# Basic collision help needed

1. Nov 21, 2008

### fluidistic

1. The problem statement, all variables and given/known data
Some days ago and specially yesterday I realized I have a strong weakness when it comes to elastic collision problems. Unfortunately I disregarded it and I couldn't finish an exercise in a test today.

2. Relevant equations
Linear momentum is conserved ($$\vec{P_{i}}=\vec{P_f}}$$) and so is the kinetic energy of the system. (Two bodies)

3. The attempt at a solution

Well in fact say you have 2 particles, $$m_1$$ and $$m_2$$. $$m_2$$ is initially at rest while $$m_1$$ has a speed of $$\frac{4m}{s}$$.
If $$m_1$$ hits $$m_2$$, what would be the speed of $$m_2$$?

I get that $$m_1v_1=m_1v_1'+m_2v_2'$$ [1]. One equation, two unknowns: $$v_1'$$ and $$v_2'$$.
Have I to work it out using the equation of the kinetic energy? $$\frac{m_1v_1^2}{2}=\frac{m_1v_1^2'}{2}+\frac{m_2v_2^2'}{2}$$. I solve for $$v_1'$$ and back substitute it in [1]. Is it the way to find out $$v_1'$$ and $$v_2'$$?
If yes then I thought about it during the test but I was too lazy to do it... (it doesn't count for my average but still, I should have done that).
Thanks for your confirmation or infirmation.

2. Nov 21, 2008

### iseidthat

3. Nov 21, 2008

### fluidistic

Thank you very much iseidthat!
That doesn't answer my question I think, but at least I've learned that $$v_1'=\left( \frac{m_1-m_2}{m_1+m_2}\right) v_1$$ and that $$v_2'$$ can be determinate by a given formula.
Ok, I'll try to answer my question by my own.