# Help with a momentum exchange please!

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.

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.

$$m_{1}u_{1}+m_{2}u_{2}=0$$ 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 $$|u_{1}|=\frac{m_{2}}{m_{1}+m_{2}}x$$

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