# KE and Momentum

1. Oct 31, 2007

### Gear300

In Classical and Mechanical Physics, the mass of an object is constant when in it is in simple motion (no collisions and so forth).

If the equation for Kinetic Energy is KE = (1/2)mv^2 and mass is constant, wouldn't that imply that the derivative of the Kinetic Energy in respect to velocity would be
dKE/dv = mv? This would also state that the derivative of Kinetic Energy is momentum. I also found from another source that when considering v and v^2, KE is not a vector, whereas momentum is. I'd simply like clarification (while considering things in terms of Mechanical Physics). If the derivative is true, then what is the physical relationship between KE and momentum?

Last edited: Oct 31, 2007
2. Oct 31, 2007

### rl.bhat

When you are taking the derivative of the KE with respect to velocity, you are assuming that the velocity is changing. And the velocity changes when force is acted on the mass.

3. Nov 3, 2007

### Gear300

When a force is acted on a mass, it gives it an acceleration, and from that, it can have a momentum. The relation still seems somewhat vague to me. If a mass were to have constant KE above 0J, then it would have 0 momentum (if the mentioned dKE/dv was valid). But that seems unlikely, because if the velocity is > 0, then KE > 0 and p > 0. I'm assuming the derivative is not valid. I would assume it might have something to do with the acceleration, because if KE is constant, then a = 0.
So, I would also assume the proper derivative would be dKE/dt = mv(dv/dt), whereas, its in respect to time. Is this right? (and, in that case, why does the dKE/dv not make as much sense?)

Last edited: Nov 3, 2007
4. Nov 3, 2007

### arildno

You may regard the three velocity components as independent variables.

The "gradient" of the kinetic energy with respect to these variables equals the (vectorial) momentum.

5. Nov 3, 2007

### Gear300

so, each component is independent, and the "gradient" KE would be in respect to these variables. I see...somewhat. Can you elaborate?

Last edited: Nov 3, 2007