# Relating Momentum to Kinetic Energy

1. Apr 5, 2013

### MostlyHarmless

We recently started going over momentum in my phyics class and we were given this bit of information: $impulse=FΔt=mΔv=momentum$ I noticed that I were to take the integral of both sides with respect to velocity, it would yield $FvΔt=Fx=(1/2)mv^2$ Which is the Work-Kinectic energy relationship. At first, I thought that integrating momentum and getting Kinetic Energy was just a coincidence, but then after thinking about how momentum and kinetic energy could be related and seeing that integrating both sides still yields a familiar formula I realize that it has to mean something. However, I can't figure it out, and my professor was unable to give me an answer. So, is this relationship similiar to the relationship between Acceleration, Velocity, and Position, where, rather than integrating with respect to time we are integrating with respect to position and time? How am I to interpret this? i.e. Acceleration can be described as the change in velocity over some time and the velocity, the change in position over some time.

Differntial equations comes to mind when I think about integrating with respect to two variables at once, but alas, the answer eludes me! I've been chewing on this for days, so any insights are greatly appreciated!

2. Apr 5, 2013

### Staff: Mentor

mΔv is a change in momentum.

If you integrate F with respect to t, you get momentum (change).
If you integrate F with respect to x, you get energy (change).
If F is constant, the first is just F*Δt, and the second is F*Δx. You can divide them, and get v Δp = ΔE, which is equivalent to your equation.

3. Apr 5, 2013

### MostlyHarmless

I'm not sure I follow you. How is $FΔx$/$FΔt$ not just velocity? Maybe I'm just misunderstanding what you're trying to say, would you mind expanding you're equations a bit? Like, could you show how whatever operation you're doing is equivalent to $∫Ftdv=∫mvdv$ which gives$Fx=(1/2)mv^2$?

Also, I'd like to draw attention to this question, as it is the one I'm most interested in. :)

4. Apr 6, 2013

### Staff: Mentor

It is just velocity, exactly. But at the same time, it is the relation between energy and momentum change.

I don't see how I could expand them. I can write them as formulas, however:
$$\int F dt = F \Delta t = \Delta p$$
$$\int F dx = F \Delta x = \Delta E$$
$$\frac{ F \Delta x}{ F \Delta t} = \frac{\Delta E}{\Delta p} \rightarrow v \Delta p = \Delta E$$

Energy, momentum and mass are related via derivatives in the velocity. With a constant external force (like gravity), it is even possible to relate them to each other:
Energy is related to the position (height)
Momentum is related to the velocity
Mass is constant - acceleration is constant

5. Apr 6, 2013

### MostlyHarmless

Ok thank you, I see.

Another question: If you were given an equation for an objects Kinetic Energy at any velocity, v, could you take the derivative of that equation and it would give its momentum for any velocity? I'm just curious if it works the same way as taking the derivative of a position function wrt time.

6. Apr 6, 2013

### Staff: Mentor

Sure, this is a direct result of the equations posted above.