A mass m moves with a velocity v and collides | vedclass

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(A) According to the law of conservation of linear momentum: $m_{1} \vec{u}_{1} + m_{2} \vec{u}_{2} = m_{1} \vec{v}_{1} + m_{2} \vec{v}_{2}$.Let the i

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$\frac{2}{\sqrt{3}}v$

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(A) According to the law of conservation of linear momentum: $m_{1} \vec{u}_{1} + m_{2} \vec{u}_{2} = m_{1} \vec{v}_{1} + m_{2} \vec{v}_{2}$.Let the initial velocity of the first mass be $\vec{u}_{1} = v \hat{i}$ and the second mass be at rest,$\vec{u}_{2} = 0$.After the collision,the first mass moves with velocity $\vec{v}_{1} = \frac{v}{\sqrt{3}} \hat{j}$ (perpendicular to the initial direction).Substituting these values into the momentum equation:$m(v \hat{i}) + m(0) = m(\frac{v}{\sqrt{3}} \hat{j}) + m \vec{v}_{2}$.Dividing by $m$,we get:$\vec{v}_{2} = v \hat{i} - \frac{v}{\sqrt{3}} \hat{j}$.The speed of the second mass is the magnitude of $\vec{v}_{2}$:$|\vec{v}_{2}| = \sqrt{v^{2} + (-\frac{v}{\sqrt{3}})^{2}} = \sqrt{v^{2} + \frac{v^{2}}{3}} = \sqrt{\frac{4v^{2}}{3}} = \frac{2}{\sqrt{3}}v$.

$A$ mass $m$ moves with a velocity $v$ and collides inelastically with another identical mass initially at rest. After collision,the first mass moves with velocity $\frac{v}{\sqrt{3}}$ in a direction perpendicular to its initial direction of motion. The speed of the second mass after collision is:

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