A helium nucleus (He-4) of mass 4 amu moving with a speed v breaks up into a neutron (mass 1 amu) and a helium isotope nucleus (He-3) of mass 3 amu. The neutron moves off at right angles to the original He-4. If the neutron speed is 3v, what is the final velocity of the He-3 nucleus?

Ok, so I drew a diagram, chose my x and y axes and I know that Momentum is conserved in both directions and that initial momentum = final momentum but I can't seem to figure it out! Here's what I have:


H4= He 4 amu Atom
N= Nucleus
H3= He 3 amu Atom

Pi = Pf

MH4VH4 = -MH3VH3 + MNVN
Subtract MNVN to get
MH4VH4 - MNVN = -MH3VH3
Divide by -MH3 to get
(MH4/-MH3)VH4 – (MN/-MH3)VN = VH3

But then what do I do with the velocity? Please help me!!

Consider conservation of momentum in two components - the original direction (let's call this the x-direction), and in the perpendicular direction (y direction).

In the x-direction, conservation of momentum says:
4au*v= 3 au* VH3x

Where VH3x is the velocity of the H3 particle in the x direction. From this you have VH3x in terms of the original velocity v. Note that the neutron has no component of momentum in this direction, so isn't included in this formula.

Conservation of momentum in the y-direction (perpendicular to the original direction of motion):

0 = 3au*VH3y + 1au*3v

So now you have the y-component of the velocity of the H3 particle, in terms of v.

Now that you have both the x- and y-components for VH3, you can calculate H3's velocity.

Hope this helps.