# Homework Help: Calculate the protons kinetic energy

1. Oct 31, 2005

### asdf1

for the following question:
a proton with the total energy E=3E0please calculat the proton's (a) kinetic energy (b) velocity (c) momentum (d)mass

my problem:
1) can someone double check my work?
(a)Ek=2EE0
(c) p=[8^(0.5)]EE0/c
(d) i've tried
(mc^2)^2 + Ek^2 +2m(c^2)Ek=(mc^2)^2+(pc)^2
=>9E0^2+(2E03E0)=8E0^2
which is weird...
the rest i'm not sure how to calculate???

Last edited: Oct 31, 2005
2. Oct 31, 2005

### quasar987

But the problem doesn't even state that the proton is moving. This is absurd.

What formula did you use for (a). I can't make sense of it. Also, you say Ek = 2EE0. But E = 3E0. So we would have Ek = 6E0², right?

3. Oct 31, 2005

### µ³

The proton must be moving by the sheer fact that the energy is greater than the rest energy (and we are not accounting for any other type of energies).
remember that
$$K = (\gamma-1) m_0 c^2=\gamma m_0 c^2 - m_0 c^2$$
but $m_0 c^2$ is the rest energy and $\gamma m_0 c^2$ is total energy
$$K= 3E_0 - E_0 = 2E_0$$
for d) Rest energy is
$$E_0 = m_0 c^2$$
so solve for m_0

Last edited: Oct 31, 2005
4. Oct 31, 2005

### quasar987

Oh, I didn't get that E0 denotes the rest energy....

5. Oct 31, 2005

6. Nov 2, 2005

### asdf1

so
(a)2Eo
(c)p=[(8)^0.5]Eo/c
(d) p^2/2m=[(8)^0.5]Eo/c => m=2Eo/c^2 but why doesn't this equal the rest mass m=E0/c^2?

7. Nov 2, 2005

### El Hombre Invisible

total energy E=3E0

calculate the proton's (a) kinetic energy (b) velocity (c) momentum (d)mass

Backwards seems easiest. Assuming E0 is rest energy, then:

(d) If by mass, rest mass is meant, then m0 = E0/c^2.
If relativistic mass is meant, then m = 3E0/c^2.

(c) pc = ROOT(E^2 - m0^2c^4)
So p = ROOT(9E0^2/c^2 - E0^2/c^2)
= ROOT(8)E0/c

(b) v = pc^2/E
= ROOT(8)/3 x c
= 0.94c

(a) Ekin = E - E0 = 2E0

Last edited: Nov 2, 2005
8. Nov 2, 2005

### asdf1

why is the way that i calculated (d) incorrect?
also, where does "v = pc^2/E" come from in (b)?

9. Nov 2, 2005

### Staff: Mentor

$p^2/2m$ is the non-relativistic kinetic energy, so your starting point for (d) says that the non-relativistic kinetic energy equals the relativistic momentum. How did you come up with that?

10. Nov 2, 2005

### El Hombre Invisible

Couldn't follow your logic, so I don't know, but this might help:

If relativistic mass is required:

m = p/v
= (ROOT(8)E0/c) / (ROOT(8)/3 x c)
= 3E0/c^2

If rest mass is required, then:

p = m0v/ROOT(1 - v^2/c^2), so...
m^2 = p^2/v^2 * (1 - v^2/c^2)
= (8E0^2/c^2)/(8c^2/9)*(1 - (8c^2/9)/c^2)
= (9E0^2/c^4)*(1/9)
= E0^2/c^4, so...
m = E0/c^2

v = pc^2/E comes from:

p = mv, so...
v = p/m

m = E/c^2, so...
v = pc^2/m

11. Nov 4, 2005

### asdf1

thank you very much!!! :)

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