1. Oct 19, 2008

crimsonn

A radioactive nucleus at rest decays into a second nucleus, an electron, and a neutrino. The electron and neutrino are emitted at right angles and have momenta of 9.30 X 10^ -23 and 5.40 X 10^ -23 kg * m/s, respectively. What is the magnitude and direction of momentum of the second (recoiling) nucleus?

Here is my attempt at a solution:
Mn -- mass of first nucleus Mn' -- mass of second nucleus
Vn -- velocity of first nucleus. Vn' -- velocity of second nucleus

Me & Ve -- electron Mne & Vne --neutrino

MnVn = Mn'Vn' + MeVe + MneVne

MnVn = 0 (because it is at rest)

so

Mn'Vn' = -MeVe - MneVne

which, when you break it down into it's components

Mn'Vn' = -MeVe (sin 90) - MneVne (sin 90)

I know that might be a mouthful. Is it even correct? I have no idea how to do this problem with the mass of the nucleus....

answer is 1.08 X 10 ^-22, 30.1 degrees from the direction opposite to the electron's

2. Oct 19, 2008

Stovebolt

Re: Momentum Conservation Problem??? Help please!

Have you drawn a diagram? Plot the given information on a basic x,y graph and it might become a lot more clear for you.

I think you have the right basic idea, but pay attention to the angles - given a 90º angle between the vectors, you don't really need to use trigonometry functions (sin 90 = 1, cos 90 = 0).

You do not need the specific mass of the nucleus - you are given units of kg*m/s and are asked for the answer in the same units. You would only need the mass if you wanted to break down the momentum into mass and velocity.

3. Oct 19, 2008

crimsonn

Re: Momentum Conservation Problem??? Help please!

I have drawn a diagram. But...what really gets to me is that this is supposed to be a collision problem, yet I don't understand quite how it actually IS one.

Could my equations work though? I'll still confused about the mass of the nucleus?

I mean...i've tried them over and over again. If my equations ARE correct, I have no idea what I'm doing wrong

4. Oct 19, 2008

Stovebolt

Re: Momentum Conservation Problem??? Help please!

This equation:
Mn'Vn' = -MeVe (sin 90) - MneVne (sin 90)

is close, but the angle of 90 is only accurate for one of the momenta.

Be careful not to make this problem more complicated than it is. Individual masses and velocities of each component are not what is given and are not what is being asked. Stick to momentum only.

When you look at your diagram, note that you should have two vectors. If you add the vectors, you will get the resultant. What you are looking for is the equal and opposite reaction to the resultant. Make sense?

5. Oct 19, 2008

crimsonn

Re: Momentum Conservation Problem??? Help please!

okay! I think i've got I just figured out the angle:

arctan = 5.40 X 10^ -23 / 9.0 X 10 ^ -23 = 30.1 degrees! (does happy dance)

but then,

sqrt((9.8 X 10^ -23 X cos 30.1)^2 + (5.4 X 10^-23 X sin 30.1)^2)

= 3.6 X 10^-12

hmmm....

Last edited: Oct 19, 2008
6. Oct 19, 2008

Stovebolt

Re: Momentum Conservation Problem??? Help please!

First part is exactly right (but you knew that )

But the second part you are approaching backwards.

If R = the resultant of the given vectors,

R * cos 30.1 = 9.8 x 10^-23

7. Oct 19, 2008

crimsonn

Re: Momentum Conservation Problem??? Help please!

{i was actually a lot more off in my answer from doing it backwards (i forgot the sqrt) }

Wow. Thanks, that actually made a lot of sense.
I get the correct answer! thank you again. I really appreciate it!