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Author Topic: How do equal and opposite reaction forces work?  (Read 5478 times)

David Cooper

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How do equal and opposite reaction forces work?
« on: 24/09/2013 17:35:43 »
If an arrow is shot from a bow which is of equal mass to the arrow and this takes place in space without anyone holding the bow, the bow and arrow should move apart in opposite directions at the same speed.

Now increase the mass of the bow without changing its power. If the bow weighs a thousand times as much as it did before, it will hardly move at all. The result of it staying almost stationary would appear to be that more energy will be transferred to the arrow because in the later stages of energy transfer from string to arrow the string will be moving forwards more quickly than on a bow which has begun to move backwards. In this case, the arrow must surely move faster and carry more energy than in the original case at the top, so the bow must move the opposite way with less energy.

How is this compatible with the law that every action has an equal and opposite reaction?
« Last Edit: 25/09/2013 19:05:43 by chris »

Pmb

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Re: Equal and opposite reaction?
« Reply #1 on: 24/09/2013 17:59:02 »
Quote from: David Cooper
If an arrow is shot from a bow which is of equal mass to the arrow and this takes place in space without anyone holding the bow, the bow and arrow should move apart in opposite directions at the same speed.
Correct.

Quote from: David Cooper
Now increase the mass of the bow without changing its power. If the bow weighs a thousand times as much as it did before, it will hardly move at all. The result of it staying almost stationary would appear to be that more energy will be transferred to the arrow because in the later stages of energy transfer from string to arrow the string will be moving forwards more quickly than on a bow which has begun to move backwards. In this case, the arrow must surely move faster and carry more energy than in the original case at the top, so the bow must move the opposite way with less energy.

How is this compatible with the law that every action has an equal and opposite reaction?
Action/Reaction pairs refers to force. The force on the bow due to the arrow is equal in magnitude and opposite in direction to force on the arrow due to the bow. Since they bow has a large mass the force on it will make it move slowly. The same force acting on the arrow will make it move quickly for the same reason.

David Cooper

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Re: Equal and opposite reaction?
« Reply #2 on: 24/09/2013 18:11:42 »
The part I'm having difficulty with is that the bow releases a fixed amount of energy. In the case where bow and arrow have equal masses, half of that energy will end up in the movement of the bow and half in the movement of the arrow. If the bow has a much higher mass though, more than half of the energy appears to end up in the movement of the arrow, so less than half of the energy should end up in the movement of the bow. That does not seem to be equal.

Pmb

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Re: Equal and opposite reaction?
« Reply #3 on: 24/09/2013 18:57:48 »
The part I'm having difficulty with is that the bow releases a fixed amount of energy. In the case where bow and arrow have equal masses, half of that energy will end up in the movement of the bow and half in the movement of the arrow. If the bow has a much higher mass though, more than half of the energy appears to end up in the movement of the arrow, so less than half of the energy should end up in the movement of the bow. That does not seem to be equal.
I think that you're confusing equal force with equal energy. It's the forces that must be equal and opposite, not the energies.

Try working out some  numbers. It'll clear some of this up. Try letting a large body collide with a small body with the large body at rest. See what happens to the energy of each body. Assume the collision is elastic meaning that kinetic energy is conserved.

David Cooper

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Re: Equal and opposite reaction?
« Reply #4 on: 24/09/2013 19:22:46 »
Maybe the mistake I've been making is to think that the momentum of the arrow should be the same as the momentum of the bow, because maybe it shouldn't. If the bow is heavier, the arrow will move off at higher speed and must also have higher momentum. So the real question is about how the forces are equal and opposite.

If I redesign the question, we could have two masses with a compressed spring sitting between them and touching both. If the spring is released such that it can extend, the two objects will be pushed apart. If they are of the same mass they will be pushed away at the same speed as each other and in opposite directions. If one has greater mass than the other, the less massive one will be accelerated more and the spring itself will end up moving after it rather than remaining still. The spring must still deliver the same force as before, and yet surely it must apply more force to the lighter of the two masses in order to make that object move at a higher speed and with greater momentum. It looks as if the forces are only equal from the perspective of the shifting frame of reference of the spring.

Pmb

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Re: Equal and opposite reaction?
« Reply #5 on: 24/09/2013 19:29:42 »
Maybe the mistake I've been making is to think that the momentum of the arrow should be the same as the momentum of the bow, because maybe it shouldn't.
No. You have it right. Before the arrow is released the total momentum of the system is zero. Suppose the system is left hanging from a string and the arrow released by an auto release switch. Then since momentum is conserved the conservation of momentum reads

m1 v1 = m2 v2 = 0   ==>   m1 v1 = -m2 v2

Play around with that for a bit.

David Cooper

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Re: Equal and opposite reaction?
« Reply #6 on: 24/09/2013 19:47:14 »
m1 v1 = m2 v2 = 0   ==>   m1 v1 = -m2 v2

Play around with that for a bit.

That doesn't appear to address the issue.

David Cooper

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Re: Equal and opposite reaction?
« Reply #7 on: 24/09/2013 20:36:06 »
There's something going on with the efficiency of the transfer of the force.

(1) If instead of two masses with a spring between them you just have one mass and a spring, the spring will release all its energy without transferring any force. That energy will presumably be lost as heat generated in the spring as its movement gradually dies down.

(2) With two equal but tiny masses around the spring, they will be flung off at high speed but most of the energy will still be lost as heat generated in the spring.

(3) With two big masses on either side of the spring, most of the energy could be transferred to them instead with very little lost as heat.

(4) With one big mass and one tiny one, again most of the energy would be lost as heat, but the tiny mass would be flung away about twice as fast as in case (2). In this case, (4), more of the energy from the spring is being converted into momentum and less lost in heat than in case (2).

What this means in the case of the bow and arrow is that more energy from the bow is being transferred into the movement of both arrow and bow in the case where the bow is made to be more massive for the same power. It apears that the more mass involved in the system, the more energy ends up being turned into movement energy rather than being lost as heat.

CPT ArkAngel

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Re: Equal and opposite reaction?
« Reply #8 on: 24/09/2013 20:45:17 »
Momentum is always conserved. Action = Reaction even with a spring, just make an experiment and you will see... Don't think in terms of energy but momentum. When you look at energy, you must take account of all mechanisms and see it as a whole.
« Last Edit: 24/09/2013 20:53:45 by CPT ArkAngel »

David Cooper

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Re: Equal and opposite reaction?
« Reply #9 on: 24/09/2013 21:02:17 »
Momentum is always conserved. Action = Reaction even with a spring, just make an experiment and you will see... Don't think in terms of energy but momentum.

I'm thinking in terms of both, and I have given you the experiment. Let me give the masses actual values this time: A=2kg and B=1kg. We put a powerful spring between them, held compressed by something that can be released in an instant. We trigger the release and the 1kg mass, B, shoots off at speed X and momentum Y. If we now repeat the experiment with the 2kg mass replaced with another 1kg mass, C, the original 1kg mass B now goes off at a speed lower than X and a momentum lower than Y.

What this means is that if the momentum of B = the momentum of A in the first case and the momentum of B = the momentum of C in the second case but the momentum of B is different in each of the two cases, then the total energy delivered from the spring to the two masses must be different in each case.

CPT ArkAngel

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Re: Equal and opposite reaction?
« Reply #10 on: 24/09/2013 21:29:17 »
If you have the same amount of energy in the spring, neglecting heat loss, you will get same energy and different momenta because the masses and velocities are different. But the total momentum is still 0.

Remember, momentum is mv and kinetic energy is mv^2.
« Last Edit: 24/09/2013 21:34:32 by CPT ArkAngel »

David Cooper

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Re: Equal and opposite reaction?
« Reply #11 on: 24/09/2013 21:31:39 »
So the energy isn't balanced both ways at all. The same energy is being delivered from the spring in each case, but when there are equal masses at either side of the spring, the same amount of energy goes both ways. With unequal masses at either side, more energy goes into the lesser of the two masses, so it goes off with higher speed and higher momentum. If the momentum of the higher mass object is to match that of the lesser mass object, then the momentum of that will be higher too, but it doesn't matter that these values are both higher because all that counts with momentum is that they cancel each other out.

David Cooper

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Re: Equal and opposite reaction?
« Reply #12 on: 24/09/2013 21:35:42 »
In the case of a spring pushing either nothing or a tiny mass, the loss of energy as heat results from the mass of the spring being accelerated and then not being allowed to move away. If the mass of the spring is eliminated, all the energy will be transferred to the tiny mass and it will shoot off much faster and with no heat being generated, so that's a different situation. Things are becoming clear now. Thanks for your help.

CPT ArkAngel

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Re: Equal and opposite reaction?
« Reply #13 on: 24/09/2013 21:38:26 »
So the energy isn't balanced both ways at all. The same energy is being delivered from the spring in each case, but when there are equal masses at either side of the spring, the same amount of energy goes both ways. With unequal masses at either side, more energy goes into the lesser of the two masses, so it goes off with higher speed and higher momentum. If the momentum of the higher mass object is to match that of the lesser mass object, then the momentum of that will be higher too, but it doesn't matter that these values are both higher because all that counts with momentum is that they cancel each other out.

Yes! The stress is higher when you have one heavier mass...
« Last Edit: 24/09/2013 21:40:05 by CPT ArkAngel »

Ethos_

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Re: Equal and opposite reaction?
« Reply #14 on: 24/09/2013 21:49:31 »
So the energy isn't balanced both ways at all.

The energy that's available is contained within the flexed bow and transmitted thru the string to both arrow and bow. The energy of the bow is thus divided between itself and the arrow. In the case where the bow is much heavier than the arrow, the energy available has not changed. But because it takes much more energy to move the bow the same distance as the arrow, more momentum is transferred to the arrow. When calculated, the action and reaction are still equal.

alancalverd

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Re: Equal and opposite reaction?
« Reply #15 on: 24/09/2013 22:55:20 »
Energy is conserved. The potential energy of the bow E is shared as kinetic energy between the bow MV2/2 and the arrow mv2/2.

Momentum is conserved. MV = -mv (pedantic minus sign - remember v is a vector)

So now you have two simultaneous equations from which, knowing M and m, you can calculate the ratio of their subsequent velocities, and if you knew the original potential energy E = (pull force x pull distance) you could calculate their actual speeds.

A neat question! Once upon a time, it could have featured in an O level exam. Nowadays I guess you could get a degree for solving the equations.
« Last Edit: 24/09/2013 22:57:37 by alancalverd »

David Cooper

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Re: Equal and opposite reaction?
« Reply #16 on: 24/09/2013 22:57:09 »
I now realise that I'm still not quite there with this.

If equal masses are used (e.g. 1kg on each side), half the energy goes into each. If unequal masses are used (e.g. 2kg and 1kg masses), more of the energy goes into the lighter one. If this lighter one is the same as the masses in the experiment with equal masses (i.e. 1kg), it will go faster than either of them did, so a greater force must have been applied to it. If the momentum is the same for it and the heavier mass and the force applied in each direction is the same, doesn't that mean that more force is applied in the case where a heavier mass is involved? It looks as if the spring can only deliver a fixed amount of force and that different amounts are applied in different directions when the masses are not equal.

David Cooper

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Re: Equal and opposite reaction?
« Reply #17 on: 24/09/2013 22:59:51 »
A neat question! Once upon a time, it could have featured in an O level exam. Nowadays I guess you could get a degree for solving the equations.

I've been tripping over this problem in different forms over many years and until today I never got round to asking for help clearing it up. Things are beginning to fall into place at last. I hadn't realised how much momentum was an abstraction.

alancalverd

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Re: Equal and opposite reaction?
« Reply #18 on: 24/09/2013 23:02:38 »
No problem. The force is the same in both directions (Newton's third law) but the acceleration and hence the final velocity differs because the masses differ.

That's why shot putters and Sumo wrestlers are really big guys. If you want to throw something a long way, you need to be a lot heavier than the thing you are throwing.

Late edit: Momentum isn't an abstraction! It's a property of anything that moves! It's the whole reason that aeroplanes and rockets fly, and the earth doesn't hurtle backwards when you walk! (it moves, but not a lot).

One of the many problems with the National Curriculum is it drivels on about energy (without defining it) from Year 1, but doesn't mention momentum because politicians can't tax it, so there's no point in teaching kids about it.
« Last Edit: 24/09/2013 23:08:59 by alancalverd »

David Cooper

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Re: Equal and opposite reaction?
« Reply #19 on: 24/09/2013 23:11:55 »
No problem. The force is the same in both directions (Newton's third law) but the acceleration and hence the final velocity differs because the masses differ.

From the point of view of the 1kg mass, in one case it's accelerated to one speed (1kg mass at other end of the spring) and in the other case (2kg mass at the other end) it's accelerated to a higher speed in the same amount of time. That has to register as a higher force being applied to it.

Quote
Late edit: Momentum isn't an abstraction! It's a property of anything that moves! It's the whole reason that aeroplanes and rockets fly, and the earth doesn't hurtle backwards when you walk! (it moves, but not a lot)

I'm not convinced that it's anything other than an abstraction. There's energy being transferred, so isn't that enough to account for all those things? They keep on going until that energy is scrubbed off.

alancalverd

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Re: Equal and opposite reaction?
« Reply #20 on: 24/09/2013 23:24:47 »
From the point of view of the 1kg mass, in one case it's accelerated to one speed (1kg mass at other end of the spring) and in the other case (2kg mass at the other end) it's accelerated to a higher speed in the same amount of time. That has to register as a higher force being applied to it.

What makes you think that? Nobody has mentioned time, and there's no reason to suppose that the acceleration time is the same for both experiments.

If you start with the known conservation equations, you can calculate the acceleration time from the final velocities. Or just suppose that M is very large, so V is very small (i.e. the earth doesn't move): obviously the spring has to move twice as far in the direction of v, so the force acts on m for longer, than if M=m. a=F/m, v = at. More t => more v.

jeffreyH

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Re: Equal and opposite reaction?
« Reply #21 on: 25/09/2013 16:49:52 »

David Cooper

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Re: Equal and opposite reaction?
« Reply #22 on: 25/09/2013 17:11:21 »
What makes you think that? Nobody has mentioned time, and there's no reason to suppose that the acceleration time is the same for both experiments.

So it takes longer then but the force is no stronger - it just adds up to more energy transferred over time. Okay - nearly there now.

The problem I've had with all this is that the way it was taught at school left me with the idea that everything was equal, but it isn't: more energy can go one way than the other, and it's only terms of momentum that there is equality, and momentum appears to be nothing more than an abstraction which is useful for calculating how the energy will be distributed during an interaction.

It's now obvious to me that the amount of energy going in different directions can be extremely unequal. If two masses of a ton are drifting together in space, almost touching but separated by a compressed spring, that spring may take hours to push them a tiny distance apart, so if you watch it for a minute you will see virtually no energy transferred as the spring is unable to release it. If you replace one of the ton masses with a little ball bearing, it will obviously pick up almost 100% of the energy from the spring and ping off into the distance while the remaining ton mass will recieve practically 0%. What has always misled me in the past was the idea that momentum is something fundamental and that there is real equality in the way energy is distributed, but it's surely energy that is fundamental with momentum being a mere abstraction, albeit an extremely useful one.

David Cooper

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Re: Equal and opposite reaction?
« Reply #23 on: 25/09/2013 17:31:12 »

http://cnx.org/content/m42073/latest/

I have a mental block with anything where symbols start to dominate because I can never remember what they are for. I have no problem with maths if it's written up in ordinary language so that the full functionality is clear at every step, but the meanings of squiggles simply refuse to fix themselves in my head and are lost within minutes every time. Near the top of that page it also makes some distinction between Fnet=ma and Fnet=ma which means nothing to me and appears to depend on some difference between ordinary letters and italics which I've never heard of.
« Last Edit: 25/09/2013 17:33:21 by David Cooper »

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Re: Equal and opposite reaction?
« Reply #23 on: 25/09/2013 17:31:12 »