Inelastic collision (finding e, don't understand concept)

Click For Summary

Homework Help Overview

The discussion revolves around the concept of inelastic collisions, specifically focusing on the coefficient of restitution (e) and its implications in calculations involving velocities before and after the collision. Participants are attempting to understand the relationships between the variables involved and the reasoning behind different approaches to the problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are examining different methods for calculating the coefficient of restitution and questioning the validity of their own and others' approaches. There is a focus on understanding the assumptions made regarding the velocities of the colliding objects and how these assumptions affect the calculations.

Discussion Status

There is ongoing debate about the correctness of various methods used to derive the coefficient of restitution. Some participants express confusion over the reasoning behind their teacher's approach, while others attempt to clarify the implications of using different signs for velocities. Multiple interpretations of the problem are being explored, with no clear consensus reached yet.

Contextual Notes

Participants are grappling with the implications of using different formulas and assumptions in the context of inelastic collisions. There is mention of specific values for the coefficient of restitution and the conditions under which they are derived, highlighting the complexity of the problem.

coconut62
Messages
161
Reaction score
1
I have problems with a(ii). (please view the first image.)

Now I have found that w = (13-8e)u/7 and v = (20e+13)u/7

What I did is like this:

(Using NEL)

v = 4ue
Where the LHS is actually v-0 because I assume Q = 0.

(20e+13)u/7 = 4ue

e = 33/28 >1 ∴ e cannot exist.

1. Is this way correct?

2. My teacher used another way. (please view second image.)

I don't understand the reason behind his workings.

Actually he like to use the formula like this :
V1- V2 = -e(U1-U2) ,even when V2 is faster than V1. Because he said in his brain, 2 always comes after 1, so he doesn't want to use V2-V1. I don't follow him, I follow the book, using speed of separation = fast - slow. But I still don't understand why he used V-W = -e(4u) to assume Q is brought to rest. (or did he even make that assumption at all?)

My interpretation is like this: Since he uses "slower-faster = negative e (speed of approach)"
Then in his workings, W should be assumed faster than V. And in the situation, both P and Q are moving to the left after collision, which makes their velocities negative. So if you want to assume Q is brought to rest, then W = 0. Since 0 is larger than negative, then his assumption is valid.

This is just simply a hypothesis. I only have a vague picture of the concept but I cannot fully grasp it. Someone please explain it to me.

3. The second part of a(ii), determining the least possible value of w. I just substituted e=1 into v-w = 4ue and got the answer. But after thinking about it, I suddenly don't understand why it should be e=1 which is substituted inside. I tried e=0.9, and the resulting w is smaller than the answer when e=1. So why is the least value obtained when e=1?
 

Attachments

  • 20130305_145127.jpg
    20130305_145127.jpg
    37.4 KB · Views: 453
  • image_1362463482838409.jpg
    image_1362463482838409.jpg
    23.7 KB · Views: 434
Physics news on Phys.org
coconut62 said:
(20e+13)u/7 = 4ue
e = 33/28 >1 ∴ e cannot exist.
You went a bit astray in there. You should have got e = 13/8. But your method is fine.
2. My teacher used another way. (please view second image.)
I'm afraid your teacher has blundered. He has the sign wrong on the RHS. His method should have given e = e.
he like to use the formula like this :
V1- V2 = -e(U1-U2) ,even when V2 is faster than V1.
That's fine. In this case, that gives v-w = -e(-u - 3u), so v - w = 4eu.
3. The second part of a(ii), determining the least possible value of w. I just substituted e=1 into v-w = 4ue and got the answer. But after thinking about it, I suddenly don't understand why it should be e=1 which is substituted inside.
w = (13-8e)u/7, so as e increases w decreases. Therefore the smallest w goes with the largest e, and the largest possible e is 1.
 
haruspex said:
I'm afraid your teacher has blundered. He has the sign wrong on the RHS. His method should have given e = e.

Why is it that when e=e, there is no solution? Isn't e equal to e? (err...)

haruspex said:
That's fine. In this case, that gives v-w = -e(-u - 3u), so v - w = 4eu.
I thought the right hand side should be -e (speed of approach) ? Why is it (-u-3u)? The particles are moving in opposite directions, which makes their speeds add up to become the total speed, no?
 
coconut62 said:
Why is it that when e=e, there is no solution? Isn't e equal to e? (err...)
Arriving at e = e does not mean there is no solution. What it means is that the method your teacher used was ineffective. This can happen quite often when manipulating equations. If you use the same equation twice you can find everything cancels out and you end up with 0 = 0.
I thought the right hand side should be -e (speed of approach) ? Why is it (-u-3u)? The particles are moving in opposite directions, which makes their speeds add up to become the total speed, no?
You must measure all the velocities in the same direction. The given u and v are in opposite directions, so you have to flip the sign of one of them.
 
haruspex said:
You must measure all the velocities in the same direction. The given u and v are in opposite directions, so you have to flip the sign of one of them.

This (-u-4u) thing only applies when using my teacher's version right? If I use W-V instead of V-W for speed of separation, then it would be just (u+4u)?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
Replies
10
Views
3K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K