Does Newton's third law violate the law of energy conservation?

Question

Consider two bodies of same mass A and B kept at rest in contact with each other. Body A exerts force $F$ on body B which is accelerated to velocity $v$ by expending its energy by $K$ units. A reaction force $-F$ acts on A, and since it has same mass, it also gains kinetic energy $K$ as it moves with velocity $v$ in the opposite direction. So, the total energy of the system is the sum of energies of A and B or, $2K$. If the energy $K$ that A expended was given from outside, an extra $K$ amount of energy is present due to the reaction force. This is contradictory and violates the law of conservation of energy, though momentum is conserved.

Where am I wrong?

Clarification:

After reading the answers, I arrived at the conclusion that if A had some potential energy $K$ which is converted to kinetic energy, then both A and B will have equal shares of the energy, i.e., $\frac{K}{2}$; where $K = mv^2$. Is this understanding correct?

Answer 1

If the energy $K$ that A expended was given from outside, an extra $K$ amount of energy is present due to the reaction force.

This assumption is your error. If the force is acting from outside the system, then the system is not closed and there is no energy conservation. This also contradicts your initial statement that

$A$ exerts force F on body B (and vice versa)

implying an internal force and closed system. If that is true, then the full energy cost $2K$ must be paid by whatever is exerting the forces. This could for example be a spring between the bodies or tension in the bodies itself from being pressed together previously.

Answer 2

Since A and B are initially at rest when in contact with one another, clearly their interaction does not involve a collision. Assuming that, in addition, there is no net external force acting on the system of A and B, then the AB system cannot gain kinetic energy due to net external work being done on the AB system.

In the absence of any other information on the nature of the force that A applies to B, it appears that the scenario would necessarily involve the conversion of some form of potential energy (elastic, chemical) stored with A, to the kinetic energy of A and B. One possibility is A is initially kept in a compressed state.

Should A then somehow expand (decompress) it would exert a force on B which would in turn exert an equal and opposite force on A per Newton's third law. Given equal masses, conservation of momentum would dictate that the resulting velocity of A would be equal and opposite to the velocity of B, with a total kinetic energy of A and B equal to the initial elastic potential energy of A.

Hope this helps.

Answer 3

No, you have misunderstood. If A accelerates B, it can only do so by decelerating, so while B gains kinetic energy, A loses it, the overall energy remaining the same.

Answer 4

What you in fact have shown here is that your initial assumption cannot be correct. It is not possible that A expends exactly the amount of energy $K$, if it also delivers that amount of energy $K$ to B. In your clarification you have fixed this wrong assumption, realizing that A in fact must exert $2K$ of energy in order to supply $K$ of energy to B (or equivalently, that A exerts $K$ of energy and supplies $K/2$ to B).

So let's consider the process here: you made a hypothesis, an expectation. You then applied it to the know laws of the universe. You realized that this leads to a contradiction. Now, either the laws of the universe are wrong, or your initial assumption was wrong. And we found the issue to be with your assumption, which was then corrected and applied to the laws once again, this time successfully. That is the scientific method in glory 🙌

Answer 5

Please note, this answer went through a complete rewrite after I found a serious mistake. I apologize for any confusion!

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Let's assume that, in accord with your assumptions, the two bodies were initially at rest and in contact with one another. Suddenly one of them "explodes", so for some period of time, there was an explosive force, creating an action/reaction pair. So we have $F_{12}=-F_{21}$ and for simplicity (and without loss of generality) let's also put $F=F_{12}$ with the assumption that $F\gt 0$.

Now, the explosive force did some positive work $F_{12}\Delta{x}$ on the right mass as it displaced it by $\Delta{x}$ to the right, and a positive work of $F_{21}\cdot(-\Delta{x})$ on the left mass as it displaced it by $\Delta{x}$ to the left (recall $F_{21}<0$). Overall then the work done by the explosive force is: $$W=F_{12}\Delta{x}+F_{21}(-\Delta{x})=2F\Delta{x}$$

Now, since this necessarily is the total energy expanded in the explosion, we will set it equal to the quantity you defined as $K$ (not denoting kinetic energy in particular). If we then by symmetry further see that both masses had to gain speed equal in magnitude and opposite in sign, the final KE of the system must be $$\text{KE} = \frac{1}{2}mv^2+\frac{1}{2}mv^2 = mv^2$$ Where $v$ is the absolute value of the final speed of the masses due to the explosion.

Since by energy conservation, $KE=W$, the conclusion is that:

$$ mv^2 = 2F\Delta{x} = K $$

Which of course means that, the kinetic energy gained by each mass is given by:

$$ \frac{1}{2}mv^2 = \frac{1}{2}K$$

We see then that in such an exchange, each mass gains only half of the released potential energy $K$, and there is no contradiction with Newton's third law which we have made use of in setting up the problem.

Answer 6

Energy conservation and linear momentum conservation are two indipendent laws, which need to be verified indipendently.

What is true is that the law of action and reaction, i.e. Newton's third law, implies the conservation of momentum. That is because the force is equal to the time derivative of momentum.

But this same law tells you nothing about energy variation. In the process you describe for example, momentum is conserved. For it to be possible, energy must also be conserved (it's another condition you see). There are many ways this could be achieved, if the force is conservative we might say that some potential energy (which was there since the beginning) gets converted into kinetic energy. If it's not conservative, energy must come from somewhere anyway. If you describe the process of an explosion for example, the two fragments' energy comes from chemical potential energy.

Answer 7

If $A$ and $B$ are initially at rest, $A$ cannot exert a force $F$ on $B$. This is because the work done on $B$ by the force $F$ has to come from the kinetic energy of $A$, which it does not have.

If $A$ was moving at velocity $v_A$ and collided head-on with $B$ moving at velocity $v_B$, then it would do work $F_{AB} \cdot d$ on $B$, where $F_{AB}$ is the force exerted by $A$ on $B$, and $d$ is the distance for which $A$ and $B$ were in contact. This will increase the kinetic energy of $B$ by the same amount $F_{AB} \cdot d$, from the work-energy theorem.

Due to Newton's third law, $B$ will exert a force $F_{BA} = -F_{AB}$ on $A$, which does work $-F_{AB} \cdot d$ on $A$, which decreases the kinetic energy of $A$ by $F_{AB} \cdot d$, keeping the total kinetic energy of the system constant.

Note that if the velocities of $A$ and $B$ after collision become $v_A'$ and $v_B'$, the following condition holds:

$$\frac{1}{2}mv_A^2-\frac{1}{2}m{v'_A}^2 = \frac{1}{2}m{v'_B}^2 - \frac{1}{2}mv_B^2 = F_{AB} \cdot d$$

Answer 8

If the energy K that A expended was given from outside, an extra K amount of energy is present due to the reaction force.

If an outside object C exerted a force on A, then the reaction force would be from A to C, not from B to A.

Consider two bodies of same mass A and B kept at rest in contact with each other. Body A exerts force F on body B which is accelerated to velocity v by expending its energy by K units.

As you've shown, for A to accelerate to v by pushing of B, it must expend 2K of energy, not K. the smaller the reaction mass something uses to move, the higher a velocity it must impart on the reaction mass, and so the more energy must be expended. This is why planes need less power to generate the same lift when they go faster: they generate lift by pushing air down, and the faster they go, the more air they have to push down, so they have to expend less energy. The power consumption of a plane is the amount of energy is take to move the plane, plus the energy needed to push the air in the opposite direction.