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Derive a formula for momentum in terms of kinetic energy 
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#1
Oct1410, 03:54 AM

P: 24

1. The problem statement, all variables and given/known data
Using: particle velocity, beta particle momentum, p total energy, E Lorentz factor, gamma kinetic energy, KE Derive an equation for momentum as a function of kinetic energy. The functions have to depend either on the variable in the bracket, p(KE), or on a constant. 3. The attempt at a solution This is what I've done so far, and I am now stuck, and unsure if the way I am doing it is correct or if there is a different approach. [tex]E^{2} = p^{2}c^{2} + m^{2}c^{4}[/tex] [tex]KE = E  m_{0}c^{2}[/tex] [tex]KE = \sqrt{p^{2}c^{2} + m^{2}c^{4}}  m_{0}c^{2}[/tex] [tex]p^{2} = \frac{KE^{2}}{c^{2}}  m^{2}c^{2}  m_{0}^{2}c^{4}[/tex] The only thing I could think of doing next is: [tex]KE = \frac{p^{2}}{2m_{0}} , m_{0} = \frac{p^{2}}{2KE}[/tex] [tex] p^{2} = \frac{KE}{c^{2}}  m^{2}c^{2}  \frac{p^{4}}{4KE^{2}}c^{2}[/tex] [tex]p^{2} + \frac{p^{4}}{4KE^{2}}c^{2} = \frac{KE}{c^{2}}  m^{2}c^{2}[/tex] [tex]p^{2}(1 + \frac{p^{2}}{4KE^{2}}c^{2}) = \frac{KE}{c^{2}}  m^{2}c^{2}[/tex] I'm not sure if this is the best or easiest way to do this, as it seems to be pretty messy, and I also have one more m in the equation that I need to get rid of but am not sure of the best way of doing so. Any help will be greatly appreciated :) 


#2
Oct1510, 03:28 AM

P: 24

Ok, so I've been working on this problem for about 24 hours and I think I'm finally getting somewhere with it. In class we were given a sheet of useful formulae, and this included:
[tex] p = \gamma \beta m_{0} c = \frac{m_{0} \beta c}{\sqrt{1  \beta^{2}}} [/tex] [tex] = \frac{\sqrt{E_{tot}^{2}  m_{0}^{2}c^{4}}}{c} [/tex] From this final equation, I noticed that [tex] KE = \sqrt{E_{tot}^{2}  m_{0}^{2}c^{4}} [/tex] So this means that I have the relation: [tex]p = \frac{KE}{c} [/tex] Which is momentum which is only dependent on KE or a constant! The only problem I have now is working out where that equation from p comes from, can anybody help? 


#3
Oct1510, 03:43 AM

P: 535

The formula for the momentum is derived from the expression for total relativistic energy. This page has some good info:
http://hyperphysics.phyastr.gsu.edu...iv/releng.html 


#4
Oct1510, 03:50 AM

P: 24

Derive a formula for momentum in terms of kinetic energy
Great, finally I've figured it all out! Thank you for your help! :)



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