Solving Momentum Transfer: 2 kg Steel Balls Colliding

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The discussion centers on the elastic collision of two 2 kg steel balls, where ball A travels west at 5 m/s and collides head-on with a stationary ball B. After the collision, ball A's velocity is 0 m/s, and ball B's velocity is 5 m/s, confirming the conservation of momentum and kinetic energy. Participants emphasize the importance of understanding both momentum and kinetic energy conservation principles in solving such problems. The initial and final momentum calculations validate the outcomes, with a total momentum of 10 kg·m/s before and after the collision.

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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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