# Collisions in two dimensions

1. Aug 7, 2010

### rvnt

1. The problem statement, all variables and given/known data

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 9.3x10^23kg*m/s and 5.40x10^23kg*m/s, respectively. What are the magnitude and direction of the momentum of the second (recoiling) nucleus?

2. Relevant equations
px conserved: mAvA=mAv'ACosΘ'A+mBv'BCosΘ'B
py conserved: mAvA=mAv'AsinΘ'A+mBv'BSinΘ'B

3. The attempt at a solution
Really confused on how to approach this one? Tried to draw a picture first of what's going on but not really working...please help

2. Aug 7, 2010

### kuruman

How big is vA?
How are the trig functions of θA and θB related?

3. Aug 8, 2010

### rvnt

I think the angles are both 45degrees...and maybe vA= v'A Cos (45degrees)+v'B Cos(45degrees)= 2v'A Cos(45degrees)??

4. Aug 8, 2010

### kuruman

There is nothing in the problem that says that they are. The problem says that the neutrino and the electron are emitted at right angles. Let

θn = neutrino angle
θe = electron angle

then θn + θe = 90o

in which case
cos(θn) = cos(90o - θe) = ______ ???
sin(θn) = sin(90o - θe) = ______ ???
What does vA represent? The velocity of what? Does it have a value that you can ascertain?

5. Aug 8, 2010

### rvnt

vA is the velocity of the electron before the collsion. p=mv the mass of an electron is 9.11*10^-31kg so v=(9.30*10^-23)/(9.11*10^-31)=1.0208*10^8m/s so I think va= 2(1.0208*10^8m/s) CosΘ?

6. Aug 8, 2010

### kuruman

You mean the nucleus. There is no "electron" before the collision. What is a number for that? Read the problem.