p+7Li->4He+4He energy of He nuclei

[EngageEngage]

1. The problem statement, all variables and given/known data
p+^{7}Li \rightarrow ^{4}He +^{4}He
Mass of 7Li = 6533.8 MeV/c^2
Calculate the kinetic energy of each helium nucleus if the incident proton has nearly zero kinetic energy. Answer: 8.68MeV
2. Relevant equations
E_{total}=\frac{mc^{2}}{\sqrt{1-(\frac{v}{c})^2}}
3. The attempt at a solution
I tried to apply conservation of momentum here:

E_{iproton}=mc^{2}=938.27MeV
E_{iLi} = 6533.8MeV

By conservation, these quantities should be conserved, so I add them which gives me the wrong answer. I'm definitely missing some crucial concept here. If anyone could help me figure this out I would appreciate it greatly.

[EngageEngage]

I tried a different approach and I got the right magnitude, but still wrong.
E_{i}=m_{p}c^{2}+m_{He}c^2
Since the protons are at rest, I assign only rest energies. I now assume that the 2 products are moving in the same direction (which is erroneous im pretty sure). Doing so allows me to define a second frame S', in which the momentum of the two product particles is zero:
In S'
p'_{x,f}=0
since they are at rest in this frame, their combined energies must equal their rest energies:
E'_{f}=2m_{^{4}He}c^{2}
Now, i convert this energy back into the unprimed frame, S.
E_{f}=\gamma (E'_{f}-Vp'_{x,f})
The second term is zero
E_{f}=\gamma E'_{f}
I set the initial and final energies equal to each other and solve for the velocity to find:
v = 0.02914660097c
Using the relativistic kinetic energy:
KE = m_{^{4}He}c^{2}(\gamma - 1)=1.58MeV
This is incorrect. once again I'm pretty sure that I shouldnt have made a new frame, since even in the unprimed S frame the momentum should be zero, because the initial momentum is zero (thus by conservation). However, that doesn't get me anywhere near the right answer. Can someone please help me out and give me some ideas of what else i could try? Thank you

[granpa]

what is the mass of 4He?

[EngageEngage]

its about 4*mp = 4*(1.67*10^-27)

[granpa]

thats not good enough. it should have been supplied in the question. or at least the book.

[EngageEngage]

the exact number in the book (not supplied in the question) is 4.002602u where 1u = 1.66054*10^-27kg

[granpa]

mass=energy
and energy must be conserved.

[EngageEngage]

Figured it out:
E_{i}=m_{p}c^{2}+m_{Li}c^{2}
E_{f}=2m_{He}c^2+2KE_{He}
KE_{He}=8.65MeV
I forgot to subtract off the rest energies in the very first post for some dumb reason!