# Problem (bump)

1. Oct 8, 2006

### byerly100

A particle of rest mass m. and kinetic energy 2m.c^2 strikes and sticks to a stationary particle of rest mass 2m.. Find the rest mass M. of the composite particle. (.=zero subscript)

Last edited: Oct 9, 2006
2. Oct 8, 2006

### quasar987

Do you know the formula for addition of speed ? The speed in S is the speed in S' "plus" the speed of S' (where the "" around "plus" refers to the law of speed "addition" that is not as simple as the "+" operation in galilean relativity)

3. Oct 8, 2006

### quasar987

Yeah, except this as seen in S' of an object moving in S at speed ux. To find the inverse formula, just change all the velocities signs:

$$u_x = \frac{u_x'+v}{1+\frac{vu_x'}{c}}$$

You know v and you know ux'. Plug and chug away!

4. Oct 8, 2006

### quasar987

This a "conservation law" question. You know that the total energy is conserved and that the total momentum is conserved.

tip: use Einstein equation E²=(pc)²+(mc²)² to find the momentum instead of toying with the gammas to retrieve v.

5. Oct 9, 2006

### quasar987

Since you destroyed the OP, I'm guessing you're no longer interested in getting help.

6. Oct 9, 2006

### quasar987

Write what the initial energy is:

$E_i$ = sum of rest energy of all particles + sum of kinetic energy of all particle. You have all the ingredients in the question to write this out.

After the particles have stuck together and form one chunck of matter of mass M_0, the total energy is

$$E_f^2 = p_f^2c^2+M_0^2c^4$$

But by conservation of energy, you know that $E_f$ must equal $E_i$. So you can combine the two equations of energy to find one equation in two unknowns: M_0 and p_f. All you need to know then, is to find p_i, since by conservation of momentum, p_f must equal p_i.

7. Oct 9, 2006

### quasar987

Not me. Show me how you got that answer.

8. Oct 9, 2006

### quasar987

The equation E²=(pc)²+(mc²)² works for one particle only. It says that the energy of a particle of mass m and momentum p is $\sqrt{(pc)^2+(mc^2)^2}$. It does not say that the energy of a system of two particles is $\sqrt{(p_1c+p_2c)^2+(m_1c^2+m_2c^2)^2}$. If you want the energy of a system of two particles, you gotta add the individual energies of each particles: $E_1+E_2=\sqrt{(p_1c)^2+(m_1c^2)^2}+\sqrt{(p_2c)^2+(m_2c^2)^2}$.

But for this problem, you only need to use E²=(pc)²+(mc²)².

Last edited: Oct 9, 2006
9. Oct 9, 2006

### quasar987

I explained how you misinterpreted the Einstein equation and how it should be interpreted, namlely, that E²=(pc)²+(mc²)² related the energy of one particle to its momentum and mass. In your problem, there is only one particle that has a momentum (the other is at rest), so you only need to apply E²=(pc)²+(mc²)² on the particle that has a momentum and solve for p.

10. Oct 9, 2006

### quasar987

Why are you deleting your posts? Suppose what I'm telling you is wrong and you're right. Then no one can follow the unfolding of our discussion to point out that I'm wrong and you're right.

11. Oct 9, 2006