Momentum conservation to kinetic energy

[rem45]

I am trying to follow this solution but I can't see how the momentum

p is equivalent to the sqrt(T^2+2Tmc^2)

I get that T=p^2/2m which yields p=sqrt(2mT) but why is there a T^2 term...??

 

[DEvens]

I'm going to need a bit more context.

 

[vela]

I am trying to follow this solution but I can't see how the momentum

p is equivalent to the sqrt(T^2+2Tmc^2)

I get that T=p^2/2m which yields p=sqrt(2mT) but why is there a T^2 term...??

$T = \frac{p^2}{2m} = \frac 12 mv^2$ is the non-relativistic expression for kinetic energy. You want to start with the relativistic expression.

 

[rem45]

I see that yields p= sqrt(2mT) but the solution is saying that p= sqrt(T^2+2Tmc^2)

where is this coming from??

 

[Orodruin]

As vela said, you are using the classical expression for kinetic energy and so you are missing terms which become relevant when the kinetic energy is comparable to or larger than the rest energy (i.e., mass). The relativistic relation is
$$
E^2 = p^2 c^2 + m^2 c^4.
$$
In addition, the definition of kinetic energy is the total energy $E$ minus the rest energy $mc^2$, i.e.
$$
T = E-mc^2 \quad \Rightarrow \quad E = T + mc^2.
$$
You should be able to take it from there.

 

[rem45]

That makes sense but I don't see how the total energy as stated in the solution is

the square root of kinetic energy squared plus two times the kinetic times potential...??

Forgive me, I've been doing too much physics today and perhaps it's obvious but I'm not getting this one.

 

[vela]

Perhaps you should take a break and come back to this problem tomorrow when you're refreshed. The derivation is literally a line or two of algebra starting with what Orodruin gave you.

If you still can't get it, you need to post your work so we can see what you're doing.