# Relativistic momentum

1. Jun 24, 2010

### fys iks!

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

A photon of momentum P strikes a nucleus at rest and s absorbed. If The final (excited)
nucleus is M calculate its velocity.

2. Relevant equations

p = mv/(gamma)

3. The attempt at a solution

P = Mv/gamma

P^2 - (Pv/c)^2 = (M^2)(v^2)

P^2 = (Mv)^2 + (Pv/c)^2

V = P / sqrt( (M^2) + ((P^2)/(C^2)) )

The answer in the back is

P(C^2) / sqrt((M^4)(C^2) + P(C^2))

2. Jun 24, 2010

### dulrich

Your first formula is wrong. Should be $p = \gamma mv$.

There are a couple of other equations to consider using also (though there is nothing wrong with your derivation):

$$E^2 = p^2 c^2 + m^2 c^4$$

$$E = \gamma mc^2$$

$$pc^2 = Ev$$

3. Jun 24, 2010

### Rasalhague

I get the same answer as you, fys iks, starting out from $p=Mv \gamma$. The units in the book's answer aren't consistent.

4. Jun 24, 2010

### Rasalhague

In this problem, we don't know the energy, but we do know 3-momentum and mass, so it should be just a matter of solving $p = \gamma mv$ for v, shouldn't it?

5. Jun 25, 2010

### fys iks!

yes, energy isn't introduced until the next chapter. I found a similar question and tried out my equation and it worked so i just think the book made an error.

Thanks

6. Jun 25, 2010

### dulrich

It's the exponent of 4 that makes me think of the first equation I posted. I'm guessing the author of the solution was probably thinking like this:

Divide numerator and denominator by c2 and you get your answer. A couple of typos in the denominator explains the mistake in the book (though that's no excuse). That's where my thinking was going when I provided the formulas. It's a bit quicker, but mathematically the same as your approach.

7. Jun 25, 2010

### Rasalhague

Oh, I see, yeah, that's simpler than messing around with the gamma.