Conservation of momentum question

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Homework Help Overview

The discussion revolves around a conservation of momentum problem involving a ball colliding with a wall. Participants are exploring the implications of momentum conservation in the context of external forces and the behavior of the system post-collision.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum in a system that includes external forces, such as a wall and the Earth. There are questions about the implications of these forces on the momentum of the ball and the system as a whole.

Discussion Status

Some participants have provided insights regarding the nature of momentum as a vector and the need to consider changes in direction. There is an acknowledgment of the complexity introduced by external forces, but no consensus has been reached on the final interpretation of the problem.

Contextual Notes

Participants are considering the constraints of the problem, particularly the assumption that the wall is stationary and the implications of including the Earth in the system. There is also mention of the difficulty in measuring the effects of the wall's force on the overall momentum.

David112234
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Homework Statement


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Homework Equations


m1v1+m2v2=m1v1+m1v1 after

The Attempt at a Solution


my answer = -1m1v1
because the wall is not moving so its velocity is 0
therefore
m1v1 = m1-v1 after
I put -v because it is the same just bounces in the opposite X direction
I found somewhere online the answer should be 2 * mv * sin(θ)
I don't understand why
 
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David112234 said:
because the wall is not moving so its velocity is 0
therefore
m1v1 = m1-v1 after
Conservation of momentum only applies to a closed system, i.e. no outside forces. The wall exerts a force. If you include the wall, the Earth exerts a force on the wall. If you include the Earth then in fact it will acquire a velocity from the impact, but so small you could not possibly measure it.
Just consider the change in momentum of the ball (as a vector).
 
haruspex said:
Conservation of momentum only applies to a closed system, i.e. no outside forces. The wall exerts a force. If you include the wall, the Earth exerts a force on the wall. If you include the Earth then in fact it will acquire a velocity from the impact, but so small you could not possibly measure it.
Just consider the change in momentum of the ball (as a vector).
It is the same as before only its x component changes, first it goes right, then it goes left
 
David112234 said:
It is the same as before only its x component changes, first it goes right, then it goes left
The magnitude is the same as before, but momentum and velocity are vectors. What is the net change in velocity?
 
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haruspex said:
The magnitude is the same as before, but momentum and velocity are vectors. What is the net change in velocity?
Oh so I do vector addition
I did the math out and got the same answer, thank you
 

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