If a photon collides with an electron not moving, and you know the momentum of the photon before and after, and the electrons momentum, is it possible to know the direction of the electron and the photon after the collision?
If you know the momentum of the electron and the photon _after_ the collision, then by definition you know their directions! Momentum is a vector quantity!
You only know only the absolute value of the momentum after the collision. You know the vector of the photon before.
In general, your situation has no solution because it is over-specified: too much initial information, with only two unknown quantities. [added later: see my next post below] All of the momentum vectors lie in a plane, therefore we need to use only two components of momentum, e.g. x and y. We have three equations for conservation of x-momentum, y-momentum, and energy. In order to guarantee a solution, we need three unknown quantities.
If we know the momentum of e we know the Ke of e, and then we you know, by difference, the energy Es of the scattered photon ps. If we know Es then we know the direction of ps and from this we can get the direction of e, isn't it so?
He already knows (is given) the momentum of the scattered photon, therefore he already knows its energy. I just now noticed that he didn't say that he knows the momentum (or energy) of the incoming photon. If indeed he doesn't know that, then that is his third unknown quantity, and he can solve the momentum and energy conservation equations to find: the momentum (and energy) of the incoming photon the direction (angle) of the outgoing photon the direction (angle) of the outgoing electron
It's complementary, jitbell, I missed the second bit and you the first. The information he has is more than enough
Oops, I missed the "before." Yes, he can use your procedure, provided that the three values of momentum are indeed consistent with conservation of energy (which he should confirm, as a first step). If you give three "random" values for the momenta, in general they won't work because energy won't be conserved.