How is V1 deduced to be zero in the last step here?

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Homework Help Overview

The discussion revolves around a perfectly elastic collision involving two billiard balls, one moving and the other at rest. The original poster seeks to understand the reasoning behind deducing that the velocity of one ball becomes zero after the collision.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of momentum conservation and energy conservation in elastic collisions. They discuss the mathematical relationships derived from the equations governing the collision, questioning how to logically deduce which ball has zero velocity.

Discussion Status

Participants are actively engaging with the problem, raising questions about the mathematical steps involved and the physical implications of the collision. Some guidance has been offered regarding the expansion of equations and the interpretation of results, but there is no explicit consensus on the reasoning for which ball has zero velocity.

Contextual Notes

There is an ongoing discussion about the assumptions involved in the collision scenario, including the implications of both momentum and energy conservation. Participants also consider the possibility of the balls passing through each other, which raises questions about the validity of physical laws in this context.

califauna
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Homework Statement


A billiard ball moving at 2 m/s collides (0 degrees) with another of the same weight and at rest, in a perfectly elastic collision. Demonstrate with equations why the balls trade velocities.

Homework Equations

The Attempt at a Solution



In the provided explanation, I can understand what the parallel equations they set up represent, and how they reduce them by cancelling, but I can't see how the final logical step is made, where they deduce that V1 = 0,

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What does the lower (momentum) equation tell you about v2?
 
Plug v=v1+v2 into v2=v12+v22 and you will get 2v1v2 = 0. So either v1 =0 or v2 = 0
 
haruspex said:
What does the lower (momentum) equation tell you about v2?

That v^2 = (v1+v2)^2
 
FactChecker said:
Plug v=v1+v2 into v2=v12+v22 and you will get 2v1v2 = 0. So either v1 =0 or v2 = 0

When I do (v1+v2)2=v12+v22, after taking the root on both sides I get get v1+v2=v1+v2. What am I doing wrong?
 
califauna said:
When I do (v1+v2)2=v12+v22, after taking the root on both sides I get get v1+v2=v1+v2. What am I doing wrong?
You cannot take the square root of the right hand side like that. The square root of a sum is not the same as the sum of the square roots.
Instead, expand the left hand side.
 
haruspex said:
You cannot take the square root of the right hand side like that. The square root of a sum is not the same as the sum of the square roots.
Instead, expand the left hand side.
Ah yes. Ok, I am left with 2V1v2 = 0.

Obviously the struck ball is going to be moving so it cannot be the one which ends up with a velocity of zero, but is there any other logical or mathematical method of deducting which of the two balls will be the one which will have zero velocity? Something which the exam question might require demonstation of in order to get the full marks for the question?
 
califauna said:
Ah yes. Ok, I am left with 2V1v2 = 0.

Obviously the struck ball is going to be moving so it cannot be the one which ends up with a velocity of zero, but is there any other logical or mathematical method of deducting which of the two balls will be the one which will have zero velocity? Something which the exam question might require demonstation of in order to get the full marks for the question?
Both momentum and energy would be conserved if the first ball manages to pass straight through the stationary ball, but some other law may violated, do you think?
 
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haruspex said:
Both momentum and energy would be conserved if the first ball manages to pass straight through the stationary ball, but some other law may violated, do you think?

Not sure if the glitch in the matrix type clue about the balls passing through each other is a hint at this particlar law, but, Newtons third law regarding equal and opposite force?
 
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califauna said:
Not sure if the glitch in the matrix type clue about the balls passing through each other is a hint at this particlar law, but, Newtons third law regarding equal and opposite force?
Yes, you could look at it that way, but I was thnking more in terms of everyday experience that solid objects cannot pass through each other. Quantum effects aside.
 
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