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Assume that I was in space, where there is no friction. Also assume that there were two objects at rest: myself and a matchbox. Does this mean that, by Newton’s Third Law, since there is no friction, if I push on the matchbox, then the matchbox and I will float apart with the same force/velocity/acceleration (not sure what the correct term is here?)?

My understanding of Newton’s Third Law is that, if I tried the same experiment on Earth, with a box of matches sitting on a table, the matchbox would move and I wouldn’t because, although it exerts an equal and opposite force, some of that force is lost due to friction, and so I remain motionless whilst the matchbox moves?

I would greatly appreciate it if people could please take the time to clarify my understanding/intuition.

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You're close, yes.

If you pushed a matchbox (or anything else actually) in empty space, you bouth would feel the same FORCE. Not acceleration, not velocity, Newton's law says ·FORCES are equal in magnitude and opposite", it talks about forces.

But, force is mass times acceleration, so $$a=\dfrac{F}{m}$$. This means that you will accelerate much less than the box (because I assume that the matchbox is considerably lighter than you)

So yes, same force on both of you (in opposite directions), but the more mass, the less acceleration, as intuition says.

See for example: a shot bullet has much more acceleration than the gun, because the gun is much more massive, so it returns with less acceleration. The force is the same however.

Wait, if the force is the same, why don't you get injured? Because all that force is divided into the whole surface of your arm/hand, whereas the bullet exerts all the force in a tiny tip, so that's much pressure.

So keep in mind:

Equal forces (in magnitude, but opposite directions). Accelerations depend inversely on masses.


Now let's go to your second question: why don't you see this on Earth? Well, you do, as shown in the gun example. But why don't you see that with the matchbox?

Well, you do see it. But, when you push the box forward, you experience a force backwards, on your finger. This force is really small, so it cannot move you. The force is "absorbed" by your muscles, the floor, and so on. No problem.

Nevertheless, if you try to push something heavy, you definitely feel the force backwards. The best example is pushing a wall, or the floor. Try to do push-ups. When you push the floor, you lift yourself, and that's why you're interchanging forces with the building. Since the building is extremely heavy, its acceleration is negligible (you won't see it moving, and, if you do, get out of it NOW haha). The force can be "absorbed" by the soil as well.

However, that force is able to create more acceleration on you, smaller mass.

This happens everywhere. In fact, now you're attracting Earth as much as it is pulling you to its centre. But there are about 22 orders of magnitude of mass difference, so the Earth's acceleration is completely negligible and strongly compensated by other forces. In contrast, you are a tiny lonely mass where nothing else can prevent you from falling down to Earth.

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Yes,both of you will feel equal and opposite force .But as we know $F = m \times a$ and your mass is very high as compared to matchbox. So the matchbox will move with a significant acceleration but the acceleration of your body will be very very less. But this force which acts upon both of you will be equal to the force applied by you on matchhbox(third law of motion) .So a very large force on matchbox will give your body a significant acceleration

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  • $\begingroup$ Tip: use dollar symbols to enclose formulas so that they display beautifully. Use laTex notation inside them. Read about MAthJax for more info. $\endgroup$ – FGSUZ Oct 2 '18 at 13:59
  • $\begingroup$ Thanx, but i m not good with those symbols $\endgroup$ – Sourabh Oct 2 '18 at 14:00
  • $\begingroup$ Haha what? It's very easy, just try. You'll soon get used to them. Have a look at the site, they're everywhere, and that's becasue you cannot do physics without math (almost). $\endgroup$ – FGSUZ Oct 2 '18 at 14:02
  • $\begingroup$ Ok i will try those symbols and representation $\endgroup$ – Sourabh Oct 2 '18 at 14:03
  • $\begingroup$ I edited your formula to show you how it looks. Edit it again to see how it works. $\endgroup$ – FGSUZ Oct 2 '18 at 14:16
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On earth we have multiple factors which will make the result appear different from space.
We have

  • Gravity
  • Friction
  • And maybe air?

    With these things around us, you may get different results.

    If we shoot a bullet in space then accoring to the first law, the bullet will forever travel with the same speed and the same direction in space unless a foregin object interacts But not on earth as we have these disturbances above.

    Your acceleration while you push the matchbox will be
    $$a = {F\over m}$$
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  • $\begingroup$ Not "after", rather "while". After your push ceases, reaction ceases to, and both accelerations go to 0. $\endgroup$ – Elio Fabri Oct 2 '18 at 14:16
  • $\begingroup$ @ElioFabri true $\endgroup$ – weegee Oct 2 '18 at 14:20
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In the first case a way to relate what happens is to realise that Newton’s third means that the matchbox exerts a force on you and you exert an equal magnitude and opposite direction force on the matchbox.
Now add that these forces act for the same time and this means that the product of force times time (actually an integration is required if the force varies with time) which is called the impulse changes the magnitude of the momentum of both the matchbox and you by the same amount.
If you and the matchbox started from rest then $$m_{\rm you} v_{\rm you,final} = m_{\rm matchbox} v_{\rm matchbox,final} $$ and so the ratio of the final speeds is inverse to the ratio of the masses.
If you have a mass of $50\,\rm kg$ and the matchbox has a mass of $50\,\rm g$ your speed will be $\frac{1}{1000}^{\rm th}$ that of the matchbox.

Now in the second case there are two Newton third law pairs to consider:

  • The matchbox exerts a force on you and the Earth which is connected to you and you, and the Earth exert an equal magnitude and opposite direction force on the matchbox.
  • The kinetic frictional force exerted by the matchbox on the table which is connected to the Earth and you, and the equal magnitude and opposite direction kinetic frictional force exerted by the table which is connected to the Earth and you, on the matchbox.

In this case if the force that you exert on the matchbox is the same as before then the net force on the matchbox is smaller than before.
If in turn the contact time is the same as before then the impulses are smaller than before and hence the change in momentum is smaller than before.
In this case
$$m_{\rm table+Earth+you} v_{\rm table+Earth+you,final} = m_{\rm matchbox} v_{\rm matchbox,final} $$
As the mass of the Earth is approximately $6\times 10^{24}\,\rm kg$ you will infer that there will be no discernible movement of the Earth (and the table and you).

You can compare accelerations by noting that acceleration is the rate of change of velocity.

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