Two spheres S_1 and S_2 of masses m_1 and m_2 | vedclass

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Question

(C) Let the initial velocity of $S_2$ be $\vec{u}_2 = v \hat{i}$ and $S_1$ be $\vec{u}_1 = 0$. After the collision, $S_2$ moves with velocity $\vec{v}

Vedclass · Accepted Answer

with a velocity whose direction makes an angle $	heta$ with the $x$-axis such that $	heta = 	an^{-1}\left(\frac{1}{2}ight)$ or $	heta = 	an^{-1}\left(-\frac{1}{2}ight)$

Vedclass · Answer

(C) Let the initial velocity of $S_2$ be $\vec{u}_2 = v \hat{i}$ and $S_1$ be $\vec{u}_1 = 0$. After the collision, $S_2$ moves with velocity $\vec{v}_2 = \frac{v}{2} \hat{j}$ (assuming it moves in the positive $y$-direction). Let the velocity of $S_1$ be $\vec{v}_1 = v_{1x} \hat{i} + v_{1y} \hat{j}$.By conservation of linear momentum along the $x$-axis:$m_2 v = m_1 v_{1x} + m_2(0) \implies v_{1x} = \frac{m_2}{m_1} v$By conservation of linear momentum along the $y$-axis:$0 = m_1 v_{1y} + m_2 \left(\frac{v}{2}ight) \implies v_{1y} = -\frac{m_2 v}{2m_1}$The magnitude of velocity of $S_1$ is $v_1 = \sqrt{v_{1x}^2 + v_{1y}^2} = \sqrt{\left(\frac{m_2 v}{m_1}ight)^2 + \left(-\frac{m_2 v}{2m_1}ight)^2} = \frac{m_2 v}{m_1} \sqrt{1 + \frac{1}{4}} = \frac{m_2 v}{m_1} \frac{\sqrt{5}}{2}$.The direction $	heta$ of $S_1$ with the $x$-axis is given by $	an 	heta = \frac{v_{1y}}{v_{1x}} = \frac{-m_2 v / 2m_1}{m_2 v / m_1} = -\frac{1}{2}$.If $S_2$ had moved in the negative $y$-direction, $	an 	heta = \frac{1}{2}$.Thus, $	heta = 	an^{-1}\left(\frac{1}{2}ight)$ or $	heta = 	an^{-1}\left(-\frac{1}{2}ight)$.

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