How to understand Newton's 3rd Law

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In summary: So, even though the car travels at constant speed, the car is accelerating because it's direction of travel is constantly changing. It's constantly "turning" because it's constantly following the curve of the earth.The first car (the one that gets hit), is not moving, so when it gets hit it's just like you getting hit by a car while standing still. Your body accelerates backwards. The car accelerates backwards.When I throw a rock I cause a force on it and a rock causes equal force on me. So if I would stay at frictionless surface would I move backward when throwing rock?Yes.
  • #1
ultrauser
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I have problem undestanding Newton's III law. When i press the wall the wall react with the same force on me but as I have less mass I have greater acceleration so I am moving away from wall.

When I hit the wall according to Newton's III law the wall should react with the same force on me (and give me acceleration) but I am not moving(away from wall). Can someone explain it to me?

Also in the first example shouldn't I give a wall a little acceleration(even if a tiny bit) as well so why is the wall not moving at all?
 
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  • #2
However you exert a force on the wall--whether pressing or hitting--it will always exert an equal and opposite force on you.

But to determine whether something accelerates or not you must look at all the forces acting on it. (That's Newton's 2nd law.) Other forces act on you besides the force exerted by the wall. Friction from the floor, for one. Similarly, the wall is attached to the floor, which exerts a force on the wall to hold it in place.
 
  • #3
ultrauser said:
I have problem undestanding Newton's III law. When i press the wall the wall react with the same force on me but as I have less mass I have greater acceleration so I am moving away from wall.
yes, and see also the answer to your last question below.

When I hit the wall according to Newton's III law the wall should react with the same force on me (and give me acceleration) but I am not moving(away from wall). Can someone explain it to me?
Acceleration is a change of speed, whether an increase or a decrease. If you are approaching the wall at some speed, when you hit the wall and your speed is reduced to zero - that's acceleration, caused by the force of the wall on you pushing you backwards.

It's confusing sometimes when people use the word "deceleration", but as far as Newton's ##F=ma## is concerned ##a## can be positive or negative.

Also in the first example shouldn't I give a wall a little acceleration(even if a tiny bit) as well so why is the wall not moving at all?
it does. But the wall is connected to the earth, so you are accelerating the entire planet... And its mass is so enormous that the resulting acceleration is far too small to detect with even the most sensitive instruments (Don't take my word for it! Google for "mass of earth" and calculate it for yourself!).
 
  • #4
But why I move(from the wall) when I press the wall and I don't move(from the wall) when I hit it(even very hard)?

Edit:Just read Nugatory answer so while hitting the wall I get acceleration that stops me from moving instead of move me away from wall (in case of pressing it) is that right?

Also Is the wall giving the same resistance or lower in case I crush it?
 
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  • #5
ultrauser said:
But why I move(from the wall) when I press the wall and I don't move(from the wall) when I hit it(even very hard)?
When you press the wall, you are exerting the force for a long time. But when you hit the wall, the force is only exerted briefly.

If you were to stand on a perfectly frictionless surface when you hit the wall, you'd be able to see a bit of movement.
 
  • #6
ultrauser said:
Edit:Just read Nugatory answer so while hitting the wall I get acceleration that stops me from moving instead of move me away from wall (in case of pressing it) is that right?
I thought by "hit" the wall you mean stand there and hit it with your fist. (D'oh!) But if you mean body slam into it, as Nugatory thought, then he's correct. The key is that a net force on you will produce an acceleration, and that might just slow your motion down, not necessarily move you away.
 
  • #7
Thanks for answers but I still have a little problem with that law.

There 3 more examples I would like answered to understand it.

When we have 2 cars with the same mass. First car is standing and second driving.
When second car hit the first one it moves it in the direction the second one was driving.
But according to Newton's 3rd law the 2 forces and accelerations of 2 colliding objects should be the same(they have same mass) with contrary directions so why they are both moving in 1 direction? (Shouldn't second car stop moving? - third example shows what I mean )


When I throw a rock I cause a force on it and a rock causes equal force on me. So if I would stay at frictionless surface would I move backward when throwing rock?


Let's say there are no gravitational forces.
We have 2 planetoids: 1 standing still and second moving at certain speed. Both have mass M=10.
The second planetoid hit the first one.In that momemnt the second planetoid had acceleration a=2.
So it will cause force F=20 (a*M) on first planetoid and the first planetoid will react with the same force.
Does it mean the first planetoid will get acceleration 2 and the second one will stop moving(2-2=0)?
 
  • #8
ultrauser said:
...When we have 2 cars with the same mass. First car is standing and second driving.
When second car hit the first one it moves it in the direction the second one was driving.
But according to Newton's 3rd law the 2 forces and accelerations of 2 colliding objects should be the same(they have same mass) with contrary directions so why they are both moving in 1 direction? (Shouldn't second car stop moving? - third example shows what I mean )

The second car did receive an acceleration, as mentioned by Nugatory earlier. Let's say the second car is traveling at 60 (kph, mph, fps, doesn't matter). Then, that is the car's constant speed. So, from that car's frame of reference, that is called "sitting still". If you were sitting in that car (and I'm not for one instant suggesting that you should!), you would be experiencing no acceleration forces. However, when your vehicle collides with the other, and very suddenly slows to 30, you'll feel enormous acceleration forces! In fact, if you were sitting in the first vehicle, and it suddenly gets launched up to 30 whatevers, you would feel exactly the same acceleration; a wrenching, retina-detaching lurch to the rear.
When I throw a rock I cause a force on it and a rock causes equal force on me. So if I would stay at frictionless surface would I move backward when throwing rock?
Doesn't even have to be frictionless. Try it while standing on ice. If you throw hard forward, you will slide backward.
Let's say there are no gravitational forces.
We have 2 planetoids: 1 standing still and second moving at certain speed. Both have mass M=10.
The second planetoid hit the first one.In that momemnt the second planetoid had acceleration a=2.
So it will cause force F=20 (a*M) on first planetoid and the first planetoid will react with the same force.
Does it mean the first planetoid will get acceleration 2 and the second one will stop moving(2-2=0)?
A little unclear what you mean by "the second planet had acceleration a=2" but no; the first planet will get an acceleration in the direction the second was moving, and the second will get an equal acceleration in the opposite direction, which will slow it down, but not stop it.
 
  • #9
To the third example if action=reaction why it slows down and not stop (doesn't the reaction force when colliding is equal to the force the second planetoid have(the force that moves it) ) ?
 
  • #10
ultrauser said:
To the third example if action=reaction why it slows down and not stop (doesn't the reaction force when colliding is equal to the force the second planetoid have(the force that moves it) ) ?
All you can say from Newton's 3rd law is that the two planetoids will exert equal and opposite forces on each other (assuming for the moment they just don't break up into a zillion pieces). Thus they will have equal and opposite changes in momentum. Only under certain conditions will the first one come to rest.
 
  • #11
It is probably my bad way of thinking but I understand that law as:
1.planetoid has force(so it moves)
2.planetoid hit other planetoid
3.reaction=force the planetoid has (but in opposite direction)
4.Now planetoid doesn't have force (0)
 
  • #12
ultrauser said:
It is probably my bad way of thinking but I understand that law as:
1.planetoid has force(so it moves)
No forces are involved when the planetoid is just moving along at constant velocity. Consider Newton's 1st law.

2.planetoid hit other planetoid
When they hit, they exert forces on each other.

3.reaction=force the planetoid has (but in opposite direction)
I don't know what you mean by "force the planetoid has". I think you are thinking that the moving planetoid "has force" because it is moving. That is not correct. Forces only happen when they hit.

But when they hit, they exert equal and opposite forces on each other.

4.Now planetoid doesn't have force (0)
The planetoids only experience forces when they are in contact. Before and after they hit, there are no forces acting. (Forces are not required to maintain velocity; forces are required to change velocity.)
 
  • #13
ultrauser said:
1.planetoid has force(so it moves)
Maybe you confuse momentum with force, which is the change of momentum over time.
 
  • #14
Let's say both have mass M=10 and one planetoid(right one) has velocity v= 10(the velocity in the moment of impact) and acceleration a=2.

So shouldn't the force (action or reaction) be equal to 20 (F= mass * acceleration).

What acceleration and velocity will both planetoid have after the impact (let's say they won't crush and mass not change) ??
(will they both get acceleration a=2 (a=F/M) so the right asteroid which had acceleration a=2 now will have a=2-2=0 and stop accelerating - in that case the left asteroid(the one which was standing) would accelerate in left direction and the right asteroid would not accelerate
 
  • #15
The two planetoids collide and exert "forces" on each other during a brief (but nonzero) amount of time. So say P1 is moving in the positive 'x' direction and hits P2, which is originally at rest. During that brief time of contact, P1 is pushing on P2 and causing it to accelerate in the 'x' direction. At the same time, Newton's third law tells us that P2 is pushing back on P1 with the opposite force, therefore it is decelerating at the same rate. When the planetoids are done being in contact, P1 has lost exactly as much velocity as P2 has gained, because they were both accelerated in opposite directions by the same amount over the same time.
 
  • #16
olivermsun said:
Newton's third law tells us that P2 is pushing back on P1 with the opposite force, therefore it is decelerating at the same rate.
The same force does not mean the same acceleration. ##F=ma##. Mass also affects acceleration. Your assumption is only true for two objects of same mass.
 
  • #17
Sorry, I probably got confused through the (several) examples above and thought the planetoids were assumed to have the same mass.

So I'll state it here: mass P1 = mass P2. :smile:
 
  • #18
ultrauser said:
Let's say both have mass M=10 and one planetoid(right one) has velocity v= 10(the velocity in the moment of impact) and acceleration a=2.
It will be a lot easier if you include the units and use a coordinate system.
Acceleration and force is a vector quantity. They have magnitude and direction.

Rephrase the question including the direction(With + for right and - for left) and units.
Keep the non-moving object at the centre of the coordinate system. And make the moving object move in the x-axis. I am sure all these will make you understand things more.
 
  • #19
ultrauser said:
Let's say both have mass M=10 and one planetoid(right one) has velocity v= 10(the velocity in the moment of impact) and acceleration a=2.
Before the collision, one planetoid is stationary and the other is moving. There is no force and no acceleration of either.

So shouldn't the force (action or reaction) be equal to 20 (F= mass * acceleration).
During the collision they exert forces on each other. If that force produces an acceleration of 2 units (while it acts), then the force must equal 20 units.

What acceleration and velocity will both planetoid have after the impact (let's say they won't crush and mass not change) ??
After the impact there's no more force or acceleration. To determine the velocity of each after the impact, you'll need more information. All you can say is that the increase in velocity of one will equal the decrease in velocity of the other. (Momentum is conserved.)

(will they both get acceleration a=2 (a=F/M) so the right asteroid which had acceleration a=2 now will have a=2-2=0 and stop accelerating - in that case the left asteroid(the one which was standing) would accelerate in left direction and the right asteroid would not accelerate
Again, once the collision is over they no longer touch and thus exert no force on each other. So the acceleration of both is zero. But they will have different speeds than before the collision.
 
  • #20
Doc Al said:
All you can say is that the increase in velocity of one will equal the decrease in velocity of the other. (Momentum is conserved.)
if velocity decrease and mass won't change how can momentum(p=mv) be conserved?

Doc Al said:
Before the collision, one planetoid is stationary and the other is moving. There is no force and no acceleration of either
Doc Al said:
To determine the velocity of each after the impact, you'll need more information
what do I need to know besides mass and velocity of objects to determine force during impact? (I will try to rewrite it and understand properly)
 
  • #21
ultrauser said:
if velocity decrease and mass won't change how can momentum(p=mv) be conserved?
The momentum of the entire system (both masses together) is conserved. The center of gravity of the system should still be moving forward at the same speed as before the collision.

Also, since momentum is a vector quantity, one of the objects can actually end up moving backward to conserve the total momentum.

what do I need to know besides mass and velocity of objects to determine force during impact? (I will try to rewrite it and understand properly)
if you want to estimate the force during impact, you'll need to know about how long (or over what distance) the objects were interacting (touching).
 
  • #22
ultrauser said:
if velocity decrease and mass won't change how can momentum(p=mv) be conserved?
The total momentum of the system--both planetoids--remains the same. But the momentum of each does change: one increases and the other decreases.

what do I need to know besides mass and velocity of objects to determine force during impact? (I will try to rewrite it and understand properly)
I don't think that's a useful approach. It's too complicated to determine the forces involved in a collision and how long they act for. The beauty of Newton's law is that you know that total momentum will be conserved regardless of how hard or soft they hit.
 
  • #23
Doc Al said:
I don't think that's a useful approach. It's too complicated to determine the forces involved in a collision and how long they act for
Would it be easier if I know Force of impact instead of mass and velocity before? I would like to somehow check the relation of impact force and velocity planetoids had before.
 
  • #24
It probably would't be easier, because then you'd need to know if the objects bounced or how much they crushed or whatever.
 
  • #25
ultrauser said:
Would it be easier if I know Force of impact instead of mass and velocity before?
Like I said before, in real life it would be too difficult to estimate such forces. And you'd need to know the mass and velocities anyway.

I would like to somehow check the relation of impact force and velocity planetoids had before.
You can just play around with some made up numbers to see what happens. Say they both have a mass of 10 kg. (Very small planetoids!) One is at rest, the other moving at 10 m/s. They collide, creating a force of 200 N for 1 ms. Assuming the collision all takes place along a straight line, you have all you need to calculate the final velocities of each.
 
  • #26
Doc Al said:
You can just play around with some made up numbers to see what happens. Say they both have a mass of 10 kg. (Very small planetoids!) One is at rest, the other moving at 10 m/s. They collide, creating a force of 200 N for 1 ms. Assuming the collision all takes place along a straight line, you have all you need to calculate the final velocities of each.
a=F/m
a=200/10=20m/s2
but acceleration act in the time of t= 1/1000 s
so the change in velocity will be 0.02 m/s ?

and the final velocity v1=0 + 0.02 m/s=0.02 m/s and v2=10m/s - 0.02 m/s=9.98m/s ??

It seems like I messed up something here
 
  • #27
ultrauser said:
a=F/m
a=200/10=20m/s2
but acceleration act in the time of t= 1/1000 s
so the change in velocity will be 0.02 m/s ?

and the final velocity v1=0 + 0.02 m/s=0.02 m/s and v2=10m/s - 0.02 m/s=9.98m/s ??
Right.

It seems like I messed up something here
No, you got it. That's what the (made up) numbers tell us. (The values were poorly chosen--by me, at random--since they imply that the first object keeps going right through the second one. Oops!)
 
  • #28
I redid it without a given force and
v1=0m/s v2=10m/s (<-flying left direction)
time of impact t=0.001s

F=m*a -> F=m*v/t -> F=10*10/0.001 = 100 000N

a=F/m= 100 000/10 = 10 000

so acceleration is 10 000m/s2 in time 0.001s meaning change in velocity would be 10m/s ?
It would mean left planetoid would travel at velocity 10m/s (flying in left direction) and the right one would stop but

it seems against a logic. (the second planetoid would always stop after impact in this case regardless of velocity before impact or the time of impact - the velocity change after impact for both planetoids would always be the same as a constant velocity of second planetoid before impact)
 
  • #29
ultrauser said:
I redid it without a given force and
v1=0m/s v2=10m/s (<-flying left direction)
time of impact t=0.001s

F=m*a -> F=m*v/t -> F=10*10/0.001 = 100 000N

a=F/m= 100 000/10 = 10 000

so acceleration is 10 000m/s2 in time 0.001s meaning change in velocity would be 10m/s ?
That's correct. And that was the example I was just going to post. (It happens to be the special case of a perfectly elastic collision--one in which kinetic energy is conserved.)

It would mean left planetoid would travel at velocity 10m/s (flying in left direction) and the right one would stop but

it seems against a logic. (the second planetoid would always stop after impact in this case regardless of velocity before impact or the time of impact - the velocity change after impact for both planetoids would always be the same as a constant velocity of second planetoid before impact)
No, the second planetoid does not always have to stop after impact. It could just slow down.

For example, it's perfectly possible for the first planetoid to end up moving at 6 m/s while the second one slowed down to 4 m/s. (Note that momentum is still conserved.)
 
  • #30
But using this method velocity after impact of second planetoid would always be zero regardless of how massive or light (smaller mass) is the object it hits
(when i throw rock at wall it is reflected not stopping)
 
  • #31
ultrauser said:
But using this method velocity after impact of second planetoid would always be zero regardless of how massive or light (smaller mass) is the object it hits
Why do you think that? What do you mean by "using this method"? All we are doing is applying conservation of momentum.

If you would like to see the second bounce off of the first, make the second very light.
 
  • #32
Doc Al said:
Why do you think that? What do you mean by "using this method"?
I am calculating it like that (different masses example):
planetoid 1 - standing ; planetoid 2 -constant velocity(<-- left direction)
m1=100kg; m2 = 10kg
v1=0m/s ; v2=10m/s
time of impact t=0.001s

F=m*v/t; F=10kg*10m/s / 0.001s = 100 000N

a2=F/m2 = 100 000/10 = 10 000 m/s2 velocity change 10 000m/s2 * 1/1000s = 10 m/s so now planetoid 2 speed (constant speed - velocity change = 10-10=0)

a1=F/m2 = 100 000/100 = 1000m/s2 velocity change 1000m/s2 * 1/1000s = 1m/s planetoid 1 speed

whatever data I always get velocity change equal to constant velocity for second planetoid (resulting in 0m/s)
 
  • #33
ultrauser said:
I am calculating it like that (different masses example):
planetoid 1 - standing ; planetoid 2 -constant velocity(<-- left direction)
m1=100kg; m2 = 10kg
v1=0m/s ; v2=10m/s
time of impact t=0.001s
Note that the impact force and resulting acceleration is unknown.

F=m*v/t; F=10kg*10m/s / 0.001s = 100 000N
You are merely assuming that the force brings planetoid #2 to a stop. Don't do that.

whatever data I always get velocity change equal to constant velocity for second planetoid (resulting in 0m/s)
That's because you are just assuming that, whether you realize it or not. So of course you get that same answer every time.

Do this instead. Assume a perfectly elastic collision. Then you have enough information to properly calculate the final velocities of both objects without having to make assumptions about the forces involved.
 
  • #34
Doc Al said:
Do this instead. Assume a perfectly elastic collision. Then you have enough information to properly calculate the final velocities of both objects without having to make assumptions about the forces involved.
Isn't it what I did in that example? What do you mean and how to do it?

Doc Al said:
You are merely assuming that the force brings planetoid #2 to a stop. Don't do that.
What do you mean by that? I thought if I have data and calculate force than calculation isn't assumption (and force is accurate)
 
  • #35
ultrauser said:
Isn't it what I did in that example? What do you mean and how to do it?


What do you mean by that? I thought if I have data and calculate force than calculation isn't assumption (and force is accurate)
You merely assumed that the force was 10kg*10m/s / 0.001s = 100,000 N. That's one possibility, but it is not required. Why not calculate the final speed of the first planetoid, given this assumption?

If you make that assumption then of course the second object comes to rest. That's what you assumed to calculate the force.

To calculate final speeds for the case of a perfectly elastic collision, apply both momentum conservation and energy conservation.
 

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