# Homework Help: Elastic collision problem involving nuclei

1. Nov 14, 2012

### Hoodoo

1. The problem statement, all variables and given/known data
Two nuclei make a head-on elastic collision. One nucleus (mass m) is initially stationary. The other nucleus has an initial velocity (v) and a final velocity of (-v/5). What is the mass of this nucleus?

2. Relevant equations
conservation of momentum (m1v1+m2v2=m1v1'+m2v2')
conservation of energy (1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2 + 1/2m2v2'^2)

3. The attempt at a solution
I've substituted in the knowns and came up with (m2v = mv1' + m2*-v/5)
I isolated for v1' though I'm not really sure why and came up with v1' = (6m2v1)/5m
This is probably totally wrong and I'm not sure why I even did it.
The examples given to me for elastic collision all involve a bunch of substitutions and isolated for variables without any explanation. This question has too many variables for me to keep it straight in my head. I'd love a thorough explanation if anyone has the time. Thanks.

2. Nov 14, 2012

### tiny-tim

Welcome to PF!

Hi Hoodoo! Welcome to PF!

(try using the X2 and X2 buttons just above the Reply box
ok, you don't know v1' (and you don't want to),

so eliminate v1' from those 2 equations to make a single equation without v1' …

show us what you get

3. Nov 16, 2012

### Hoodoo

I got:

m = (6m2v)/5)

Now I'm supposed to substitute into the equation for conservation of energy..

I end up at the end with:

0 = v((6m2)/5) - ((24m2v)/25))

Therefore, v = 0 or v = 5/4

Sub back into original equation:

m = (6m2v)/5
m = (6m25)/5*4
m2 = 2m/3

Now, my main question is.. how was I supposed to know to NOT include v1'? My instructors seem to have left this little leap out of the written materials and lectures. It's not mentioned at all in the question.. is that enough of a reason or am I supposed to know this due to the laws?

Last edited: Nov 16, 2012
4. Nov 16, 2012

### haruspex

That's dimensionally wrong. Typo in the post?
I get the ratio the other way around. I think I'm right because in an elastic collision with a stationary mass, the first mass bounces backwards if and only if it is the lighter of the two. Check the signs in your working.
It's not that you don't include it. Whenever you have multiple equations with shared variables, you'll be wanting to use some of the them to eliminate variables from the others. A question is, which to eliminate? That is directed by knowing which you want in the final equation. In this case, v1' is a good candidate for elimination.

5. Nov 16, 2012

### Hoodoo

No typo. I get the answer that is given in the text doing what I did. I'm looking for m2. If you get it the other way around, you've already solved the question, haven't you?

0+m2v = m + m2(-v/5)
m = m2v + m2v/5
m = 6m2v/5

I removed v1' entirely instead of eliminating it because I misunderstood the instructions but it worked out in the end.

I still don't know how to KNOW what I'm supposed to eliminate and solve for and this doesn't help.

6. Nov 16, 2012

### haruspex

Dimensionally wrong. m is a mass, the other terms are momenta. The equation is
0+m2v = mv1' + m2(-v/5)
My mistake there - I misread which mass had mass m. I now agree with your answer, if not with the details of how you got there.
Is it possible you don't know what is meant by 'eliminating a variable'?
Given m2v = mv1' + m2(-v/5)
and m2v2 = mv1'2 + m2(v/5)2, you know you want an equation involving m2 and m, and you don't mind if it involves v. m and v were given you, so could validly turn up in the answer. But you certainly know you don't want v1'. So that's the one to eliminate. To do that, you rearrange one equation in the form v1' = something. The first equation is the more convenient since that won't result in a square root:
v1' = m2v/m - m2(-v/5)/m = m2(6v/5)/m
Now use that to substitute for v1' in the other equation:
m2v2 = m(m2(6v/5)/m)2 + m2(v/5)2
A factor m2v2 cancels:
1 = (m2(36/25)/m) + 1/25
m2=2m/3
Is that clearer?

7. Nov 16, 2012

### Hoodoo

That is a big mistake.. yes I see what you're saying. Completely removing v1' was totally incorrect but I lucked into the answer in the end.

So I should originally isolate the variable that is both unknown and unwanted in the final solution? Then substitute that into the other equation and solve for what I REALLY wanted. That 'eliminates' it?

I find that people that 'get' physics sometimes have a hard time relating simple concepts because they are immediately apparent to them and they have a hard time putting themselves into the shoes of someone that doesn't just 'get it'. Half of my struggle in this subject is explaining what I don't understand to people who find answers and methods simply apparent.

Last edited: Nov 16, 2012
8. Nov 16, 2012

### haruspex

Exactly. This is what you do in solving simultaneous equations, and it's the same process as diagonalising a matrix.