# SR problem, howd they get here?

1. Mar 25, 2005

### 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 isnt part of the equation but if you add the two masses up to find out what the original particle was, why wouldnt 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

2. Mar 25, 2005

### 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...

3. Mar 25, 2005

### 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, dont you have to square 0.893 too?

And i know that m1+m2 cant be conserved... but where would the extra mass from the initial particle go? Converted to energy?

4. Mar 25, 2005

### dextercioby

I did square 0.893.

Compute the "mass defect".

Daniel.

5. Mar 25, 2005

### Pengwuino

How does

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

Because wouldnt the c^2 make it different?

Wouldnt it be 0.797?

6. Mar 25, 2005

### 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.

7. Mar 25, 2005

### 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?

8. Mar 25, 2005

### 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.

9. Mar 25, 2005

### Pengwuino

what do you mean by variation in the rest mass

10. Mar 25, 2005

### 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.