How does ball A come to rest and Ball B remain stationary?

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In the collision involving three identical steel balls A, B, and C, Ball A comes to rest due to the forces of gravity and friction acting on it during the collision. Ball B remains stationary because it does not receive enough momentum transfer from Ball A, which is consistent with the principles of an elastic collision where momentum and kinetic energy are conserved. The conservation equations indicate that the total kinetic energy and momentum before and after the collision remain constant. The discussion highlights the importance of understanding the forces at play and the implications of friction on energy conservation. Ultimately, the interaction between the balls illustrates key concepts in physics regarding motion and collision dynamics.
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Homework Statement


In a collision between three identical steel balls A B C

A comes to rest and B remains stationary, while C rolls off.

In terms of the Forces acting, explain how Ball A came to rest and why Ball B remained stationary.

Homework Equations

The Attempt at a Solution


Well what I know here is that Ball A experiences the downward pull of gravity and also friction upon collision? Maybe that is why it stops.

I know that momentum is conseved and transfers through ball B but I can't think of the ''Forces acting here''

I do have the general idea of what's going on and that this collision is a good representation of an elastic collision where momentum and kinetic energy are conserved.

How can I represent this answer in a concise way without getting too complicated? Just the fundamental reasons
 
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Give ball A some velocity and mass.
Ball B and ball C have the same mass but zero velocity.
Assume an elastic collision.
Conserving kinetic energy gives something like:
##\left(v^2_A m_A + v^2_Bm_B+v^2_Cm_C\right)_{pre}=\left(v^2_A m_A + v^2_Bm_B+v^2_Cm_C\right)_{post} ##
Since all the balls are the same weight, you can factor them out of the equation.
##\left(v^2_A + v^2_B+v^2_C\right)_{pre}=\left(v^2_A + v^2_B+v^2_C\right)_{post} ##
Also, you can use conservation of momentum to get:
##\left(v_A + v_B+v_C\right)_{pre}=\left(v_A + v_B+v_C\right)_{post} ##
It's not very clean to solve all three together, so imagine a tiny gap between balls B and C...work out the relation from A to B (ignoring ball C), then the new B to C.
https://en.wikipedia.org/wiki/Elastic_collision
 
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Richie Smash said:
also friction upon collision? Maybe that is why it stops.
Richie Smash said:
kinetic energy are conserved.
You can't have it both ways. If friction makes it stop then KE will not be conserved.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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