2D elastic collision in CM Frame, velocities antiparallel?

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In a 2D elastic collision, the Center of Momentum (CM) velocities of the masses are anti-parallel due to the conservation of momentum and energy principles. After the collision, the total momentum in the CM frame remains zero, which necessitates that the velocities of the colliding masses point in opposite directions. This relationship ensures that the vector sum of their momenta cancels out, maintaining the system's overall momentum. Understanding this concept is crucial for analyzing the dynamics of collisions in physics. The discussion seeks a proof of this phenomenon, indicating a gap in accessible resources on the topic.
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Homework Statement


Why are the Center of Momentum velocities of masses after a 2D elastic collision anti-parallel? (as in the following diagram)
ZjhFlcz.png

Homework Equations


CM_Velocity = (∑m_i*v_i)/(∑m_i)

The Attempt at a Solution


This is not actually a problem I have to do. I am just looking for a proof of this fact. I have not been able to find one online, but have probably been looking in the wrong places!
 
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hsbhsb said:

Homework Statement


Why are the Center of Momentum velocities of masses after a 2D elastic collision anti-parallel? (as in the following diagram)
ZjhFlcz.png

Homework Equations


CM_Velocity = (∑m_i*v_i)/(∑m_i)

The Attempt at a Solution


This is not actually a problem I have to do. I am just looking for a proof of this fact. I have not been able to find one online, but have probably been looking in the wrong places!

What is the velocity of the CM in the centre-of-mass frame of reference? What does it mean for the colliding particles, how are their velocities related?
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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