Potentially painful

You have just disproven the theory of relativity and thus destroyed the entire foundation of modern physics. Bad boy!

John

The relative speed difference is all that matters. Consider that both of you are zipping along at over 1000 mph on the surface of the Earth, and the Earth is whizzing along at 19 miles per second, etc, etc, none of which matters here.

Best regards,

Bob Masta dqatechATdaqartaDOTcom D A Q A R T A Data AcQuisition And Real-Time Analysis

Yeah, and since the Earth is moving at 30km/s around the Sun, we have so much energy stored in us that on every step we explode like a nuclear bomb. I can't imagine what should happen if we take the speed of our solar system into account, not to mention the galaxy...

No, as I recall, kinetic energy is 1/2 * m * v^2.

No, it's 200 times less dangerous if you hit something stationary. However, you haven't yet hit anything stationary.

The dissipation can also be measured from the lorry's reference frame. From that frame, the system has only your kinetic energy measured in relation to it's own velocity (1/2 m * 1^2). When you hit the lorry, the system has a tiny bit of kinetic energy (it's now moving slightly in the pre-collision reference frame), and a tiny bit of energy from the collision (probably in heat, or in radiation). It isn't possible to determine the relative percentages in general.

From the road's reference frame, the difference in energy you perceived is translated into that small amount of heat, plus a change in the kinetic energy of the lorry + you system. You only dissipated a tiny bit of the energy into heat (which is what hurts); you transferred the rest of it to the kinetic energy of the lorry + you system.

No. It's the same amount of energy. Assume you are on a train travelling at 100 mph. Now, accelerate 1 mph in the direction the train is travelling. How much energy do you need to expend?

I recall seeing that claim from a high school physics teacher when I was a smart-ass twerp. I posed the following puzzle to him: A rocket car starts at rest, accellerating at a constant rate because its thrust is constant. It is burning fuel at a constant rate to produce that constant thrust. The kinetic energy of the rocket car is allegedly M * V^2 / 2, so it is increasing quadratically versus time. But the fuel consumed increases only linearly with time. How can this be?

I would be interested in your take on this. My physics teacher could not resolve it, (but, to his credit, that bothered him).

I think the rest of the universe might not go along with your theory.

Jim

You've put that behind you then. ;)

Beats me. May have something to do with the exhaust, as Don says. Might also have to do with the heat generated in the rocket. Probably something to do with reference frames, and the velocity of the fuel exhaust changing with regards to a stationary observer. That is all it really can be, since the closed system is rocket, fuel, exhaust. Inside the system, the energy is constant. However, working it out is another matter.

Energy is a funny thing. You have to be careful about how you count it up.

The classic riddle here on s.e.b is 'given a capacitor charged up to V volts, the energy is 1/2 * C * V^2. If you connect an equal uncharged cap in parallel, the charge will equalize such that the voltage is 1/2 what it was. Thus, the energy is now 2 * (1/2 * C * (V/2)^2) = 1/2 what it was before. Where did the energy go?'

At low velocities, a rocket is a very inefficient source of propulsion; at near-zero velocity, it's using its usual amount of fuel but hardly delivering any kinetic energy to the vehicle. As velocity increases, efficiency improves (or rather becomes less terrible) and vehicle energy accumulates faster. That trend continues until you run out of fuel.

In a system that goes from extremely inefficient to only rather inefficient, it's not hard to shape the efficiency curve into a quadratic. A cog railway can be nearly 100% efficient, so it will need increasing amounts of fuel if it accelerates at constant gees, but anywhere on the path it will be a lot more efficient than a rocket.

That make sense?

John

Here is a web page written by somebody who has been driven insane by this problem:

His solution is that kinetic energy is really mv, not 1/2 mv^2. He has equations and everything. Pretty wierd stuff. Who knows?

I have resolved it. Consider the kinetic energy of the rocket exhaust along with that of the car. Assuming none of the exhaust loses kinetic energy due to turbulent mixing with air that converts kinetic energy to heat, the combined sum of kinetic energies of the car and exhaust will increase linearly with respect to time during the time period that rocket exhaust is being generated.

--------------------------------------------------

Back to the original: If a car moving 1 MPH hits a stationary one, or a car moving 101 MPH rear-ends one doing 100, the amount of kinetic energy loss from causing a tiny dent or a "bump" noise from the bumper or slight heating of any cushioning in the bumper is the same. But if the cars doing 101 and 100 go out of control, their remaining kinetic energy is a big problem if they hit enything else. If their wheels lock, the remaining kinetic energy can abrade a lot of rubber from the tires. Although I doubt a car doing 101 tapping the rear bumper of one doing 100 will result in any catostrophe unless at least one driver is incompetent in ways besides one tapping another at high speeds.

- Don Klipstein ( snipped-for-privacy@misty.com)

"John Larkin" wrote in message news: snipped-for-privacy@4ax.com...

Yes. I should have known this little puzzle would not take you folks long to sort out. The key is to consider that starting condition. All the energy ends up in the exhaust at the limit of zero rocket speed. The problem gets a little easier to see if it involves chucking cannonballs from a railway car.

"Jon Hoyle" wrote in message news:vhvXd.27926$ snipped-for-privacy@fe3.news.blueyonder.co.uk...

Hi.

free to correct things, lol.

false?

reached?

speed is reached?

You have assumed something into the problem that complicates it without affecting its nature. The "terminal velocity" concept requires some kind of liquid or gaseous medium. No such medium was postulated in the original problem.

increasing quadratically versus time' True or false?

This is a reasonable point, but again a complication that does not change the nature of the problem. One can imagine an ion engine whose mass usage is so low that it can be neglected for a limited time that still fully exposes the quadratic/linear "dilemma".

fuel burn rate (thrust) also increase?

Another complication. The original problem is not about startup characteristics of rocket motors.

I'm not ready to say "flawed", but I wonder if you plan to go into engineering. If so, you will have to learn how to neglect certain phenomena and how to decide that such neglect is appropriate.

Hello!

Just wanna add a couple of my thoughts after reading this one. Please feel free to correct things, lol.

'Accelerating at a constant rate because its thrust is constant' True or false? I think false. Does drag increase with speed up until terminal velocity is reached? If thrust is constant, will drag eventually equal the thrust once a certain speed is reached? Therefore does the thrust need to increase to match the increase in drag?

'The kinetic energy of the rocket car is allegedly M * V^2 / 2, so it is increasing quadratically versus time' True or false? False? Does 'M' get smaller as the rocket car uses fuel? What shape would the graph be if it does?

'But the fuel consumed increases only linearly with time' True or false? False? As pressure and temperature builds inside a combustion chamber, does the fuel burn rate (thrust) also increase?

Well that was fun. Please tell me my understanding of physics is phlawed ;)

Jon

if there is not resistance in circuit and cap, circuit will oscillate forever; otherwise energy dissipate in heat and oscillation eventually end regards sergio

I think he demonstrate the exact opposite; if you jump from 2nd floor, speed at arrival is not double that a jumping from 1st floor as it take less time from floor 1 to 0 than from 2 to 1 , but energy is double as you need double energy to go up 2 floor than 1 , so energy increase more rapidly than speed i/e Energy>mv; regards sergio

The voltage changes, and you do mechanical work on the system to change it. Electrostatics causes a force to be exerted, and you need to move against or with that force, thus doing work.

Same thing.

Where is the oscillation coming from again? I know that an LC circuit near absolute zero will oscillate, but I didn't say anything about an inductor.

But you are right, the transfer of energy requires some dissipation. That is the answer, the extra energy is dissipated. It's easier to see if you put in resistors instead of wires between the caps (but wires are just resistors with lower values, right?). Then, the integral of power through the resistor from 0 to infinity gives the solution, which is 1/2 the original energy.

Start with cap A charged to V, connected to an uncharged cap B through one terminal. The other terminal has a switch and a resistor. B is uncharged. To the circuit, the two resistors are simply series caps, so the total is 1/2 C. Thus, the voltage across the resistor at t is

V(t) = V * exp (-t*2/RC)

Thus, instantaneous power through the resistor is

P(t) = V(t)^2/R

The integral of this from 0 to infinity is

C*V^2/4 = 1/2 * (1/2 C * V^2)

You didn't read the page. His thesis is that energy should be measured as mv, not 1/2 mv^2. He claims that mv, and not 1/2mv^2 is conserved. He uses rockets in a stationary reference frame to prove this (but doesn't seem to take the kinetic energy of the fuel into account).

Ok. I guess I'm feeling thick today, but it sounds like your solution is to attach a reference frame to the accelerating rocket. I don't know enough physics to determine if that's reasonable.

The rate of change of kinetic energy of the rocket when the rocket isn't moving is still the time derivative of 1/2mv^2, which is

F^2*t/M

where M is the (assumed unchanging) mass of the rocket, and F is thrust. This is a function of t, meaning that the rate of energy at t=0 is 0, so all the energy goes into the fuel. That is nice, but what happens at t=1? We are still increasing our 'rate of change' of energy...

If you were to say that, for example, the rate of change of energy of the thrust (from the point of view of a stationary observer) was decreasing with t, and the sum of these was constant, I'd buy that. ;)