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(B) According to the law of conservation of linear momentum,the total momentum before collision is equal to the total momentum after collision.Let the

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

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(B) According to the law of conservation of linear momentum,the total momentum before collision is equal to the total momentum after collision.Let the direction towards the west be along the negative $x$-axis $(-\hat{i})$ and the direction towards the south be along the negative $y$-axis $(-\hat{j})$.The initial momentum of the first particle is $\vec{p}_1 = m v(-\hat{i}) = -m v \hat{i}$.The initial momentum of the second particle is $\vec{p}_2 = m v(-\hat{j}) = -m v \hat{j}$.The total initial momentum is $\vec{P}_i = \vec{p}_1 + \vec{p}_2 = -m v \hat{i} - m v \hat{j}$.After the collision,the particles stick together to form a single particle of mass $2m$ moving with velocity $\vec{V}$.The final momentum is $\vec{P}_f = (2m) \vec{V}$.By conservation of momentum,$\vec{P}_i = \vec{P}_f$,so $(2m) \vec{V} = -m v \hat{i} - m v \hat{j}$.Dividing by $2m$,we get $\vec{V} = -\frac{v}{2} \hat{i} - \frac{v}{2} \hat{j}$.The speed is the magnitude of the velocity vector: $|\vec{V}| = \sqrt{(-\frac{v}{2})^2 + (-\frac{v}{2})^2} = \sqrt{\frac{v^2}{4} + \frac{v^2}{4}} = \sqrt{\frac{v^2}{2}} = \frac{v}{\sqrt{2}}$.

$A$ particle of mass $m$ moving towards west with speed $v$ collides with another particle of mass $m$ moving towards south. If the two particles stick to each other,the speed of the new particle of mass $2m$ will be ..........

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