Help with a momentum exchange please!

1. Apr 12, 2009

Quant ummm?

Hi there,

I am dealing with an inelastic, collinear momentum exchange of the form:

m1v1+m2v2=m1u1+m2u2

where m1 & m2 are known.

v1 & v2 are both 0

u1 & u2 are both unknown, however

MOD(u1)+MOD(u2) is known (ie the sum of modulus of each speed, I don't know how to do straight brackets here.........)

which initial speed is regarded as + or - is irrelevant (to me).

I know that the exact speeds for u1 & u2 can be calculated, but I can't quite get me head around how (I'm more used to knowing one or the other, not their sum).

Although I can find the answer by gradually increasing one of the values on a spreadsheet, I'd like to see the actual solution. I'm guessing it can be solved either simultaneously or with a bit of calculus, but I'm not very good and working these things out.

Any help would be much appreciated.

p.s. I know this looks like homework, but its not. It really does though doesn't it. Real world though, honest.

2. Apr 12, 2009

mathman

Since v1 and v2 are both 0, you immediately have:
u1=-[m2/m1]u2

You the substitute into your expression for|u1| + |u2| to get |u1|

u1 = the result with a sign ambiguity.

3. Apr 12, 2009

haaj86

$$m_{1}u_{1}+m_{2}u_{2}=0$$ can be rearragned to $$\frac{u_{1}}{u_{2}}=-\frac{m_2}{m_1}$$and taken the modulus we get $$\frac{|u_{1}|}{|u_{2}|}=\frac{m_{2}}{m_{1}}$$
Now, Lets call $$|u_{1}|+|u_{2}|=x$$ where x is known, and if we divide by $$|u_{2}|$$ and rearrange we get $$\frac{|u_{1}|}{|u_{2}|}=\frac{x}{|u_{2}|}-1$$
Therefore $$\frac{x}{|u_{2}|}-1=\frac{m_{2}}{m_{1}}$$ and rearragning for $$u_{2}$$ we get $$|u_{2}|=\frac{m_{1}}{m_{1}+m_{2}}x$$ and similarly for $$|u_{1}|=\frac{m_{2}}{m_{1}+m_{2}}x$$

4. Apr 13, 2009

Quant ummm?

That's brilliant, thank you very much for your help.