# Homework Help: Mass and Momentum Question

1. Dec 30, 2008

### DukeLuke

1. The problem statement, all variables and given/known data
Determine how accurately you will need to measure the speed of a particle that has energy 2 GeV to determine whether it is a pion (mass = 0.14 Gev/c^2) or a Kaon (mass = 0.49 Gev/C^2). *This question is in a relativity section, so I'm assuming relativistic equations are neccesary*

2. Relevant equations
$$E^2 = (p^2)(c^2) + (m^2)(c^4)$$
$$p = (\gamma)mv$$

3. The attempt at a solution
I started by substituting $$p= (\gamma)mv$$ into the other equation so velocity can be related to mass, but I'm lost as to what the basic approach to this problem would be.

2. Dec 30, 2008

### Dick

That looks like a good start. I'd now try to solve for v in terms of E and m and get the value of v for a kaon and a pion and see what the difference is. It looks to me like it becomes a quadratic equation in v^2.

3. Dec 31, 2008

### SimonZ

use gamma r = E/(mc^2) = 1/sqrt(1 - v^2/c^2)
so v/c = sqrt[1 - (mc^2/E)^2]
for pion, mc^2/E = 0.14/2 = 0.07
v/c = 0.9975

4. Jan 2, 2009

### DukeLuke

Thanks, the only thing is when I solved it I didn't end up with a quadratic.

5. Jan 3, 2009

### Dick

What DID you get?

6. Jan 4, 2009

### DukeLuke

I started by solving for p and substituing $$p= (\gamma mv)$$ for it.

$$\frac{mv}{\sqrt{1 - v^2/c^2}} = \sqrt{\frac{E^2 - m^2c^4}{c^2}}$$

then I squared both sides and moved gamma and $$c^2$$ to opposite sides

$$m^2v^2c^2 = (E^2 - m^2c^4) (1 - v^2/c^2)$$

multiply out right hand side

$$m^2v^2c^2 = E^2 + m^2c^2v^2 -m^2c^4 - \frac{E^2v^2}{c^2}$$

group $$v^2$$ on same side and factor it out

$$v^2 (c^2m^2 + E^2/c^2 - m^2c^2) = E^2 - m^2c^4$$

solve for v

$$v = \sqrt{\frac{E^2 -m^2c^4}{c^2m^2 + E^2/c^2 - m^2c^2}}$$

7. Jan 5, 2009

### Dick

That looks ok to me. You can cancel the two m^2*c^2 terms. It is even easier than I thought.