A particle of mass m_1 collides with a partic | vedclass

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(A) Let the initial velocity of $m_1$ be $\vec{u}_1$ and final velocities be $\vec{v}_1$ and $\vec{v}_2$. Since the collision is elastic,kinetic energ

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(A) Let the initial velocity of $m_1$ be $\vec{u}_1$ and final velocities be $\vec{v}_1$ and $\vec{v}_2$. Since the collision is elastic,kinetic energy and momentum are conserved.Conservation of momentum: $m_1 \vec{u}_1 = m_1 \vec{v}_1 + m_2 \vec{v}_2$.Squaring both sides: $m_1^2 u_1^2 = m_1^2 v_1^2 + m_2^2 v_2^2 + 2 m_1 m_2 \vec{v}_1 \cdot \vec{v}_2$.Since the particles move at $90^{\circ}$ to each other,$\vec{v}_1 \cdot \vec{v}_2 = 0$,so $m_1^2 u_1^2 = m_1^2 v_1^2 + m_2^2 v_2^2$.Conservation of kinetic energy: $\frac{1}{2} m_1 u_1^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$,which implies $m_1 u_1^2 = m_1 v_1^2 + m_2 v_2^2$.From energy conservation,$m_1(u_1^2 - v_1^2) = m_2 v_2^2$.From momentum conservation,$m_1^2(u_1^2 - v_1^2) = m_2^2 v_2^2$.Dividing the two equations: $m_1 = m_2$. Thus,the ratio $\frac{m_2}{m_1} = 1$.

$A$ particle of mass $m_1$ collides with a particle of mass $m_2$ at rest. After the elastic collision,the two particles move at an angle of $90^{\circ}$ with respect to each other. The ratio $\frac{m_2}{m_1}$ is

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