# Mass frame.

1. Oct 24, 2007

### azatkgz

1. The problem statement, all variables and given/known data

Nucleus A,which has rest mass $$m_A$$,collides with nucleus B,which has rest mass $$m_B$$.In the laboratory frame,nucleus A has energy $$E>m_Ac^2$$
abd nucleus B is stationery.Find the total energy of the system in center of mass frame in terms of $$m_A,m_B and E$$.

3. The attempt at a solution

What's the center of mass frame?Is it a frame of nucleus A?

2. Oct 24, 2007

### robphy

Consider the frame in which the total spatial momentum is zero.

3. Oct 24, 2007

### azatkgz

Is it a center of two nuclei?

4. Oct 24, 2007

### azatkgz

Then check my solution.
Velocity of A in labframe is
$$v=c\sqrt{1-(\frac{m_Ac^2}{E-m_Ac^2})^2}$$
velocity of center of mass
$$v_c=\frac{m_Av}{m_A+m_B}$$

In frame which moves with velocity v_c,nucleus A has velocity

$$v'_A=\frac{v-v_c}{1-\frac{vv_c}{c^2}}$$.And nucleus B has velocity $$v'_B=v_c$$.So the total energy in this frame is

$$E'=\gamma(v'_A)m_Ac^2+\gamma(v'_B)m_Bc^2$$

5. Oct 24, 2007

### azatkgz

I think from the frame moving with velocity v_c,velocities for A and B arer true.But as seems to me $$m_Av_A\neq m_Bv_B$$ in this frame.

6. Oct 30, 2007

### azatkgz

Here I tried another solution:

$$E'=\sqrt{(E+m_Bc^2)^2-E^2+m_A^2c^4}$$

7. Oct 30, 2007

### learningphysics

Looks right to me.

8. Oct 30, 2007

### azatkgz

Thanks very much.

And ,if after colission reaction occurs $$A+B\rightarrowC+D+Q$$.$$m_A+m_B=m_C+m_D+Q$$.Is it

$$E'^2=(m_Cc^2+m_Dc^2+Q)^2$$ ? For E minimum

9. Oct 30, 2007

### learningphysics

I don't understand. What is the question asking for?

10. Oct 30, 2007

### azatkgz

Oh,sorry.After collision reaction occurs $$A+B\rightarrow C+D$$.Suppose that

$$m_A+m_B=m_C+m_D+Q$$.Find the minumum E for reaction to occur?And I think that in center of mass frame after reaction total momentum is zero.

11. Oct 30, 2007

### learningphysics

sorry. still a little confused. what does Q represent here?