A particle of mass m moving eastward with a s | vedclass

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(C) Let the velocity of the first particle be $\vec{v}_1 = v \hat{i}$ and the velocity of the second particle be $\vec{v}_2 = v \hat{j}$.According to

Vedclass · Accepted Answer

$\frac{v}{\sqrt{2}}$ North-East

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(C) Let the velocity of the first particle be $\vec{v}_1 = v \hat{i}$ and the velocity of the second particle be $\vec{v}_2 = v \hat{j}$.According to the law of conservation of linear momentum,the total initial momentum must equal the total final momentum:$\vec{P}_i = \vec{P}_f$$m \vec{v}_1 + m \vec{v}_2 = (2m) \vec{V}$$m(v \hat{i}) + m(v \hat{j}) = 2m \vec{V}$$m v (\hat{i} + \hat{j}) = 2m \vec{V}$$\vec{V} = \frac{v}{2} (\hat{i} + \hat{j})$The magnitude of the final velocity is $V = \sqrt{(\frac{v}{2})^2 + (\frac{v}{2})^2} = \sqrt{\frac{v^2}{4} + \frac{v^2}{4}} = \sqrt{\frac{v^2}{2}} = \frac{v}{\sqrt{2}}$.The direction is along the vector $(\hat{i} + \hat{j})$,which is North-East.Thus,the new particle moves with a velocity of $\frac{v}{\sqrt{2}}$ in the North-East direction.

$A$ particle of mass $m$ moving eastward with a speed $v$ collides with another particle of the same mass moving northwards with the same speed $v$. The two particles coalesce on collision. The new particle of mass $2m$ will move with velocity:

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