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So I get the general idea behind Newton's Third Law. But sometimes it becomes confusing. The thing i have observed about newtons laws are that they define something that is sometimes inately logical.

So imagine there are two balls in vacuum with the mass of 1 unit. Ball a hits ball b with a force of 10 N. As per conservation of momentum, the total momentum must remain the same. If ball a transfers all of its energy to ball b, then how does ball b applies the same force to ball a and it stays the same. Shouldnt according to newtons law, the ball a reflect back at the same speed. Doesnt that violate conservation of momentum and energy.

I understand that newtons third law is part of some deeper concept of symmetry. Is Newton law simply a way of thinking, that when ball a hits ball b that means that ball b is also hitting ball b.

Also since i know that newtons third law doesnt violate anything, how can force applied on two objects be same and their energy be different

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  • $\begingroup$ Look up Newton’s cradle or buy one. $\endgroup$
    – my2cts
    Commented Aug 31 at 16:59
  • $\begingroup$ Re your final paragraph, momentum change is force times time (impulse) while energy change is force times distance (work). That's why the two can be different. $\endgroup$
    – John Rennie
    Commented Aug 31 at 17:49
  • $\begingroup$ what does it mean to hit something with 10 newtons? $\endgroup$
    – JEB
    Commented Aug 31 at 18:16
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    $\begingroup$ Is Newton law simply a way of thinking, that when ball a hits ball b that means that ball b is also hitting ball b This is basically it, in my opinion. It is defining what a force or interaction is, and that it is inherently binary. If A pushes on B with 10 N, then inherently B also pushes on A with 10 N. $\endgroup$
    – RC_23
    Commented Aug 31 at 21:06
  • $\begingroup$ It is absolutely wonderful that two of the three answers below correctly point out that forces are the wrong way to look at Newtonian mechanics. The awkwardness of forces leads to the necessity of bringing in N3L, when in fact the conserved quantity is momentum, and then we would not need N3L. The answer that is not considering momentum, is also correctly pointing out a confusion by the OP, so all three deserved my upvotes. $\endgroup$
    – naturallyInconsistent
    Commented Sep 1 at 3:50

3 Answers 3

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If ball a transfers all of its energy to ball b, then how does ball b applies the same force to ball a and it stays the same.

The force ball b exerts on ball a does negative work on ball a to make it stop. Per the work energy theorem that work equals the change in kinetic energy of ball a, which is $-\frac{1}{2}mv^2$ where $v$ is the velocity of ball a prior to the collision. The negative work by ball b takes the kinetic energy away from ball a.

At the same time, the equal and opposite force ball a exerts on ball b does positive work on ball b, per Newton's 3rd law, doing positive work on ball b equal to the change in kinetic energy of ball b, $+\frac{1}{2}mv^2$, where $v$ is now the velocity of ball b following the collision.

Thus both momentum and kinetic energy are conserved for this perfectly elastic collision.

Hope this helps.

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  • $\begingroup$ so, from what i have understood. The whole point of newtons third law is the relationship of forces. when object a hits object b, energy is transfered, but object b also hits object a back, which is what causes it to stop. And if a human was part of newtons third law, it views the opposite stoping force as a reaction. So when i push wall continously, the wall pushing on me stops me, and i feel that force, i push against the ground, not solely because of the reaction, but because i also pushed against the ground to push against the wall $\endgroup$
    – Ahmed Nadeem
    Commented Sep 1 at 8:22
  • $\begingroup$ but wait when some ball hits a wall then it should stop why does it reflect back $\endgroup$
    – Ahmed Nadeem
    Commented Sep 1 at 8:39
  • $\begingroup$ It reflects back because the mass of the wall (plus the earth to which it is fixed) is so much greater than the ball. In your example only reason ball a stops is because it has the same mass as ball b. The “ball” could stop (stick to the wall) if the collision was perfectly inelastic, for example if the ball was made of putty. In all cases momentum has to be conserved. In the case of the wall/earth it moves but its movement is infinitesimal because it’s so massive. $\endgroup$
    – Bob D
    Commented Sep 1 at 10:41
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Nowadays a much favored approach is to port new insights back to Newtonian dynamics.

As you mention, there is the approach of emphasizing conservation principles and recognition of symmetry.

The assertion 'an object not subject to any force will travel in a straight line with uniform velocity' can be thought of as expressing a conservation: conservation of momentum of a single object. It can be thought of as an assertion of symmetry: everywhere space has the same symmetries as euclidean geometry.


The core assertion of newtonian mechanics is of course the force-acceleration relation $F=ma$. The relation is linear in the following sense: when you double the mass of the object then the same force results in half the acceleration.


The case of two objects exerting a force upon each other, changing each other's momentum, can also be thought of as an instance of conservation.

When two objects are exerting a force upon each other, for each the amount of change of momentum is the same.

With respective masses $m_1$ and $m_2$:

$$ \Delta m_1 \vec{v_1} + \Delta m_2\vec{v_2} = 0 \tag{1} $$

(1) can be thought of as expressing a conservation principle. And it can be thought of as expressing a symmetry, in the following sense: $m$ represents the inertial mass of the object. If inertia is the same everywhere then (1) will hold good



There is no need to put the historical formulation of newtonian mechanics front and center. I prefer to reformulate the concepts in terms of conservation principles and symmetries.

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Idealized examples of collisions can be thought as happening at one instant $t$. In this case we can say that the momentum is conserved, but as force is $F = \frac{dp}{dt}$, and the derivative is not defined when the function is not continuous, it is not appropriate to talk about forces.

Of course in a real situation there is a finite time interval of contact, where some elastic deformation happens. One way to realize that the magnitude of the force is equal is that, by the symmetry of the situation, the elastic deformation must be the same in each object if they are identical.

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