# Finding the final energy of a particle after relativistic collision

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A ball of rest mass $M$ and relativistic energy $E$ collides elastically with a stationary ball of rest mass $m$, such that the direction of motion is along the line joining the centers of the two balls.

I am asked to show that the final energy of $M$ is given by (when defining c=1 for simplicity)

$E$=(2$m$$M^2+E(m^2+M^2))/(2m$$M$+$m^2$+$M^2$)

I then tried to square it and make use of the identity:
$E^2$=$m^2$+$p^2$ and equate the momentum components before and after the collision.

However, the equations became exhaustively long and I am unable to reach the final equation I'm trying to proof (in the above). May I ask for some hints or tips of what I'm supposed to do? Perhaps is there another (much simpler) method to tackle this problem?

Thanks.

edited Jan 1, 2017
You are using $E$ for both the initial and final energies of $M$. These are not necessarily equal, so this is confusing.

You cannot start from the expression given, because that is the answer which you are being asked to find.

You should use the usual procedure as in classical mechanics - ie write equations for the conservation of total energy and linear momentum. The fact that the collision is described as elastic means that the rest mass of each particle is the same after the collision as it was before.

Label the particles $1$ and $2$. Then :
$E_1+E_2=E_1'+E_2'$
$P_1+P_2=P_1'+P_2'$
where
$E_2=m$ and $P_2=0$, and $E$ and $P$ are related by $E^2=P^2+M^2$.

Rearrange these 2 equations to get $E_2'$ and $P_2'$ then substitute into $E_2'^2=P_2'^2+m^2$. The final equation contains $E_1, E_1'$; rearrange to find $E_1'$.

The calculation is quite involved, as you can see from the complexity of the result for $E_1'$ the final energy for the incoming particle $1$.

After a lot of algebra I get
$E_1'=[(m^2+M^2)E_1+2mM^2]/[(m^2+M^2)+2mE_1]$.

answered Jan 2, 2017 by (26,678 points)
selected Jan 2, 2017