[TextBook] Question Mechanics

[mjlogan]

This question is from the M2 UK maths textbook. None of my set seem able to complete it nor can any of the teachers so far. If any would like to give it a go, please do.

Note: There is also the chance that the question was printed wrong and it will not work.

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A sphere of mass m1 moving with speed of u1 collides directly with a similar sphere of mass m2 moving with speed u2 in the same direction (u1 > u2). The coefficient of restitution between the two spheres is e. Show that the loss of kinetic energy E due to the collision satisfies the equation:

2(m1+m2)E=m1m2(u1-u2)^2(1-e^2)

Thanks

Mark

[A$$Bandit]

CLM: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ --> v₂ = (m₁u₁ + m₂u₂ - m₁v₁) / m₂

NIL: v₂ - v₁ = e(u₁ - u₂) --> v₂ = e(u₁ - u₂) + v₁

So:

e(u₁ - u₂) + v₁ = (m₁u₁ + m₂u₂ - m₁v₁) / m₂

Since nothing else can be eliminated, you can tell that the question is doable only with those variables given, i.e. you can't do it because you don't have another equation giving v₁ in terms of e, the initial velocities and the masses.

My teacher is top cheese in EdExcel A-Level Mechanics, Chief Examiner and the likes (and no, he wasn't involved in the scandal). He's missing like 3 lessons this week because of deciding on grade boundaries. I'll show him and see what he thinks about it, he writes a lot of the papers for that board at AS/A2 so if it's doable, he'll know. Also if it's infamously non-doable, he'll know that too. We'll see next monday evening!

[mjlogan]

Thank you A$$Bandit

[A$$Bandit]

He took a quick look at it and decided it can be done. Unfortunately for you and me, he decided I should be able to do it, and said he wasn't telling me how!

[mjlogan]

Hummm, ok thank you.

Does he actually know the solution? or is he just saying it is possible to do. As I have known a few people say it is possible then have tried it and failed.

Thanks anyway.

[A$$Bandit]

Well I got the impression he knows how to do it, but as usual he left it up to me to figure it out, which I haven't (primarily because I haven't really tried). So I'd keep trying!

[mjlogan]

I have tried an teachers have tried again, but still no one seems to be able to get it.

[Allegro]

Hello mjlogan,

The first thing I did was writing your equation with E on the lefthandside:

E=m_1*m_2(u_1-u_2)²(1-e²)/(2(m_1+m_2))

What I see now is :

The equation for the Kinetic Energy :½mv²,where v=u_1-u_2 ? ,

m_2/(m_1+m_2) and

(1-e²)

The problem is that I dont see yet the relationship between the 3 parts of the equation.
I have been looking at the following website , maybe you or someone else will find here the missing "link":

http://theory.uwinnipeg.ca/physics/mom/node7.html

Good luck!

[A$$Bandit]

Before you start, it's a proof which means you have to start with the information given in the question and arrive at the result, not the other way around.

[Allegro]

Starting with the given information is exactly what I intend.
However , first I'am looking to the relationship in the given equation and then using this relationship to derive the desired equation.

If you look to the examples of the web link that I gave ,you will see the difference between an elastic and an inelastic collission.

Here we have something between these extreme cases, that's where the coëfficiënt "e" comes in the picture.
So one way or another this coëfficiënt should be related to m_1 and m_2.

I think this schedule will , if you find the relations , provide you a method to prove the disired equation. If I have time this week I will certainly try to spend some more time to the question.

This is my idea , maybe someone else comes up with other idea's ,
the more the better I would think.

[TheAlchemist]

i've been thinking very seriously about this problem guys,
i recently came across this problem in my own text book:
2 spheres of mass m are traveling toward each other at speed u, find the loss in k.e. after the collision in terms of u, m and e.

this is what i came up with: E = mu^2(1-e^2)

applying the same techniques im sure you guys can work out the answer now,i'll post it tomorrow!

happy thinking!

[mjlogan]

you have the answer to my problem? waits to see.

[TheAlchemist]

hey guys,
very embarrased! ,
i still haven't managed to find the answer. ok thats it i said it.
if you do please do post( does it really have an answer?)

[mjlogan]

I have been told it is possible, but I have got no idea what it is.