Calculating Impulse Ratio for Colliding Molecules A and B

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Homework Help Overview

The discussion revolves around calculating the impulse experienced by two identical molecules, A and B, during their collision with a vertical wall. The problem involves analyzing the momentum changes of both molecules, which approach the wall at an angle of 20 degrees.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of impulse for both molecules, questioning the differences in momentum changes due to one molecule rebounding and the other sticking to the wall. There are inquiries about the velocity components before and after the collision, particularly for molecule B.

Discussion Status

The discussion is active, with participants raising questions about the definitions of impulse and momentum in the context of the collision. Some guidance has been offered regarding the velocity components of the molecules, but there is still uncertainty about the correct interpretation of the impulse experienced by molecule B.

Contextual Notes

Participants are grappling with the implications of the collision dynamics, particularly regarding the assumptions about velocity components and the nature of the collision (elastic vs. inelastic) for the two molecules.

Fuzzykatecake
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Two identical molecules A and B each of mass M travel with the same velocity towards a vertical wall at an angel 20 degrees with respect to the normal to the wall. A rebounds from the wall with the same speed while B sticks to the wall.

Determine the ratio of the implus experience by A to that by B during collision.

I know the inpluse experienced by A is
final momentum - initial momentum=
Mvcos20-(-mvcos20)

But for B, why is the impluse mv? And not 0- mvcos20
 
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What is the velocity of B before the collision? What is the velocity after? As a result, what is the difference in B's momentum before and after?
 
Orodruin said:
What is the velocity of B before the collision? What is the velocity after? As a result, what is the difference in B's momentum before and after?
Isn't the velocity of B before collision vcos20. And velocity after is 0 since is stick to wall. Therefore the momentum is 0-mvcos20.
But the correct answer is mv which I don't understand why
 
Fuzzykatecake said:
Isn't the velocity of B before collision vcos20.
No, that is the velocity component orthogonal to the wall.
 
Orodruin said:
No, that is the velocity component orthogonal to the wall.
But why for A is vcos20, what's the difference?
 
Fuzzykatecake said:
But why for A is vcos20, what's the difference?
A does not stick to the wall. The momentum component of A orthogonal to the wall is simply reflected while that perpendicular to it is unchanged.
 

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