How does the center of mass affect collisions in theoretical mechanics?

[allistair]

There's something in my theoretical mechanics handbook that i don't quite understand. Say you have an observer in a labratory watching 2 particles, one of which is in a state of rest (m_2) and the other one (m1) is moving towards it.

the placevector of the center of mass of the system is
$$R = \frac{m_{1}.R_{10}+m_{2}.R_{20}}{m_{1}+m_{2}}$$
wich moves with a speed
$$V = \frac{m_{1}.v_{10}}{m_{1}+m_{2}} = \frac{p_{10}}{m_{1}+m_{2}}$$
v_10 and R_10 and R_20 seen by the labratory observer, say you have an observer at the centre of mass of the system, he observes the particles with speeds v_1 and v_2

for the observer at the centre of mass before the particles collide the impules p_1 and p_2 are the same but in oppsing directions. Now my handbook states that
$$p_{1}=p{2} = m_{2}.V = \frac{m_{2}.p_{10}}{m{1}+m{2}}$$
wich i don't fully understand

 

[arildno]

1.Compute the velocities of the particles relative to C.M.
2. Then figure out their momenta.

 

[allistair]

i worked it out like you said and i was able to get the answer, thanks alot, intuitivly the answer just felt a bit wrong.

 

[arildno]

i worked it out like you said and i was able to get the answer, thanks alot, intuitivly the answer just felt a bit wrong.

Well, then your next task is to improve your intuition as follows:

The total momentum of a system is MV, where M is the total mass, and V is a velocity determined so that MV equals the sum of the constituent particles' momenta.

That is, we regard momentum as an additive quantity, and the C.M velocity is just the average velocity with masses as weights.

But, if your system consists of two particles, 1 and 2, and we choose to regard it in the system where the C.M is at rest, that is V=0; then we must have:
$$0=p_{1}+p_{2}$$ since the systems total momentum (i.e, 0) must equal the sum of the constituent particles' momenta.

all right?