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

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(C) Let the mass of each particle be $m$. The initial velocity vectors of the particles $A, B, C, D$ directed towards the center are $\vec{v}_A, \vec{

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$\frac{v}{2}$

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(C) Let the mass of each particle be $m$. The initial velocity vectors of the particles $A, B, C, D$ directed towards the center are $\vec{v}_A, \vec{v}_B, \vec{v}_C, \vec{v}_D$. Since the particles are at the corners of a square and moving towards the center with speed $v$,their initial total momentum is $\vec{P}_i = m(\vec{v}_A + \vec{v}_B + \vec{v}_C + \vec{v}_D) = 0$ because the vectors cancel each other out.By the law of conservation of linear momentum,the final total momentum must also be zero: $\vec{P}_f = m(\vec{v}'_A + \vec{v}'_B + \vec{v}'_C + \vec{v}'_D) = 0$.Given:$1$. Particle $A$ comes to rest: $\vec{v}'_A = 0$.$2$. Particle $B$ traces its path back with speed $v$: $\vec{v}'_B = -\vec{v}_B$.$3$. Particles $C$ and $D$ move with equal speeds $v'$: $\vec{v}'_C = \vec{v}'_D$.Substituting these into the conservation equation:$0 + (-\vec{v}_B) + \vec{v}'_C + \vec{v}'_D = 0$Since $\vec{v}'_C = \vec{v}'_D$,we have $2\vec{v}'_C = \vec{v}_B$.Taking the magnitudes,$2v' = v$,which gives $v' = \frac{v}{2}$.Thus,the velocity of $C$ after the collision is $\frac{v}{2}$.

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