Conservation of momentum (neuclear decay)

[avenkat0]

Homework Statement

A certain nucleus at rest suddenly decays into three particles, two of which are charged and can be easily detected. The data gathered for these two particles is:
- Particle 1 has mass 2.93e-25 kg moving 5.69e+06 m/s at 69.5o.
- Particle 2 has mass 1.33e-25 kg moving 9.31e+06 at 221o.

Find the Momentum of the third particle
_____i km m/s + _____ j km m/s

The Attempt at a Solution

I tried first calculating the Momentum of each of the particles and then split the 2D momentum vectors into i and j

and then i set that the sum of all the i's = 0 and that the sum of all the j's = 0 since there aren't any external forces and momentum is conserved and then i tried solving for the missing i and the missing j.

I keep getting the answer wrong...

Thanks for your help
Aditya

 

[nasu]

You method sounds OK. I cannot tell where is your mistake unless you show your work (or at least results)

 

[avenkat0]

Thanks for your reply...

This is what i have done:
my results came out to be
-2.38e-20 i kg m/s + -1.741e-18 j kg m/s

 

[nasu]

I don't get your results. As I said, if you show your work I could figure out where is your mistake.
Maybe you calculator is set on the "wrong" mode (radians instead of degrees)

 

[avenkat0]

well this is what i got
2.55e-18 j + 8.06e-19 j + yj = 0
y= -1.74e-18 j kg m/s

9.52e-19 i + (-9.28e-19) i + xi = 0
x= -2.37e-20 i kg m/s

 

[nasu]

I don't know how you calculate your components but I can say this much:
Something is not right. The second momentum (221 Deg) is in the third quadrant. Both x and y components must be negative. I don't see this in your components.
And your y component of the first momentum (2.55 e-18) seems to be larger than the whole momentum...

 

[avenkat0]

this is how I am getting the 1D values
m3v3 i = - (m1v1cosß +m2v2cosΘ)
m3v3 j= - (m1v1sinß +m2v2sinΘ)

 

[avenkat0]

hey i re-did the calculations and is this the right answer:
-2.11e-19 i kg m/s + -1.72e-18 j kg m/s