An unknown particle with rest energy of $1 \text{ MeV}$ is traveling with a speed of $0.6c$ along the $x$-axis in our rest frame when it decays into two photons, also travelling along the $x$-axis. What are the energies of the photons in $\text{MeV}$?
I've been trying to learn special relativity recently and came across this question, but all I really know is basic length contraction and time dilation. Would solving this use similar concepts and how should I go about it?
You need to apply conservation of energy & momentum.
Based on your comment it seems the difficulty is that "photons are massless". They are indeed massless, but they can still carry momentum. Once you know the momenta of the photons, it's easy to convert to energy via $E = pc$.
You know that $m = 1 MeV/c^2$. Your calculation of the initial total energy of 1.25 MeV is correct. The initial momentum should be 0.75 MeV/c. You can now find 2 equations for the momenta of the 2 photons, one equation from conservation of energy, the other equation from conservation of momentum.
Note that energy is a positive scalar, but momentum is a vector. In this one dimensional problem, momentum can be positive or negative (or zero). So to calculate photon energy from momentum you need to use the magnitude of the momentum, that is $E=|p|c$.
In the rest frame of the initial particle, the momentum is obviously zero. So in that frame the sum of the photons' momenta must also be zero to conserve momentum.
Hopefully, you now have enough clues to solve this.
