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

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(D) The momentum of the particle moving towards the east is $\vec{p_1} = mv \hat{i}$.The momentum of the particle moving towards the north is $\vec{p_

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The momentum of the particle moving towards the east is $\vec{p_1} = mv \hat{i}$.The momentum of the particle moving towards the north is $\vec{p_2} = mv \hat{j}$.Let the velocity of the combined mass $2m$ after collision be $\vec{V'} = v_x \hat{i} + v_y \hat{j}$.According to the law of conservation of linear momentum,$\vec{p_1} + \vec{p_2} = \vec{p_{final}}$.$mv \hat{i} + mv \hat{j} = (2m) \vec{V'}$.Dividing by $2m$,we get $\vec{V'} = \frac{v}{2} \hat{i} + \frac{v}{2} \hat{j}$.The magnitude of the resultant velocity is $V' = \sqrt{(\frac{v}{2})^2 + (\frac{v}{2})^2} = \sqrt{\frac{v^2}{4} + \frac{v^2}{4}} = \sqrt{\frac{v^2}{2}} = \frac{v}{\sqrt{2}}$.

$A$ particle of mass $m$ moving eastward with a speed $v$ collides with another particle of same mass moving northward with same speed $v$. The two particles coalesce after collision. The new particle of mass $2m$ will move in north-east direction with a speed (in $m/s$) of:

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