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(B) For an elastic collision between two particles of equal mass $(m_1 = m_2 = m)$, where one particle is initially at rest $(u_2 = 0)$: By conservati

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(B) For an elastic collision between two particles of equal mass $(m_1 = m_2 = m)$, where one particle is initially at rest $(u_2 = 0)$: By conservation of linear momentum: $\vec{p}_1 = \vec{p}_1' + \vec{p}_2'$. Squaring both sides: $p_1^2 = p_1'^2 + p_2'^2 + 2\vec{p}_1' \cdot \vec{p}_2'$. By conservation of kinetic energy: $\frac{1}{2}mu^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2$, which implies $p_1^2 = p_1'^2 + p_2'^2$. Comparing the two equations, we get $2\vec{p}_1' \cdot \vec{p}_2' = 0$. Since the momenta are non-zero, the dot product is zero only if the angle between the final velocity vectors is $90^\circ$. Thus, $	heta_1 + 	heta_2 = 90^\circ$.

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