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(A) Using the law of conservation of linear momentum,the initial momentum is $\vec{p}_i = m v \hat{i} + m v \hat{j}$.After the collision,the particles

Vedclass · Accepted Answer

$v/\sqrt{2}$ in North-East direction

Vedclass · Answer

(A) Using the law of conservation of linear momentum,the initial momentum is $\vec{p}_i = m v \hat{i} + m v \hat{j}$.After the collision,the particles stick together to form a mass of $2m$ moving with velocity $\vec{v}'$.By conservation of momentum: $m v \hat{i} + m v \hat{j} = (2m) \vec{v}'$.Dividing by $2m$,we get $\vec{v}' = \frac{v}{2} \hat{i} + \frac{v}{2} \hat{j}$.The magnitude of the 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 resultant of $\hat{i}$ and $\hat{j}$,which is the North-East direction.

$A$ particle of mass $m$ moves with velocity $v$ towards the East. It collides with another particle of the same mass and same speed moving towards the North and sticks to it. What will be the velocity of the combined particles?

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