Delta K}{K_i} = \frac{m_2}{m_1 + m_2} \).
