Calculating Recoil Momentum in Two-Dimensional Nuclear Decay

[rvnt]

Homework Statement

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?

Homework Equations

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

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

 

[kuruman]

How big is vA?
How are the trig functions of θ_A and θ_B related?

 

[rvnt]

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

 

[kuruman]

I think the angles are both 45degrees...

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

θ_n = neutrino angle
θ_e = electron angle

then θ_n + θ_e = 90^o

in which case
cos(θ_n) = cos(90^o - θ_e) = ______ ?
sin(θ_n) = sin(90^o - θ_e) = ______ ?

and maybe vA= v'A Cos (45degrees)+v'B Cos(45degrees)= 2v'A Cos(45degrees)??

What does vA represent? The velocity of what? Does it have a value that you can ascertain?

 

[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?

 

[kuruman]

vA is the velocity of the electron before the collsion.

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