Derive a formula for momentum in terms of kinetic energy

In summary, the student is trying to find the equation for momentum as a function of kinetic energy. They have found that momentum is only dependent on KE or a constant. They are now stuck trying to work out where that equation comes from.
  • #1
martinhiggs
24
0

Homework Statement



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.

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 :)
 
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  • #2
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?
 
  • #4
Great, finally I've figured it all out! Thank you for your help! :)
 
  • #5




Your approach is on the right track, but there are a few corrections and simplifications that can be made. First, it is important to remember that momentum is a vector quantity, so we should be looking for a formula that gives us the magnitude of the momentum. This can be done by taking the square root of both sides of your equation:

p = \sqrt{\frac{KE}{c^{2}} - m^{2}c^{2} - \frac{p^{4}}{4KE^{2}}c^{2}}

Next, we can use the fact that the kinetic energy is equal to the total energy minus the rest energy (m_0c^2):

KE = E - m_{0}c^{2}

Plugging this into our equation for momentum, we get:

p = \sqrt{\frac{E - m_{0}c^{2}}{c^{2}} - m^{2}c^{2} - \frac{p^{4}}{4E^{2}}c^{2}}

Now, we can use the definition of the Lorentz factor (gamma) to simplify this equation. Remember that gamma is equal to 1/\sqrt{1 - \beta^2}, where \beta is the particle velocity divided by the speed of light. So we can rewrite our equation as:

p = \sqrt{\frac{E - m_{0}c^{2}}{c^{2}} - m^{2}c^{2} - \frac{p^{4}}{4E^{2}}c^{2}} \cdot \frac{1}{\sqrt{1 - \beta^2}}

This can be further simplified by using the fact that the total energy is equal to the rest energy plus the kinetic energy (E = m_0c^2 + KE). Substituting this into our equation, we get:

p = \sqrt{\frac{m_{0}c^{2} + KE - m_{0}c^{2}}{c^{2}} - m^{2}c^{2} - \frac{p^{4}}{4(m_{0}c^{2} + KE)^{2}}c^{2}} \cdot \frac{1}{\sqrt{1 - \beta^2}}

Simplifying further, we get:

p = \sqrt{\frac{KE}{c^{2}} - m^{2}c^{
 

1. What is momentum and kinetic energy?

Momentum is a measure of an object's motion, taking into account its mass and velocity. It is represented by the symbol p. Kinetic energy is the energy an object possesses due to its motion, and is represented by the symbol K.

2. What is the formula for calculating momentum in terms of kinetic energy?

The formula for momentum in terms of kinetic energy is p = √(2mK), where m is the mass of the object and K is the kinetic energy.

3. How is this formula derived?

This formula is derived from the equation for kinetic energy, K = 1/2 * mv^2, and the definition of momentum, p = mv. By substituting the equation for kinetic energy into the definition of momentum and solving for p, we arrive at the formula p = √(2mK).

4. What are the units for momentum and kinetic energy?

The SI unit for momentum is kg*m/s, and the unit for kinetic energy is joules (J). However, in some cases, momentum can also be expressed in terms of newton-seconds (N*s).

5. How is momentum related to kinetic energy?

Momentum and kinetic energy are closely related, as both are measures of an object's motion. However, they are not the same. While momentum takes into account an object's mass and velocity, kinetic energy only considers its velocity. In a closed system, where there is no external force acting on the object, both momentum and kinetic energy are conserved.

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