Built on Facts

Here is a question. It’s a sort of subtle question, but one that can be answered with freshman-level physics. But it’s an excellent test of understanding. I’m not promising that the question itself is not in some sense a “trick” question, but the trick is in how you might think about the physics, not the question itself. Which is to say, I’m not promising that what the question assumes happens can actually happen. But it might – it’s up to you to tell me!

Ok. You have a perfectly efficient car, which transmits all of the energy in its fuel into kinetic energy of forward motion. Kinetic energy is related to the forward velocity by the famous equation

We’ll make up our own unit of energy (the zap) so the concept is more clear and we don’t have to worry about the particular mass of the car. The driver of the car pours 1 zap worth of fuel into the engine and the car accelerates from rest to 1 m/s. To double the speed you have to quadruple the kinetic energy. So you pour three more zaps of fuel into the car, and with a total kinetic energy of 4 zaps the car is moving at 2 m/s. To go 100 m/s, the car needs 10,000 zaps of fuel.

But what if you’re watching the experiment from a bus which is itself traveling at 100 m/s alongside the car, then you observe the car standing stationary. One more zap of energy should by your prediction therefore increase its speed in your reference frame by 1 m/s – or equivalently 101 m/s with respect to the ground. But from the reference frame of the ground, the speed of the car should be sqrt(K), where K is the 10,001 zaps of energy the car has used in total. That should put the car at 100.005 m/s, not 101 m/s.

Who is right, given that we know both inertial reference frames are equivalent?

Comments

#1 Kobra
March 18, 2009

The distance between the two vehicles is increasing at 101 m/s, but the car only has enough energy to move at 100.005 m/s. Therefore, its velocity is 100.005 m/s.

Ah, the things I learn from random emails.
#2 Eric Lund
March 18, 2009

There is a flaw in your statement of the problem: whence comes the momentum of the car?

Let’s say that if the car is moving 1 m/s it has one zip of momentum. It acquires this momentum from a standing start in one of two ways: (1) it’s a rocket, so it expels one zip of fuel backwards, keeping the net momentum at 0 (we will assume that the fuel is expelled at a much higher velocity than the car moves, so that we can neglect the car’s changing mass), or (2) friction between the tires and the road propels the car forward. (Both effects can come into play, e.g., at the Bonneville track in Utah.) Real cars operate by scenario #2, which means that some of the energy is lost to friction and wind resistance (100 m/s is 360 km/h, which is triple the highest speed limit–75 MPH, or ~120 km/h–on any American freeway), and the ground observer will be closer to correct. However, under the rocket scenario the bus observer will be correct. I can’t tell which of those two scenarios was your intent.
#3 beebeeo
March 18, 2009

Is it possible that the inertia in the two reference frames is changing as well? If we are changing the V in 1/2*m*V*V is it maybe wrong to assume that m is staing the same?
#4 Carl Brannen
March 18, 2009

It would take a lot less fuel to accelerate if you could accelerate yourself by running your tires against the bus instead of against the pavement. Another way of saying this is to note that W = F d, work is force times distance. Since the bus is relatively stationary to you, you can apply a force to it with little change in distance. The problem with the ground is that it is receding.
#5 beebeeo
March 18, 2009

no wrong, my previous post is wrong.
Now I got it, maybe.
momentum stays the same. I order to accelerate you have to throw something backwards. In case of the car you are actually throwing the road backwards (of course the earth is too big to really start rotating in different direction or so, so we don’t realise it)
In the moving reference frame of the bus, you have to throw back the road. Of course the road is already moving at 100 m/s (compared to the bus reference frame) so you need more energy to actually accelerate and 1 zap only brings you 0.005 m/s. In case of a rocket I guess the difference would be the speed of the air in which we move.
I wonder what would happen if everybody started accelerating in a certain direction simultaneously. Could we actually, at least temporarily make the earth spin faster/slower ?
I am not a physicist but I think I got it right this time …
I still can’t think of what would be if we were talking about two rockets in a vacuum …
#6 Tom
March 18, 2009

The key concept here is understanding that kinetic energy, while conserved, is not invariant, and it’s important to understand the distinction. IOW, the speed increase as the result of a zap is not the same in the two coordinate systems.
#7 Tom
March 18, 2009

I should clarify that energy, in general, is conserved. Not necessarily kinetic energy, though.
#8 Chris P
March 18, 2009

Work is force times distance. When you are travelling faster the work has to be done over a longer distance.

This comes out in the problem of high speed bicycles. Now that the top speed is just over 80 mph one of the major problems is the time and distance to actually get to that speed.

Chris P
#9 Anonymous
March 18, 2009

Tom has the right key point (although the second post about KE not being conserved confuses the issue): KE is NOT INVARIANT, and will depend on your frame of reference.

Invariance has little to do with conservation here. The question of invariance is: If you change frames of reference, what happens to the quantities m, v, KE?

* 1/2 is invariant, i.e. both observers agree that numbers have the same value
* m is invariant, i.e. both oververs will measure the same mass (remember, freshman-level physics here)
* v is NOT invariant, almost by definition, i.e. one observer will say the speed is v1, while the other will say the speed is ( v1 – u ), where u is the relative speeds of their reference frames.

Ergo, 1/2 m v^2 cannot be invariant. That’s OK, that’s no more controversial than stating that v itself is not invariant.

So, what are we left with? (I’ll make the mass equal to 2 to cancel the 1/2 in KE). Let’s start with the reference frame of the person on the bus; I’ll label the appropriate quantities with a prime (‘). At t=0, she sees the car moving at v = -100 m/s. At t=t1, she sees the car moving at 0 m/s. At t=t2, she see the car moving at 1 m/s. That gives:

KE_0′ = 10,000
KE_1′ = 0
KE_2′ = 1

As for the person on the side of the road, we have:

KE_0 = 0
KE_1 = 10,000
KE_2 = 10,201

Well, here it might get uncomfortable: it seems that differences in KE are not invariant either, i.e. the observers disagree about how much KE the car gained between t1 and t2. But again, no shock: a quantity that depends on (v1^2 – v2^2) should not be invariant.

In other words, the misleading (or “tricky”) part of the problem is to state that if the observer on the bus sees an increase in KE of 1 zap, the observer on the ground should see that same increase. This is not true, and this is where the logic in the problem statement leads to a “connundrum”.
#10 Mark Harrison
March 18, 2009

The frames are equivalent, but an observer on the ground and an observer in the bus won’t agree on how much work was done, because they won’t agree on how much distance the car covered while accelerating. The adding of one zap of work is ambiguous.

Let’s say a force of one zork over a distance of one meter results in one zap of work. In order to observe one zap of work in the bus frame, an observer on the bus would have to see a one zork force applied to the car as it moved one meter ahead of it. This would result in the car moving at 1 m/s in the bus frame.

Now let’s switch to an observer on the ground. That same zork of force would be applied over a distance much longer than a meter, namely, 100t + 1, where t is the time over which the one zork force is applied (classically, the two observers should agree on the force applied). The observer on the ground sees the force applied over a greater distance, and so would measure more work done. In this case, the final speed is 101 m/s in the ground frame, 1 m/s in the bus frame.

If the observer on the ground measures one zap of work, the observer on the bus would observe less work done because the car wouldn’t move as far ahead of it. In this case, the final speed is 100.005 m/s in the ground frame, 0.005 m/s in the bus frame.
#11 Nemo
March 18, 2009

Eric Lund has it right.

You cannot push something forward without pushing something else backward. The energy in the fuel goes not only into the car, but also into the ground (or into the exhaust if it is a rocket engine).

Once you take this into account and calculate the delta-v for BOTH the car and the thing it pushes against, it turns out the reference frame does not matter.

To reduce to a simple example…

Consider two identical balls of mass 1kg with a coiled spring between them. Suppose the spring contains an amount of energy such that when you release the spring the balls fly apart in opposite directions at 1 m/s.

From the point of view of someone stationary, the initial kinetic energy is zero and the final kinetic energy of each ball is .5 * 1kg * 1^2, which is 0.5, so the total final kinetic energy of both balls is 1 and the change in KE is 1.

From the point of view of someone moving at velocity 1 m/s, the initial kinetic energy of the two balls is .5 * (2kg) * (-1)^2, or 1. The final kinetic energy of one ball is zero (because it is moving with you) and the final kinetic energy of the other ball is 0.5 * (1kg) * (2^2), which is 2. So again the change in KE is 1.

This will work out no matter what reference frame you use or what masses the balls have. You just need to make sure you take into account Newton’s Third Law.

Thanks for the brain teaser!
#12 Max Fagin
March 18, 2009

“Who is right, given that we know both inertial reference frames are equivalent?”

But they’re not equivalent, are they? The car is accelerating, the bus is not. The car is not in an inertial reference frame.
#13 Boris Legradic
March 18, 2009

The answer clearly is: One n too much
#14 rex27
March 18, 2009

Hmm…

I dunno, I think in the description of the situation at the beginning of the post, we’re talking about getting the car to move at a certain speed from rest…i.e. 10,000 zaps are required to make it move 100m/s when it is initially not moving…

In the case of the bus moving alongside the car: isn’t this equivalent to asking how much faster it moves when one zap is added when it is initially moving at 100m/s as opposed to when it is initially at rest?

it’s initial K.E. is (0.5*m*v^2)
when one zap is added surely it’s K.E. changes to (0.5*m*v^2 + 1 zap)

if we call it’s new speed c, then

0.5*m*c^2 = (0.5*m*v^2 + 1z)…

if 1 zap is equivalent to 0.5*m*(1)^2, and v=100,

0.5*m*c^2 = (0.5*m*(100)^2) + 0.5*m = 0.5*m*(100^2 + 1)

therefore c^2 = 10 001
and so c = 100.005 m/s
#15 rex27
March 18, 2009

Oh, i think i realised the mistake in my last post… i was using K.E. as absolute, compared to the ground… compared to the bus the K.E. of the car is zero…

is it possible that the car moves 0.005 m/s faster relative to the ground and 1 m/s faster relative to the bus? After all, a car moving at 80m/s to the ground moves at 30m/s relative to a bus moving at 50 m/s….
#16 rex27
March 18, 2009

Oh, I think I realised the mistake in my last post… I was using K.E. as absolute, compared to the ground… compared to the bus the K.E. of the car is zero…

Is it possible that the car moves 0.005 m/s faster relative to the ground and 1 m/s faster relative to the bus? i.e. both are right? After all, a car moving at 80m/s to the ground moves at 30m/s relative to a bus moving at 50 m/s….
#17 rex27
March 18, 2009

Oh, I think I realised the mistake in my last post… I was using K.E. as absolute, compared to the ground… compared to the bus the K.E. of the car is zero…

Is it possible that the car moves 0.005 m/s faster relative to the ground and 1 m/s faster relative to the bus? i.e. both are right? After all, a car moving at 80m/s to the ground moves at 30m/s relative to a bus moving at 50 m/s….
#18 Duae Quartunciae
March 18, 2009

I found I could think of this problem more easily in terms of sleds on a frictionless ice pan. A sled moves by pushing on another sled.

The stated problem involves a light weight sled (car) pushing on a heavy sled (the highway), and this can be observed from any other sled. As a wrinkle, if you observed from the recoiling sled, you can’t assume an inertial frame. If we regard the bus as a third sled, and a truly inertial frame, then the push of the car on the highway-sled will give a small change in the velocity of the highway-sled wrt the bus-sled as well. I’ll presume that the highway on which the car is traveling is able to recoil independently of the bus. That is, the bus is coasting without friction, and can represented as a third inertial sled on the ice-pan.

The car moves at 100 m/s wrt to the road. The car pushes on the road, until it moves at 101 m/s wrt the initial frame of the road. But road has a new inertial frame. It recoils at a speed v wrt to its original frame. Assume that the road is X times more massive than the car. Then v = 1/X.

Observed from a busstop initially at rest wrt to the road, the car changes from 100 to 101 m/s, and has additional energy of 101^2-100^2 = 201 zaps. The road also gets a tiny bit of energy which is X times (1/X)^2, or 1/X.

Total additional energy is 201 + 1/X, where X is large.

Observed from a bus originally sliding alongside the car… the car changes from 0 to 1 m/s, and has additional energy of 1 zap. But the road! The road starts out with a massive amount of energy. It’s receding at 100 m/s at first, and has energy X * 100000. With the tiny additional recoil of 1/X, it gets more energy at X*(100+1/X)^2, or X * 1000000 + 200 + 1/X. When observed from the bus, the recoiling highway has picked up 200 + 1/X additional zaps of energy!
#19 ppnl
March 18, 2009

I ran into this conundrum years ago from a different direction. Say a rocket uses one unit of chemical energy to accelerate to a velocity of one meter per second. It then uses the same amount of fuel again a doubles its velocity. But its energy has increased 4 times. What gives? And why is it different than a car that has to use four times the fuel to double its velocity?

It turns out that the Galilean transform can be as counter intuitive as the Lorentz transform.
#20 Duae Quartunciae
March 18, 2009

With the rocket, most of the energy ends up the exhaust; not in the moving rocket. If the rocket accelerates by 1 m/s, with exhaust that is expelled at 1000 m/s (it’s actually a bit more than this, usually) then it must have expelled an amount of exhaust weighing about 1000 times less than the rocket. Hence only 0.1% of the kinetic energy is in the rocket. The rest is the exhaust. With the second pulse of the rocket engine, the expelled exhaust ends up moving at 999 m/s. 999^2 = 998001.

So. With the first pulse, the rocket gets energy 1 and the exhaust gets energy 1000

With the second pulse, the rocket gets energy 3 (from 1 to 4) and the second exhaust gets energy 998. Sum is the same in both cases. (It remains the same when you consider that the energy of the exhaust is a bit extra 998.001 and the rocket is a bit less, because it weighs less with less fuel on board).
#21 T
March 19, 2009

So if a spaceship burns 1 fuel container to reach a speed of 1m/s, which it then stays at constantly, what’s to stop the spaceship from declaring itself at rest (since it’s moving at constant velocity) thus requiring it to burn only 1 more fuel container to reach a speed of 2m/s (since it’s 1m/s w.r.t his new “rest” frame).

Had he not declared himself at rest once he was at 1m/s he would have needed 4 containers of fuel to reach 2m/s.

I can see that KE is frame dependent, but when you put energy into fuel containers it’s harder to understand how that works out in reality when you can’t just say that different observers will count different numbers of containers on board the spaceship…

So what’s the definitive answer?
#22 Duae Quartunciae
March 20, 2009

The definitive answer is already in the comment immediately above yours. Comment #20.

The kinetic energy of a rocket’s exhaust is many orders of magnitude larger than the kinetic energy of the rocket. That’s your definitive answer.

Your estimation of 4 containers of fuel for 2 m/s is based on the notion that the fuel goes into energy of the rocket. It doesn’t. It almost all goes into energy of the exhaust. Just calculate energy, and momentum, in any frame you like. Include exhaust and rocket, and you’ll get the right answer for a rocket working in empty space.

The second m/s for the rocket takes almost exactly the same amount of fuel as the first, no matter what frame you use. You need a bit less fuel because the rocket is a bit lighter; but that’s it.
#23 yesfm
March 20, 2009

both are valid. you can’t compare different frames of reference. it moves 0.005 m/s faster with respect to the ground and 1 m/s faster with respect to the bus.
#24 Anonymous
March 20, 2009

“So if a spaceship burns 1 fuel container to reach a speed of 1m/s, which it then stays at constantly, what’s to stop the spaceship from declaring itself at rest (since it’s moving at constant velocity) thus requiring it to burn only 1 more fuel container to reach a speed of 2m/s (since it’s 1m/s w.r.t his new “rest” frame).”

It does exactly that. The rocket is different from the car in that the car reacts against the ground. If the ground is moving it is harder to react against it. The rocket is always reacting against fuel that is motionless in the rockets frame.

It looks like the rocket is getting the better deal here. But that is negated by the fact that the rocket has to carry a huge mass of fuel with it in order to have fuel in its frame of reference.
#25 Duae Quartunciae
March 20, 2009

You can certainly compare frames of reference; and if you do the calculations you find energy and momentum conserved in any frame you like to consider.

In both cases (rocket, and car) the system gains a certain amount of energy from consuming fuel. You calculate the same CHANGE in total energy no matter what frame is used; but in different frames the distributions of energy may be different, and the total energy is different, because the velocities depend on the frame.

In the car example, the car is pushing on a very massive object (the highway) that is moving relative to the car. The rocket, however, is pushing on a very light object (its fuel) that is initially at rest with respect to the rocket.

In all cases and in any frame, momentum is conserved; and a transfer of equal momentum but opposite is made between car and road, or rocket and exhaust. When equal amounts of momentum are involved, a lighter object has the greater change in kinetic energy, and a rapid object gets the greater change in kinetic energy. The classical invariant for an object is 2Em = p^2, where E is kinetic energy, m is mass, and p is momentum. This follows from p = mv and E = 0.5*mv^2.

Hence: from the point of view of the car (initial frame), the rapidly moving highway gets nearly all the energy (200 zaps) while the car gets 1 zap. From the point of view of the road (initial frame), the rapidly moving car gets 201 zaps, with pretty much nothing for the road as it recoils because it is so massive and at rest.

From the point of view of a rocket (the frame in which it is initially at rest) with exhaust expelled at 1000 m/s, accelerating to 1 m/s gives the rocket a zap of energy, and the exhaust gets 1000 zaps. (1000 times lighter)

But in a frame where the rocket moves from 1m/s to 2m/s, the rockets gains 3 zaps (2^2 – 1^2), while the exhaust gains (999^2 – 1)/1000 = 998 zaps of energy.

Same in any other frame. The total energy of the fuel is all accounted for is independent of the frame, for both the car, and the rocket.