Four particles A, B, C and D of equal mass m | vedclass

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(D) Let the center of the square be the origin $(0,0)$. Let the diagonals lie along the $x$ and $y$ axes. The initial velocities are $\vec{v}_A = v(\f

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$v$

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(D) Let the center of the square be the origin $(0,0)$. Let the diagonals lie along the $x$ and $y$ axes. The initial velocities are $\vec{v}_A = v(\frac{\hat{i} + \hat{j}}{\sqrt{2}})$,$\vec{v}_B = v(\frac{-\hat{i} + \hat{j}}{\sqrt{2}})$,$\vec{v}_C = v(\frac{-\hat{i} - \hat{j}}{\sqrt{2}})$,and $\vec{v}_D = v(\frac{\hat{i} - \hat{j}}{\sqrt{2}})$.The total initial momentum is $\vec{P}_{initial} = m(\vec{v}_A + \vec{v}_B + \vec{v}_C + \vec{v}_D) = 0$.After the collision,$\vec{v}'_A = 0$ and $\vec{v}'_B = -\vec{v}_B = v(\frac{\hat{i} - \hat{j}}{\sqrt{2}})$.Let $\vec{v}'_C$ and $\vec{v}'_D$ be the final velocities of $C$ and $D$. Since they move with equal speed $v'$ and by symmetry,they must move in a direction that balances the momentum of $B$.By conservation of linear momentum: $m(0) + m\vec{v}'_B + m\vec{v}'_C + m\vec{v}'_D = 0$.$\vec{v}'_C + \vec{v}'_D = -\vec{v}'_B = v(\frac{-\hat{i} + \hat{j}}{\sqrt{2}})$.Since $|\vec{v}'_C| = |\vec{v}'_D| = v'$,and by symmetry,the resultant must be along the diagonal $BD$. The magnitude of the resultant is $2v' \cos(	heta)$,where $	heta$ is the angle with the diagonal. Solving for the components,we find $v' = v$.

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