Suppose we have an object in space. say gravity is negligible, and the thing has a rocket on both sides. the thing fires one rocket for, say, 5 seconds. now it's moving. The object has kinetic energy. Now the other rocket fires for 5 seconds. Then it's stationary. where does the energy go? I know that some of the energy from the rockets would just go into space as heat and kinetic energy of the hydrogen particles flying out, but how does there seem to be more energy in the situation if the first rocket just goes for 10 seconds than if one goes for 5 and the other goes for 5?

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    $\begingroup$ The energy is in the exhaust of the rocket engine. $\endgroup$
    – HDE 226868
    Sep 6, 2015 at 18:30
  • $\begingroup$ Can you explain this better: "but how does there seem to be more energy in the situation if the first rocket just goes for 10 seconds than if one goes for 5 and the other goes for 5?" What do you mean by seem? and What exactly is happening in the second situation? $\endgroup$ Sep 6, 2015 at 19:15
  • $\begingroup$ I am with you. In both cases it seems like the same amount of work is done on the rocket but the rocket does not end up with the same amount of kinetic energy. $\endgroup$
    – paparazzo
    Sep 6, 2015 at 19:46
  • $\begingroup$ @Ernie No kidding you have different kinetic energy if rockets same versus opposite. If the same amount of work is done how do you explain different kinetic energy? $\endgroup$
    – paparazzo
    Sep 6, 2015 at 20:12
  • $\begingroup$ @Frisbee You have same energy in both cases. You are assuming a symmetry that isn't there. Work out the exact velocities (strictly momenta) of all objects in both cases. $\endgroup$ Sep 8, 2015 at 8:39

4 Answers 4


I think that previous answers ignore the real element of the question - where does the KE of the object go. The exhaust KE is different in acceleration and deceleration - and the KE of the object is transferred into KE of the exhaust.

Consider a very simplistic rocket that is at rest and fires out particles at velocity V - that V is an invariant - but the velocity of the particles compared to an object that is also at rest is V - v where v is the velocity of the object and increases from 0 to the final velocity of the object. During the deceleration, the particles are fired out at velocity V + v where v decreases from that final velocity to 0. Particles that are fired out in acceleration have less energy (compared to an object at rest) than particles fired out in deceleration - because they are carrying some of the KE of the rocket.

In a real rocket - the exhaust gases will have a range of different velocities - but the average velocity of the exhaust in acceleration will be < the average velocity of the exhaust in deceleration (when compared to a body at rest at the start of the acceleration).

  • $\begingroup$ I don't know this is correct but that is my best guess also $\endgroup$
    – paparazzo
    Sep 6, 2015 at 21:10
  • $\begingroup$ If the rockets fire simultaneously, the object has zero kinetic energy. The same total amount of force is applied if the rockets fire sequentially. In each case, the rocket ends with zero kinetic energy. In my answer, kinetic energy of deceleration goes into negative work. You're accounting for kinetic energy differently, but I think our answers may be saying the same thing in different ways. Please correct me if my understanding is wrong. $\endgroup$
    – Ernie
    Sep 7, 2015 at 12:51
  • $\begingroup$ This is along the right track, but there might be a bug in the velocities... (argument is a bit hard to follow). I try a simpler case answer below. $\endgroup$ Sep 7, 2015 at 12:58

Rocket engines work by conservation of momentum. Particles are accelerated and thrown out the back, and as a result the rocket is thrown the other way. The energy of the rocket is the same, except that the chemical potential of the rocket fuel has been converted to the energy of the particles flying away from the rocket.


Another way to account for this case is to consider that the work done by the exhaust KE has a different sign in acceleration and deceleration.

The work-energy theorem says that energy associated with work done on a rigid object by a net force, is transferred to the object as kinetic energy. The work that appears in the work-energy theorem is, thus, net work.

Net Work = Final KE - Initial KE

Let's say the object is at rest. The first rocket imparts kinetic energy to the object. If the object's velocity increases, final KE must be greater than initial KE. So far, no problem. The work done is positive. Work = force * displacement * cos 0 degrees = ∆ KE. In this case the sign of cos 0 is +1.

Now the 2nd rocket fires in the opposite direction. The energy associated with the force exerted by the 2nd rocket reduces the velocity of the object and brings it to its initial at-rest state. In this case the final KE is zero, so the work done is negative. The same work equation used when the first rocket fires applies, but the angle theta between force vector and displacement vector is 180 degrees, so the cos is -1, and the work done has a negative sign.

The net work done on the object by both rockets sums to zero, but the energy in the exhaust of the rockets didn't just disappear. It accelerated and decelerated the object, and the sum of the work done by the kinetic energy imparted to the object is zero.

  • $\begingroup$ Negative work? There is no such thing as negative work. $\endgroup$
    – paparazzo
    Sep 7, 2015 at 13:37
  • $\begingroup$ @Frisbee: Negative work is addressed in this question. Please scroll down to the answer by joshphysics. What I mean by negative work is that the angle between force and displacement is 180 degrees, so the cosine is -1, and the work formula W = f * d * cos theta, winds up with a negative sign. This is common and well-known terminology. $\endgroup$
    – Ernie
    Sep 7, 2015 at 14:15

Let's simplify the system. A large block (mass $M$) ejects small blocks (mass $m$) using springs (so instantaneous $\Delta v$). Let's consider all velocities in the Lab frame (Newtonian physics will do fine for this).

We will work out the energies in the two cases (both blocks in same direction versus blocks in opposite directions) and see if the energy is the same or different.

Case 1: Opposite directions

We eject the first small block and it moves off to the right with velocity $v_1$. The big block (with a second small block still attached, so mass $= M + m$) recoils to the left with velocity $v_0$.

Now we eject the second small block to the left. It shoots off with velocity $v_2 = v_2' + v_0$, where $v_2'$ is its velocity relative to the block. Note that $v_2' \ne v_1$ because the mass of the big block is now $M$ so it's easier to stop (we had to reduce the spring's force). The big block stops.

We now have a stationary big block and two small blocks moving off in opposite directions. However... the two small blocks do not have the same velocity! (The original question assumed a symmetry that, in fact, is not there).

If we calculate the energy, we now have:

$E_1 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 $

$= \frac{1}{2}m(v_1^2 + v_2^2)$

$= \frac{1}{2}m(v_1^2 + (v_0 + v_2')^2)$

$= \frac{1}{2}m(v_1^2 + v_0^2 + 2v_0v_2' + v_2'^2)$

Case 2: Same direction

Consider now what happens when the blocks are shot off in the same direction. The first block shoots off as before at $v_1$. The big block recoils at $v_0$.

Now we shoot off the second block. The second block now has a reduced velocity in the lab frame, because the big block is moving in the opposite direction, so $v_2 = v_2' - v_0$.

Now the energy calculation is:

$E_2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 + \frac{1}{2}M(2v_0)^2$

$= \frac{1}{2}m(v_1^2 + v_0^2 - 2v_0v_2' + v_2'^2) + 2Mv_0^2$

Now we equate the $E_1$ and $E_2$ case equations and see where that leads us. If it leads to something absurd, we know the energies are not equal (as the OP suspects).

$\frac{1}{2}m(v_1^2 + v_0^2 + 2v_0v_2' + v_2'^2) = \frac{1}{2}m(v_1^2 + v_0^2 - 2v_0v_2' + v_2'^2) + 2Mv_0^2$

$\therefore mv_0v_2' = -mv_0v_2' + 2Mv_0^2$

$\therefore mv_2' = Mv_0$

This is simply the equation of conservation of momentum for the 2nd small block shooting off the big block. Since we know this must be true, if must be that the energies in the two cases are actually the same.

I think the OP's mistake was to assume that the small blocks (in his case, the rocket exhausts) had the same momentum (and energy) in both cases. They don't.

  • $\begingroup$ Yes, this makes the problem clearer for me. My answer is deficient because it doesn't account for KE, only for work done on the object. $\endgroup$
    – Ernie
    Sep 7, 2015 at 14:54
  • $\begingroup$ Thank you, this makes more sense. so the KE that isn't in the rocket goes into faster exhaust particles. Thanks a ton, Owen Boyle! $\endgroup$ Sep 16, 2015 at 21:21

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