SR problem, howd they get here?

[Pengwuino]

So the question is basically an unstable particle at rest breaks into 2 fragments.
m1 = 2.50 x 10 ^ -28
m2 = 1.67 x 10 ^ -27
v1= 0.893c

whats v2?

So it gets to a point where they ahve...

yxm2v2+ ((2.50x10^-28kg)/(sqr(1-0.893^2)))(0.893c)=0

Now why is that square root "1-0.893^2" instead of 1-(0.893^2/c^2) ?

Also... this isn't part of the equation but if you add the two masses up to find out what the original particle was, why wouldn't it be m1+m2? Does the extra mass turn into the energy used to send the particles on their way? And how would you calculate that energy used if that is the case.

And what program do you use to use the stupid equations :P

 

[dextercioby]

So u imposed the conservation of relativistic momentum.I can't follow where did that equation come from and who's "y".As for the square root,i'm sure that "c" was simplified,since u had

$$\sqrt{1-\frac{(0.893c)^{2}}{c^{2}}}=\sqrt{1-0.893^{2}}$$

Daniel.

P.S.U can't use that "m_{1}+m_{2}" is conserved,because it isn't.


EDIT:We use LaTex...

 

[Pengwuino]

Oh crap, yx = y2. But hwo was that c simplified? is that possible?... ohh wait, damn it, now i understand... wait wait no. How can (0.893c)^2/c^2 turn into 1-(0.893)? If you square c, don't you have to square 0.893 too?

And i know that m1+m2 can't be conserved... but where would the extra mass from the initial particle go? Converted to energy?

 

[dextercioby]

Oh crap, yx = y2. But hwo was that c simplified? is that possible?... ohh wait, damn it, now i understand... wait wait no. How can (0.893c)^2/c^2 turn into 1-(0.893)? If you square c, don't you have to square 0.893 too?

I did square 0.893.

And i know that m1+m2 can't be conserved... but where would the extra mass from the initial particle go? Converted to energy?

Compute the "mass defect".


Daniel.

 

[Pengwuino]

How does

$$\sqrt{1-\frac{(0.893c)^{2}}{c^{2}}}=\sqrt{1-0.893^{2}}$$

Because wouldn't the c^2 make it different?

Wouldnt it be 0.797?

 

[dextercioby]

Why?It's simplified with the one in the denominator.Oh,and the tex tags are
[ tex ] ...[ /tex ] (without the spacings,of course)

Daniel.

 

[Pengwuino]

Oh nevermind... i just realized the square sign was still there...

So what is this mass defect? How do you calculate it and what is it conceptually?

 

[dextercioby]

Well,the total energy is conserved and not the rest energy (which would correspond to rest masses).So by "mass defect" I'm referring to the variation in the rest mass.

Daniel.

 

[Pengwuino]

what do you mean by variation in the rest mass

 

[dextercioby]

$$m_{1}+m_{2}-m$$,

where "m" is the mass of the decaying particle...Anyway,i hope u got the point with c^{2}...

Daniel.