Problem Concerning Inelastic collisions.

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Homework Help Overview

This discussion revolves around a problem concerning inelastic collisions, specifically focusing on the conservation of linear momentum and kinetic energy. The original poster expresses difficulty in analyzing the second situation of the problem, which involves two gliders colliding.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the variables involved in the equations presented, questioning the definitions and dimensional consistency of the terms used. There is an exploration of different reference frames and their implications on the analysis of the collision. The original poster seeks clarification on the conservation of kinetic energy and expresses frustration over the complexity of the problem.

Discussion Status

Several participants have provided feedback on the equations and approaches used by the original poster. There is an ongoing exploration of the relationship between the velocities in different reference frames, with some participants suggesting transformations to simplify the problem. While there is no explicit consensus, productive dialogue is occurring around the interpretations of the equations and the setup of the problem.

Contextual Notes

Participants note that the problem is part of an exam and that certain assumptions, such as the fraction of lost kinetic energy, are being discussed. The original poster has indicated that they are struggling with the analysis due to the complexity of the scenarios presented in the problem.

HermitOfThebes
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Homework Statement

http://www.pdx.edu/physics/sites/www.pdx.edu.physics/files/Nov-19th_version_Comprehensive%20Exam%20I_Nov-2006.pdf

Homework Equations

The Attempt at a Solution


This problem is driving me nuts. Its the problem number 1 in the afternoon session exam provided in the link above. It's about conservation of linear momentum. I have analysed the first situation and got the equation V2=V(1-F/2)
*given that (1-f)^1/2 = 1-f/2 as f << 1. This is given in the Feynman's exercise book.
I can't for the life of me analyse the second situation. Could someone please at least tell me what the equation for the conservation of Kinetic Energy be?
 
Last edited:
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HermitOfThebes said:
I have analysed the first situation and got with the equation V2=V(1-F/2)
Your equation is meaningless unless you state what the variables are. F is the applied force, but what are "V" and "V2"? The velocities of the respective masses? If so then it's dimensionally wrong, and why are you even considering the velocities? Please show your work you want to get help.

HermitOfThebes said:
I can't for the life of me analyse the second situation. Could someone please at least tell me what the equation for the conservation of Kinetic Energy be?
For problem 2, also try to show your work. When the block first transitions from the ground to the triangle there is an inelastic collision.
(Unless you meant part (b) of problem 1?)
 
Nathanael said:
Your equation is meaningless unless you state what the variables are. F is the applied force, but what are "V" and "V2"? The velocities of the respective masses? If so it's dimensionally wrong, and why are you even considering the velocities? Please show your work you want to get help.

For problem 2, also try to show your work. When the block first transitions from the ground to the triangle there is an inelastic collision.

Nathanael I am asking about the other problem, not this one. I am talking about problem 1 in mechanics in the afternoon exam, if you scroll down you will find it. It's about two gliders colliding.
F is defined as the fraction of lost K.E.. I would really appreciate it if you could help me with this problem as its killing me.
V2 is rebound velocity for both masses in Case 1, and V is the velocity by which they approach each other.
 
HermitOfThebes said:
Nathanael I am asking about the other problem, not this one. I am talking about problem 1 in mechanics in the afternoon exam, if you scroll down you will find it. It's about two gliders colliding.
F is defined as the fraction of lost K.E.. I would really appreciate it if you could help me with this problem as its killing me.
V2 is rebound velocity for both masses in Case 1, and V is the velocity by which they approach each other.
Okay. Your equation is good then. For the second case, if you switch to a certain reference frame then it is the same as case 1. So you can use your equation to calculate the speeds in that reference frame and then transform back to the original reference frame.
 
Nathanael said:
Okay. Your equation is good then. For the second case, if you switch to a certain reference frame then it is the same as case 1. So you can use your equation to calculate the speeds in that reference frame and then transform back to the original reference frame.
I did exactly that but I keep getting vf/2 not vf/4 :S
 
HermitOfThebes said:
I did exactly that but I keep getting vf/2 not vf/4 :S
I haven't done it but I'd like to see your work.

Edit:
Okay now I've done it and I got vf/2 as well.
 
Nathanael said:
I haven't done it but I'd like to see your work.

Edit:
Okay now I've done it and I got vf/2 as well.
I switched to a frame of reference where the speed of both boxes becomes v/2 instead of one having v and the other being at rest.
Now if I substitute into the equation(V-rebound=V-approach*(1-f/2)) I get V-rebound=V/2(1-f/2)
How do I revert back to the original frame of reference?
 
HermitOfThebes said:
I switched to a frame of reference where the speed of both boxes becomes v/2 instead of one having v and the other being at rest.
Now if I substitute into the equation(V-rebound=V-approach*(1-f/2)) I get V-rebound=V/2(1-f/2)
How do I revert back to the original frame of reference?
Well to get into that reference frame, you had to start moving +v/2 with respect to the original reference frame. So now to get back to the first frame you must move -v/2 relative to this second frame.

I solved the problem using their hint as well, and I still get vf/2

Edit: No I don't! I get vf/4 both times! Silly me I was using V2=V(1+f) :redface:
 
Nathanael said:
Well to get into that reference frame, you had to start moving +v/2 with respect to the original reference frame. So now to get back to the first frame you must move -v/2 relative to this second frame.

I solved the problem using their hint as well, and I still get vf/2
yea I did that too, still :S Even in Feynman's exercise book it says it's vf/4
 
  • #10
HermitOfThebes said:
yea I did that too, still :S Even in Feynman's exercise book it says it's vf/2.
No no I get vf/4 with both methods. I made a mistake of using V2=V(1-f) instead of V2=V(1-f/2) by accident.

You get this too, check your equation.

[edited to fix typo]
 
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  • #11
Nathanael said:
No no I get vf/4 with both methods. I made a mistake of using V2=V(1+f) instead of V2=V(1+f/2) by accident.

You get this too, check your equation.
Can I please see your solution?
 
  • #12
HermitOfThebes said:
Can I please see your solution?
You already told me this equation: Vrebound=V/2(1-f/2) ... You pretty much had it, but your equation has a small mistake: it is off by a factor of -1, because you see if you imagine which block we are interested in, it is the one which is traveling backwards in this new reference frame.

To get into this new reference frame we just subtracted v/2 from all the velocities right? The first block went from v to v/2 and the second block went from 0 to -v/2...
Well now to get back to the old reference frame we just add v/2 to all velocities. Doing so gives you vf/4
 
  • #13
Nathanael said:
You already told me this equation: Vrebound=V/2(1-f/2) ... You pretty much had it, but your equation has a small mistake: it is off by a factor of -1, because you see if you imagine which block we are interested in, it is the one which is traveling backwards in this new reference frame.

To get into this new reference frame we just subtracted v/2 from all the velocities right? The first block went from v to v/2 and the second block went from 0 to -v/2...
Well now to get back to the old reference frame we just add v/2 to all velocities. Doing so gives you vf/4
Oh. That bloody -1 :D thank you very much :D I wasted a lot of time trying all kinds of crazy things, should have paid more attention. Thanks :D
 
  • #14
No problem.

Ditto about the paying attention haha
 

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