Solving Momentum Transfer: 2 kg Steel Balls Colliding

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In an elastic collision involving two 2 kg steel balls, ball A traveling west at 5 m/s collides head-on with stationary ball B. Post-collision, ball A's velocity becomes 0 m/s, while ball B moves east at 5 m/s. The initial and final momentum remains consistent at 10 kg·m/s, confirming conservation of momentum. The discussion highlights the need for both momentum and kinetic energy conservation equations to solve for the velocities. Clarification is sought on how to approach the problem given the two variables involved.
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A 2 kg steel ball A traveling west at 5 m s–1 collides elastically head-on with a
stationary ball B also of mass 2 kg. Without doing any calculations, state the
velocities (including directions) of the two balls after collision.

Initial momentum = final momentum = 10

I got 10/4 since I assumed it was coupled but it says that V of a is 0 and V of b is 5.
Even so, there would be two variables; how would I solve this? Could anyone please calrify for me?

When I did it separately I got 5 = mv(a) + mv(b)
 
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Procrastinate said:
A 2 kg steel ball A traveling west at 5 m s–1 collides elastically head-on with a
stationary ball B also of mass 2 kg. Without doing any calculations, state the
velocities (including directions) of the two balls after collision.

Initial momentum = final momentum = 10

I got 10/4 since I assumed it was coupled but it says that V of a is 0 and V of b is 5.
Even so, there would be two variables; how would I solve this? Could anyone please calrify for me?

When I did it separately I got 5 = mv(a) + mv(b)

You can easily state the direction, but they didn't give too much info about a and b, and well they said elastically so kinetic energy is conserved. You'll get another equation there.
 
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