Elastic collision between unequal masses Steps: - Apply conservation of momentum: $m_X u = m_X v_X + m_Y v_Y$. - Apply conservation of kinetic energy: $\frac{1}{2} m_X u^2 = \frac{1}{2} m_X v_X^2 + \frac{1}{2} m_Y v_Y^2$. - Solve using standard elastic formulas: $v_X = u \frac{m_X - m_Y}{m_X + m_Y}$, $v_Y = u \frac{2 m_X}{m_X + m_Y}$. - Since $m_X < m_Y$, $v_X < 0$ so X rebounds with speed $|v_X| < u$, and $v_Y > 0$ with speed $v_Y < u$. Why B is correct: - Elastic collision law requires relative velocity reversal; with $m_X < m_Y$, formulas confirm X rebounds slower than u while Y gains speed less than u (option aligns with rebound < u and forward motion). Why the others are wrong: - A: Rebound speed of X cannot exceed u, as $|v_X| = u \frac{m_Y - m_X}{m_X + m_Y} < u$. - C: X does not stop unless masses are equal; here it …