2-D Momentum Problem -- Elastic collision of two spheres

AI Thread Summary
The discussion focuses on the elastic collision of two spheres, specifically addressing the assumptions required to analyze the momentum of the spheres post-collision. It emphasizes that for ball A to recoil entirely in the horizontal plane, it must not acquire any vertical momentum during the collision. Participants clarify that the spheres are considered "smooth," implying no friction, which simplifies the analysis but may not reflect real-world scenarios. The conversation also touches on the implications of the collision angle and the forces involved, suggesting that a deeper understanding requires experimental analysis. Ultimately, the problem hinges on the assumption that the momentum component normal to the center line remains unchanged.
Jimmy87
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Homework Statement
Determining velocity after a 2-D impact.
Relevant Equations
p = mv
Hi,

Here is the problem
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What is required to answer this question is two assumptions. Firstly, the component of the momentum normal to the centre line is the same before and after. Therefore, secondly, A must recoil entirely in the horizontal plane. This is the only way to answer this question and I can get the answer (0.34 m/s which I know to be correct). How do we know after this collision A is confined to the horizontal plane? Thanks.
 
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Jimmy87 said:
How do we know after this collision A is confined to the horizontal plane? Thanks
We don't. Snooker balls sometimes jump up off the table after a collision.

You just have to assume that doesn't happen in this case.
 
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PeroK said:
We don't. Snooker balls sometimes jump up off the table after a collision.

You just have to assume that doesn't happen in this case.
Ah sorry I wasn’t clear. The only way to answer this question is if ball A recoils along the centre line. How do we know it doesn’t have any vertical component after the collision (vertical in terms of the plane of the surface).
 
Jimmy87 said:
The only way to answer this question is if ball A recoils along the centre line. How do we know it doesn’t have any vertical component after the collision (vertical in terms of the plane of the surface).
The balls are "smooth". What is the direction of the force that B exerts on A?
 
Jimmy87 said:
How do we know it doesn’t have any vertical component after the collision (vertical in terms of the plane of the surface).
A better term is lateral or tangential, rather than vertical.
 
Jimmy87 said:
Ah sorry I wasn’t clear. The only way to answer this question is if ball A recoils along the centre line. How do we know it doesn’t have any vertical component after the collision (vertical in terms of the plane of the surface).
To acquire a vertical (tangential) component of momentum, a vertical force is needed during the collision. What entity could be the origin of that force? See post #4 by @TSny.
 
kuruman said:
To acquire a vertical (tangential) component of momentum, a vertical force is needed during the collision. What entity could be the origin of that force? See post #4 by @TSny.
Yeh I’m still struggling to visualise it. Ball B changes direction by moving through an angle of 90 degrees yet ball doesn’t changes it angle at all.

If the velocity is at an angle of 30 degrees will there not be a component normal to the line imparted on ball A?
 
Ball A can only exert a force perpendicular to its surface which is the radial direction from the center of A to the point of the collision. This means that the velocity of ball B can change only in that radial direction. Draw a vector diagram of the momentum of ball B before and after the collision showing this.
 
Jimmy87 said:
Yeh I’m still struggling to visualise it. Ball B changes direction by moving through an angle of 90 degrees yet ball doesn’t changes it angle at all.

If the velocity is at an angle of 30 degrees will there not be a component normal to the line imparted on ball A?
It says "smooth" spheres in the question, which is introductory physics code for "no friction". In reality, there would be some friction between the surfaces and some exchange of lateral momentum. A super slow-motion film of the collision of two snooker balls would be interesting from this perspective. To see how much (if any!) lateral momentum is imparted to the object ball.
 
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kuruman said:
Ball A can only exert a force perpendicular to its surface which is the radial direction from the center of A to the point of the collision. This means that the velocity of ball B can change only in that radial direction. Draw a vector diagram of the momentum of ball B before and after the collision showing this.
Thanks I think I can visualise it now. So it’s because they collide exactly edge on with no friction. I guess if ball B collided with some of its radius within ball A’s radius it would impart a force perpendicular to the centre line into A? Is that right? Thanks
 
  • #11
Jimmy87 said:
Thanks I think I can visualise it now. So it’s because they collide exactly edge on with no friction. I guess if ball B collided with some of its radius within ball A’s radius it would impart a force perpendicular to the centre line into A? Is that right? Thanks
This problem makes the simplifying assumption that the spheres have a frictionless contact. If you want to take the next step, I'd look online for an experimental analysis of these sorts of collisions. Introductory physics is essential in teaching you the basics and getting experience with the mathematics, but it only takes you so far in terms of understanding the real world.
 
  • #12
Jimmy87 said:

What is required to answer this question is two assumptions. Firstly, the component of the momentum normal to the centre line is the same before and after. Therefore, secondly, A must recoil entirely in the horizontal plane. This is the only way to answer this question and I can get the answer (0.34 m/s which I know to be correct). How do we know after this collision A is confined to the horizontal plane? Thanks.
What horizontal plane are you referring to?
The plane on which the balls are sliding or rolling on, or the one perpendicular to the shown picture and containing the trajectory of A?
 
  • #13
Jimmy87 said:
Homework Statement:: Determining velocity after a 2-D impact.
Relevant Equations:: p = mv

0.34 m/s which I know to be correct
Should be 0.6 tan(30), which is nearer to 0.35, no?
 
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