51. (a) Before looking at our solution to part (a) (which uses momentum conservation), it might be ad-visable to look at our solution (and accompanying remarks) for part (b) (where a very diFerentapproach is used). Since momentum is a vector, its conservation involves two equations (along theoriginal direction of alpha particle motion, thexdirection, as well as along the ±nal proton directionof motion, theydirection). The problem states that all speeds are much less than the speed of light,which allows us to use the classical formulas for kinetic energy and momentum (K=12mv2and~p=m~v, respectively). Along thexandyaxes, momentum conservation gives (for the componentsof~voxy):mαvα=moxyvoxy,x=⇒voxy,x=mαmoxyvα≈417vα0=moxyvoxy,y+mpvp=⇒voxy,y=−mpmoxyvp≈−117vp.To complete these determinations, we need values (inferred from the kinetic energies given in theproblem) for the initial speed of the alpha particle (vα) and the ±nal speed of the proton (vp). One
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