The amount of work done by a force on an object at rest can be expressed as $$\frac{1}{2}m{v_0}^2$$

Where $v$ is the final velocity of the object as measured by a reference frame A

Frame B is moving in the opposite direction with a velocity $v_1$

Before any work is done on the object, frame A measures the object to have zero kinetic energy. Frame B measures it has a kinetic energy of $$\frac{1}{2}m{v_1}^2$$

It shouldn't matter that they have different kinetic energies in different frames as energy is frame-dependent.

However, let's say that a force does work on an object to change its velocity by $v_0$. Frame A measures the work needed to make this change in velocity as $$\frac{1}{2} m {v_0}^2$$

But frame B will measure that the work needed to make this change in velocity as $$\frac{1}{2}m(v_1+v_0)^2 -\frac{1}{2}m{v_1}^2$$ (which is the difference in kinetic energy)

There is a difference of $2v_0v_1$ between the amount of work required for the same change in momentum in each frame? This seems completely absurd, have I made an error in my experiment/logic?

  • $\begingroup$ Even in a single reference frame, we don't expect there to be a given amount of work corresponding to a given change in momentun, do we? Work is non-linear in momentum (i.e. velocity) after all. $\endgroup$ Jun 7, 2021 at 12:35

2 Answers 2


I think in this case it is necessary to do a closer examination of the concept of kinetic energy.

Kinetic energy attributed to an object is kinetic energy with respect to some specified coordinate system.

The kinetic energy attributed to an object is the amount of work that must be done to bring the object to a standstill.

Let's say we have an object, moving at velocity, and the object is slowed down by ribbons that are positioned perpendicular to the direction of velocity. (Think of the type of ribbon at the finish line of a running event.) The ribbons stretch, and then snap.

For the sake of the thought demonstration we treat the ribbons in a highly idealized way. We treat the mass of the ribbons as negligable, so that only the elasticity of the ribbons changes the velocity of the object. Just as each ribbon snaps the next ribbon is touched so that the decelerating force is effectively constant, resulting in uniform deceleration. Let the ribbons be spaced one meter apart.

To make the numbers simple we use a deceleration of 2 $m/s^2$. If the object comes in with a velocity of 4 m/s then it takes 2 seconds to come to a standstill, and over the course of those 2 seconds 4 ribbons will be snapped.

If the object comes in with double that velocity, 8 meters per second, then the total duration until standstill will be double too: 4 seconds. On the other hand: the amount of ribbons snapped will be quadrupled: 16 snapped ribbons.

Also: in the first half of the time 3/4th of the total number of snapped ribbons is snapped, in the second half of the time the remaining 1/4 of the total number of snapped ribbons is snapped.

Acceleration/deceleration is the second derivative of position, that is what gives rise to that quadratic relation between coming to a standstill and distance traveled.

To your question about frame A and frame B:

For the object to come to a standstill with respect to frame A requires a certain deceleration distance, and to come to a standstill with respect to frame B requires a certain deceleation distance, and those two distances are not the same.

Incidentally, I think the following is another way of casting the apparant paradox that you describe:

Take a train of flatbed cargo wagons, with connecting ramps such that a car can travel over the train of cargo wagons, as if travelling over a road.

Bring the train to, say, 4 m/s velocity. So: work is done, giving the car kinetic energy relative to the ground. Then start giving the car a velocity with respect to the train, accelerating to 4 m/s. So: work is done giving the car kinetic energy relative to the train.

That car will run out of train, and will hit the ground. the kinetic energy relative to the ground is not the sum of the two kinetic energies mentioned above. Instead it's the kinetic energy of a relative velocity of 8 m/s


This seems completely absurd, have I made an error in my experiment/logic?

You have not made an error, the result that you have found is indeed correct and the only mistake is the mistaken feeling that it is absurd. Indeed, it is correct that the same $\Delta v$ corresponds to different $\Delta KE$ in different frames.

The reason that this most likely feels "absurd" is that it is difficult, on its face, to reconcile such a result with the conservation of energy. To understand this it is useful to consider a “toy model” that is easy to analyze. Specifically, let’s consider an assembly consisting of a 10 kg base and a 1 kg projectile which is attached to a massless spring.

Suppose this spring has enough energy stored that when the assembly is initially at rest it can propel the base to 1 m/s, and by conservation of momentum the projectile is propelled to -10 m/s. Conversely, if the assembly starts at 5 m/s then after launching the projectile the base is propelled to 6 m/s and the projectile moves at -5 m/s.

So now let’s check energy. In the first case the KE of the base increased from 0 J to 5 J, while in the second case it increased from 125 J to 180 J. The spring stores the same amount of energy in both cases, so why does the KE increase by 5 J at the low speed and by 55 J at the high speed?

Notice that we forgot to calculate the energy that went into the projectile. This is the pivotal mistake of most such analyses. In the first case the KE of the projectile increased from 0 J to 50 J, while in the second case the KE was 12.5 J before and after. So in both cases the total change in KE (both the base and the projectile) was 55 J.

At low speeds most of the spring's energy is “wasted” in the KE of the projectile. At higher speeds more goes into the base and less into the projectile. Both energy and momentum are conserved, and in fact more power is delivered to the base as the speed increases under constant thrust.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.