# Pushing Off On Ice

1. Jun 20, 2005

### Vexxed

Okay, this is something that's been bugging me for a long time, and I should have asked my physics teacher this past year way back when we were discussing mechanics, but I never could remember to do it.

Let's say I have an identical twin, and we're both on a big frozen pond, standing still relative to the pond and the earth. I push him with a force F, and through Newton's 2nd law I'm pushed back with a force of F. He goes one way with velocity V, I go with velocity -V (relative to the earth). But what if he has special boots, or is nailed to the ground, or is stuck in the ice, or through whatever means is fixed to the earth? Would I still have velocity -V relative to the pond, or would it be -2V or something else entirely?

Back when I first started turning this over in my head, I thought that I exert F on him (and he F on me from Newton's 2nd), but since he's attached to the ground, he exerts F on the ground, which exerts F on him, which gets transferred to me, which means 2F total is exerted on me, and I'd move backward at -2V.

But I was thinking today about conservation of momentum... in the first scenario, since our masses are equal, our speeds would be equal, but in the opposite direction. But in the second scenario, the only difference is that I'm not just pushing him, I'm pushing him and the earth, which has a bit more mass and wouldn't have any significant change of motion - and I'd still have just the regular old F put back on me, and still move -V relative to the earth.

Now my problem is that the first solution doesn't seem right with all the transfers of force and whatnot, but the second seems kind of counterintuitive since I'm exerting a force on the frame of reference itself, and it seems that my speed should be different because of the different setup. But since the frame (the earth) won't be moving much due to the force, it can still be used as a frame of reference... Sorry if this all sounds kinda confusing, it's hard to put into words, especially considering I've only had one year of high school physics, but I'm hoping someone here can shed some light on the subject...

EDIT: It's the 3rd law, not the second... yeah, it's been a while for me.

Last edited: Jun 20, 2005
2. Jun 20, 2005

### whozum

If you two are standing upright, then the forces you exert on each other will actually turn into torques about your feet. Your natural impulse when being pushed is to restabilize yourself, and this alters the transfer of momentum completely. If we ignore this though, then as you said, the force he pushes you with will be the force your body applies to him. (2nd Law). You will both float away at the same speed (same mass).

If he is nailed to the ground however, then you can relate the situation to pushing against a wall. The quantity you are interested in for this problem is called impulse. It's defined algebraically by:

$$F\Delta t = \Delta mv$$ and it actually follows from Newton's second law.

Impulse is a change in momentum, so lets say you have mass 50kg, apply a force of 50N for 1 second. You then apply a total impulse of 50kgm/s. You now have momentum of 50kgm/s the instant the force is done being applied, and will float away at 50kgm/s / 50 kg = 1m/s.

However, the wall is a different story. It will absorb the force and depending on its physical characteristics, like structure and material, will dissipate this force into the ground and as a result move a miniscule and negligible amount. During this time though, the wall applies no force or impulse on you other than the reaction force you applied to it. If you draw force diagrams for yourself and the wall, then you will see why this holds. You experience only one force from the wall while you are pushing at it (which in turn gives you velocity 'v = 1m/s'). Once you are no longer touching the wall, there are no forces bothering you, and you experience no change in momentum.

3. Jun 21, 2005

### shyboy

There is some difference between "push" and force. Whether you or your twin will be "pushing", the force acting on you will be the same. Just imagine that there is a solid wall between your palms. Thus if the wall is "pushing" you, you will apply some force to the wall in any case. Or the wall may be steady and you will "push" yourself.

There will be different story if the wall is "pushing" and you are trying to "push" the wall as well. I mean that you are trying to accelerate with respect to the accelerating wall. Now the force may be doubled, and your final speed will be doubled as well.

As a matter of fact, it is often more easy to consider the conservation of energy in such cases (grounded walls). It is because for the case of an interaction an object of the infinite mass will not change its energy, but it will get some impulse to conserve the total momentum. Thus if you and your twin will make the same job to "push" , then your speed will be bigger if your twin is fixed to the ground.

4. Jun 24, 2005

### Danger

Very simply, if he is fixed to the ground and you do all of the pushing, you would move away at -2V. There would be energy lost through being absorbed by his knees, ankles etc. If he is pushing back equally to your push, then you would move away at -4V.

5. Jun 24, 2005

### ka123

I think the same force acts on you both ways, that is, -F. And therefore, your acceleration will be the same and so will the final velocity (assuming that the contact time is the same).

This is consistent with the law of conservation of linear momentum which states that:
The total momentum of an isolated system remains constant.
According to this, the earth will be in the system as stated correctly in the first post.

6. Jun 24, 2005

### quasi426

What if the twin is twice your mass? Then what?

7. Jun 24, 2005

### whozum

Then he's not your twin :tongue2:

8. Jun 24, 2005

### shyboy

As I said, the energy should be conserved. If you are both on ice, he will have a speed twice as smaller then yours. If he is fixed to Earth, then your speed depends on how much job you did together, and all this job will go to your kinetic energy (sweating neglected)