# Kinetic energy and initial velocity should be simple, right?

1. Nov 28, 2009

### elegysix

2. Nov 28, 2009

### Staff: Mentor

In rockets you always have to consider the exhaust. If you do the same calculation for the exhaust you find that the energy of the exhaust goes down as Vi goes up.

3. Nov 28, 2009

### SystemTheory

Your assumptions should lead to a simpler calculation. Try calculating the initial KE. Then calculate the change in KE. Then add or subtract terms based on whether velocity adds or subtracts from the problem statement in the reference frame.

Initial velocity does not impact the change in kinetic energy but it does enter into the solution for final KE with respect to the reference frame.

4. Nov 28, 2009

### Staff: Mentor

5. Nov 28, 2009

### SystemTheory

That looks like a great reference which I intend to study. However, upon quick review, I don't see any variable for initial velocity in the equations. I'll look again later.

In any case the simplifying assumption above appears to be constant acceleration at the rate of 1g and constant mass m over the 10{s} duration of acceleration.

v = at + vI

delta-v = at = 10{m/s}

Under these assumptions there is no rocket exhaust to consider, the change in velocity is independent of the initial velocity, so you find KE initial, then add delta-v + vI to find KE final.

delta-KE = KE2 - KE1

Solve for KE2:

KE2 = delta-KE + KE1

Expand the equation and solve.

Last edited: Nov 28, 2009
6. Nov 29, 2009

### elegysix

I sort-of did that already. I solved for delta-KE, but as you said that is equal to the difference in final and initial KE ... we know that KE1= 1/2mVi^2, but it must be that KE2=KE1 + delta-KE, as you said. KE1 is just some constant which doesn't depend on the rockets engines or fuel supply. KE1 comes from the train.

7. Nov 29, 2009

### Staff: Mentor

Yes elegysix, your math is correct, the change in KE of the rocket depends on the initial velocity. This is expected since you are neglecting the KE of the exhaust. The change in KE of the rocket + exhaust system does not depend on the initial velocity.

8. Nov 29, 2009

### SystemTheory

Do we agree that the mass m and force F are held constant under the assumptions above?

If so then in an ideal frictionless environment the impulse-momentum integral reduces to an equation:

$$v = \frac{F}{m}t + v_{I} = \Delta v + v_{I}$$

where the initial velocity is set to zero to find the change in velocity under the assumptions as a function of time. Since the change in velocity does not depend on the initial velocity, the change in kinetic energy cannot depend on the initial velocity.

This is not a rocket equation as stated under the simplifying assumptions since there is no change in mass dm.

9. Nov 29, 2009

### Staff: Mentor

SystemTheory, I appreciate your enthusiasm, but please stop spreading misinformation and confusing the OP. His math is correct, the change in KE does depend on the initial velocity. Specifically, this is wrong:
The change in velocity does not depend on the initial velocity, but KE is not a linear function of velocity, it is quadratic in velocity. So, if you differentiate KE wrt velocity you get the following relationship:

$$\frac{dKE}{dv}=\frac{d(m/2 v^2)}{dv}=mv$$

So the change in KE due to some small change in velocity does, in fact, depend on the velocity.

Take for example, m = 2, and Δv = 10, and vi = 10
KEi = 2/2 (10)² = 100
KEf = 2/2 (20)² = 400
ΔKE = KEf - KEi = 300

Now consider vi = 100
KEi = 2/2 (100)² = 10000
KEf = 2/2 (110)² = 12100
ΔKE = KEf - KEi = 2100

So the change in KE for some given change in velocity does, in fact, depend on the velocity. I apologize for my bluntness, but I think you are confusing the OP who has a correct question.

10. Nov 29, 2009

### SystemTheory

DaleSpam,

Thanks for being blunt and providing a compelling example.

It is clear to me now that the change in kinetic energy depends on the initial velocity!

I also notice now that the starting power of the force PI = F*vI must be greater for a greater initial velocity.

Since energy is the area under the power-time curve, it obviously takes more power and energy to accelerate the body by the same net velocity change when the force is applied at a greater initial velocity.

Hope that last comment adds some insight and does not contribute more confusion!

11. Nov 30, 2009

### Staff: Mentor

Yes, for a given force the power does depend on the velocity as you have stated, and I think that is indeed helpful to point out!

12. Nov 30, 2009

### rcgldr

The initial velocity is an issue even for a simple case like a car on a train. You don't need the special case of a rocket.

Although power = force x speed, it would seem appropriate to choose a proper frame of reference for the speed. In most cases the 'appropriate' speed should be the relative speed at the point of application of force. In the case of a car it would be the relative speed between car and the surface the car propels itself on.

If a car was in a very long enclosed box on a moving train, the car's power could seem to be greater when the train was moving faster, but note that the car's engine is converting potential chemcial energy into mechanical energy at some rate, minus some losses, and the car's calculated power should correspond to the rate of consumption of potential chemical energy (minus the losses), when an appropriate frame of reference is used. In this case the train itself would be the appropriate frame of reference.

A rocket also converts potential chemical energy into mechanical energy, minus some losses, so the calculated power should match the power calculated based on fuel consumption rate and efficiency. As Dale pointed out, this requires taking the KE of both exhaust and rocket into account. If in an atmosphere, the KE of the affected air needs to be taken into account, along with the fact that the efficiency is less in an atmosphere.

13. Dec 1, 2009

### elegysix

I spoke with my professor about this, and came up with a good analogy. I've done the math on this if someone wants to see it, but I have a few questions.

ok so say you're riding on a train and eating an apple. When you finish you toss the apple core into the waste basket near you. Suppose when it leaves your hand, you have given it a speed 2m/s in whatever direction you toss it, relative to you, called deltaV. Let the apple core have a mass of 1/2kg, and the train's speed be 20m/s in some constant direction, called Vt.

If you then calculate the kinetic energy from the point of view of the person in the train, initial velocity is zero, and the change in kinetic energy is 1J for any direction you throw it (neglecting air resistance and other things). However, if we say that I observe you tossing it from outside the train, stationary, the initial velocity of the apple core is the same as the velocity of the train, 20m/s, and the final velocity is the vector sum of Vt and deltaV, which gives 1J if you toss it perpendicular to the direction the train moves, 21J if you toss it in the direction the train moves, and -19J if you toss it in the direction opposite the train moves.

Now, if work is the change in kinetic energy, then this must mean you did more or less work depending on which way you tossed the apple core. You might say I need to do a calculation for the change in kinetic energy of the train, but, the only thing which is in contact with the apple core is your hand - so it must be that the change in the kinetic energy of the train system is from your hand tossing the apple. It seems reasonable to think that the amount of work my hand does on an object is proportional to "how hard it is to do". That would mean its harder or easier to toss the apple core based on the direction - which i'm sure is wrong.

On a side note, if velocity has components in three dimensions, then kinetic energy is given by KE=1/2m(Vx^2+Vy^2+Vz^2), and is equipotential for R^2=Vx^2+Vy^2+Vz^2. this means that the velocity can change direction without changing the kinetic energy, which would mean without doing work - which i'm sure is wrong. any thoughts?

14. Dec 1, 2009

### SystemTheory

Physics by Halliday and Resnik (1978), p.131-32, provides a clue.

Consider two observers, one whose reference frame is attached to the ground and another whose frame is attached, say, to a train moving with uniform velocity u with respect to the ground. Each observes a particle, initially at rest with respect to the train, is accelerated by a constant force applied to it for time t in the forward direction.

(a) Show that for each observer the work done by the force is equal to the gain in kinetic energy of the particle, but that one observer measures these quantities to be:

$$\frac{1}{2}ma^{2}t^{2}$$

the other observer measures these quantities as:

$$\frac{1}{2}ma^{2}t^{2} + maut$$

where a is the commonly observed acceleration.

(b) Explain the differences in work done by the same force in terms of the different distances through which the observers measure the force to act during the time t. Explain the different final kinetic energies measured by each observer in terms of the work the particle could do in being brought to rest relative to each observer's frame.

15. Dec 2, 2009

### Staff: Mentor

Remember, the conservation of energy only applies to a "closed" or "isolated" system. You determine if your system is closed by looking to see if any external forces are acting on it. So, (neglecting gravity) we see that there is an external force on the apple, so the apple is not an isolated system and the apple's energy is not conserved.

So next we expand our system to include the thrower. Now, what forces are acting on the thrower? We are neglecting gravity so that eliminates weight and the normal force, but there is still the reaction force from the apple and friction with the ground. For simplicity we can make the cabin slippery so there is no friction. That would mean that the apple+thrower system is an isolated system where energy is conserved.

So if X J of chemical energy is converted to mechanical energy the total KE of the apple+thrower system will increase by the same X J regardless of the direction thrown or the coordinate system used. Some directions more will go to the thrower and some directions more will go to the apple.
That is correct.

Last edited: Dec 2, 2009
16. Dec 2, 2009

### SystemTheory

The problem might be visualized like this. Alladin rides a magic carpet at intial velocity vI with respect to observer B. There is no friction except at the carpet surface and no external forces act. He tosses an apple. This is now similar to a rocket equation (discrete delta-m shrinks to continuous dm for a rocket) and DaleSpam's comments make more sense to me in this context.

Edit: An observer moves in reference frame B with velocity -vI relative to Alladin's zero velocity in reference frame A.

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17. Dec 3, 2009

### SystemTheory

This Wiki page on "working mass" is an interesting interpretation of the rocket equation.

http://en.wikipedia.org/wiki/Working_mass

In the attached sketch Aladdin considers himself at rest in his own reference frame and does work to project the working mass m at relative velocity u. (a) Standing on Earth the large M makes $$\Delta v$$ = 0. (b) Standing on his magic carpet the ratio M/m is finite so $$\Delta v$$ is significant. This may be the best approach to such problems in terms of conservation of momentum and kinetic energy. The calculation of work and energy from different frames of reference is a secondary challenge.