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

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Question

(C) Let the initial velocity of the $1^{st}$ mass be $\vec{v}_1 = v\hat{i}$. The $2^{nd}$ mass is at rest,so $\vec{v}_2 = 0$.By the law of conservatio

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the initial velocity of the $1^{st}$ mass be $\vec{v}_1 = v\hat{i}$. The $2^{nd}$ mass is at rest,so $\vec{v}_2 = 0$.By the law of conservation of linear momentum: $m\vec{v}_1 + m(0) = m\vec{v}_1' + m\vec{v}_2'$.Given that the $1^{st}$ mass moves with velocity $\frac{v}{\sqrt{3}}$ in a direction perpendicular to the initial motion (let this be $\hat{j}$),we have $\vec{v}_1' = \frac{v}{\sqrt{3}}\hat{j}$.Substituting these into the momentum equation: $mv\hat{i} = m(\frac{v}{\sqrt{3}}\hat{j}) + m\vec{v}_2'$.Dividing by $m$: $\vec{v}_2' = v\hat{i} - \frac{v}{\sqrt{3}}\hat{j}$.The speed of the $2^{nd}$ mass is the magnitude of $\vec{v}_2'$:$|\vec{v}_2'| = \sqrt{v^2 + (-\frac{v}{\sqrt{3}})^2} = \sqrt{v^2 + \frac{v^2}{3}} = \sqrt{\frac{4v^2}{3}} = \frac{2}{\sqrt{3}}v$.

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