Elastic And Inelastic Collision Questions

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Homework Help Overview

The discussion revolves around two problems related to elastic and inelastic collisions, specifically focusing on the coefficient of restitution and the dynamics of colliding objects. The first problem involves a steel ball dropped onto a steel plate and seeks to determine the rebound height. The second problem examines the mass ratio of two particles involved in a head-on elastic collision.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the definitions of variables related to the collisions, such as initial and final velocities. There is an attempt to clarify the meaning of the coefficient of restitution and its application to the problems. Questions arise about the assumptions made regarding the motion of the steel plate and the interpretation of the velocities involved.

Discussion Status

The discussion is ongoing, with participants providing insights and clarifications on the definitions and relationships between the variables. Some participants express uncertainty about their understanding of the problems, while others offer guidance on how to approach the calculations. There is no explicit consensus yet, as multiple interpretations are being explored.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is also a noted typo in the second problem regarding the recoiling speed, which may affect the interpretation of the problem.

spikefreeman
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Homework Statement


Problem 1: The coefficient of restitution for steel colliding with steel is 0.95. If steel is dropped from a height ho above steel plate, to what height will the ball rebound.

Problem 2: in an elastic head on collision with a stationary target particle, a moving particle recoils of its incident speed. What is the ration of the particles masses (m/M)

Homework Equations


1. v2-v1=-e(v2o-v1o)
v=sqrt(2gh)

2. mv= mv + MV

The Attempt at a Solution


1. -e= (v2-v1)/(v2o-v1o)
-.95=(v2-v1)/(v2o-v1o)
From there I honestly have no clue what to do.

2. mv = m(-1/3V) +MV
4/3mv = MV
The book answer is 1/2, and I'm working with thirds? Should I be using a KE equation?

Any help is appreciated.
 
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Part 1.
What do v1, v2, v10 and v20 represent? Can you find expressions for them?

Part 2.
What is (-1/3 V) ?
 
v1o and v2o are the speeds before the collision, v1o at the top and v2o at the bottom. v1 is speed after collision and v2 is speed at max height after collision. So v1o and v2 are zero, and v2o is sqrt(2gh). Is any of that right?

I'm not really sure what you mean by what is -1/3V. Its a negative velocity.
 
spikefreeman said:
v1o and v2o are the speeds before the collision, v1o at the top and v2o at the bottom. v1 is speed after collision and v2 is speed at max height after collision. So v1o and v2 are zero, and v2o is sqrt(2gh). Is any of that right?
The speeds of what before and after the collision? In your particular problem, what is object 1 and what is object 2?

I'm not really sure what you mean by what is -1/3V. Its a negative velocity.
It is a negative velocity representing the velocity of what object? Why did you pick 1/3 as opposed to 1/5 or 1/7?
 
Object 1 is the ball and object 2 is the steel plate. But I have speeds all for object one.

O v1o
l
l
l
l
V v2o
------

The o at the top is the ball and the l and v are an arrow, then the dashes are the steel plate. Then the V1 and V2 are the coresponding speeds except after the collision on the way back up. That is how I understood the problem

For the second problem, the 1/3 is given as the recoiling speed of the moving particle.
 
spikefreeman said:
Object 1 is the ball and object 2 is the steel plate. But I have speeds all for object one.

O v1o
l
l
l
l
V v2o
------

The o at the top is the ball and the l and v are an arrow, then the dashes are the steel plate. Then the V1 and V2 are the coresponding speeds except after the collision on the way back up. That is how I understood the problem
You understood the problem correctly. I believe that in this problem you are to assume that the steel plate is (and remains) at rest on the floor. What does this mean for v20 and v2? Also, can you find an expression for v10?

For the second problem, the 1/3 is given as the recoiling speed of the moving particle.
There is nothing in the statement of the problem that indicates that. Can you post the statement exactly as it is given?
 
v1o = 0 right? because its dropped from rest.
That means that v2o = sqrt(2gh) and that v2 is 0 because it is the maximum of the bounce.

Problem 2, that was a typo. It recoils at 1/3 of the incident speed. Sorry about that.
 
spikefreeman said:
v1o = 0 right? because its dropped from rest.
That means that v2o = sqrt(2gh) and that v2 is 0 because it is the maximum of the bounce.
No, no, no. As I said, the steel plate is and remains at rest on the floor. If object 2 is the steel plate then what should v20 and v2 be?

Also, v10 is the speed of the ball just before it hits the floor and v1 is its speed just after the collision. If you know v1, you can find how high the ball will rise.
Problem 2, that was a typo. It recoils at 1/3 of the incident speed. Sorry about that.
Look in your textbook. It must have an expression giving the final velocities of two colliding masses in terms of their initial masses when the collision is elastic. Use that expression.
 
ok that makes more sense. So v2 and v2o are zero. so v1o = Sqrt(2gh) and v1 is solved using mv=mv?
 
  • #10
spikefreeman said:
ok that makes more sense. So v2 and v2o are zero. so v1o = Sqrt(2gh) and v1 is solved using mv=mv?
No. You find v1 using the given coefficient of restitution.
 
  • #11
kk, so i just v1o in set the v2 as zeroes and i should be able to find v1. Is that it?

For part two I will check in the book.,

Thanks for you help.
 
  • #12
spikefreeman said:
kk, so i just v1o in set the v2 as zeroes and i should be able to find v1. Is that it?
I am not sure what you mean by this so I cannot say that it is or isn't OK.
 
  • #13
I meant to say, I plug v1 in as sqrt(2gh) and set the v2 as zeroes...
 
  • #14
spikefreeman said:
I meant to say, I plug v1 in as sqrt(2gh) and set the v2 as zeroes...
v10 = sqrt(2gh), not v1.
 
  • #15
right, and then solve for v1, and then use mgh. Is that it?
 
  • #16
That's it.
 
  • #17
Great, thanks a lot, that was very helpful.
 

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