Help with relativistic momentum question

DunWorry
Messages
40
Reaction score
0

Homework Statement



a D meson is at rest and decays into a Kaon and a Pion. The Kaon moves with speed 0.867c and has a mass of 0.494 GeV/C^2. The pion has a mass of 0.140 GeV/C^2. use conservation of momentum to calculate the speed of the Pion.

Homework Equations


Relativistic Momentum P = \gammamV

where \gamma is \frac{1}{\sqrt{1 -\frac{v^{2}}{c^{2}}}}



The Attempt at a Solution



So if the D meson is initally at rest, initial momentum = 0, which means

\gamma_{v1}m_{1}v_{1} = \gamma_{v2}m_{2}v_{2}

Where particle 1 is the Kaon and particle 2 is the Pion, we want the speed of Pion so we solve for v_{2}

After some rearrangement I got v_{2}^{2} = \frac{1} {\frac{m_{2}^{2}}{(\gamma_{v1}m_{1}v_{1})^{2}} + \frac{1}{c^{2}}}

After plugging in the numbers m2^{2} = (\frac{0.140x10^{9}}{(3x10^{8})^{2}})^{2}

and m1^{2} = (\frac{0.494x10^{9}}{(3x10^{8})^{2}})^{2}

and \gamma_{v1} = \frac{1}{\sqrt{1 - 0.867^{2}}}
I get an answer faster than light, where have I gone wrong?
 
Physics news on Phys.org
I would expect that your rearrangement is wrong. Even if you use wrong numbers for the masses, the speed has to be below the speed of light in every relativistic calculation.

Edit: After a closer look at your equation, v2 calculated there should always be below c. The formula might be right, but then your evaluation is wrong.
 
In special relativity problems, you'll find it generally better to stick to working with energy and momentum rather than velocities as it simplifies the algebra quite a bit. Try finding E and p for the pion. Once you have those, you can find its speed using the relation v/c = pc/E.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...

Similar threads

Back
Top