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(A) In a perfectly inelastic collision,the two particles stick together after the collision and move with a common velocity $v_f$.By the law of conser

Vedclass · Accepted Answer

$\frac{m_2}{m_1 + m_2}$

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(A) In a perfectly inelastic collision,the two particles stick together after the collision and move with a common velocity $v_f$.By the law of conservation of linear momentum: $m_1 u_1 + m_2(0) = (m_1 + m_2)v_f$.Thus,$v_f = \frac{m_1 u_1}{m_1 + m_2}$.The initial kinetic energy is $K_i = \frac{1}{2} m_1 u_1^2$.The final kinetic energy is $K_f = \frac{1}{2} (m_1 + m_2) v_f^2 = \frac{1}{2} (m_1 + m_2) \left( \frac{m_1 u_1}{m_1 + m_2} \right)^2 = \frac{1}{2} \frac{m_1^2 u_1^2}{m_1 + m_2}$.The loss in kinetic energy (which is converted to heat) is $\Delta K = K_i - K_f = \frac{1}{2} m_1 u_1^2 - \frac{1}{2} \frac{m_1^2 u_1^2}{m_1 + m_2}$.$\Delta K = \frac{1}{2} m_1 u_1^2 \left( 1 - \frac{m_1}{m_1 + m_2} \right) = \frac{1}{2} m_1 u_1^2 \left( \frac{m_2}{m_1 + m_2} \right)$.The fraction of initial kinetic energy converted to heat is $\frac{\Delta K}{K_i} = \frac{m_2}{m_1 + m_2}$.

$A$ particle of mass $m_1$ undergoes a perfectly inelastic head-on collision with another particle of mass $m_2$ at rest $(m_2 > m_1)$. What fraction of the initial kinetic energy is converted into heat energy during this collision?

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