# Homework Help: Momentum of ice cube problem

1. Apr 11, 2007

### neelakash

1. The problem statement, all variables and given/known data

Consider a gravity free hall in which a tray of mass M,carrying a cubical ice block of mass m and edge L,is at rest in the middle.If the ice melts,what would be the displacement of the system?

2. Relevant equations

3. The attempt at a solution

I think it has to do with the conservation of momentum,since no external force is acting.But i cannot think further.Please help.

2. Apr 11, 2007

### Päällikkö

Could you draw a simple diagram and upload it to eg. http://imageshack.us I'm not quite sure if I've understood what's going on.

From conservation of momentum it follows that if there are no external forces, the center of mass remains constant.

3. Apr 11, 2007

### Kolahal Bhattacharya

I agree with Päällikkö.
F=M(d^2R_cm/dt^2)=dP/dt=o
To have even a uniform motion of CM,what we need is a motion of CM from the rest.That must involve an acceleration.
Since acceleration of CM is zero,we must conclude that CM did not move at all.

4. Apr 11, 2007

### andrevdh

Since no gravity is present the ice cube will most likely melt into a ball of water. That is cube of ice --> ball of water. What also need to be considered is the change in volume from ice to water (will the ball push the tray away?). One would think that since only internal forces is present the objects in the system will reposition in such a way that their common com stays fixed in space, but the fact that the ice melts implies that energy is entering the system.

5. Apr 11, 2007

### neelakash

Thank you for this additional insight.

6. Apr 11, 2007

### Robert J. Grave

Where does the increase in energy to melt the ice come from? How does it disperse? Is there an atmosphper present? The water is a ball because of surface twension? That surface tension will also cause the water to cling to the tray too won't it? What about the gravitational attraction pan to water? The water, if melted, is smaller since ice is an expansion of water when froz. This may be more complexe a problem than first thought. -Robert.

7. Apr 11, 2007

### neelakash

The problem doesnot say whether there is atmosphere or not.However,we are told that the ice melts...
It is a good point regarding the surface tension you have mentioned.The water is a kind of substance for which the angle of contact --->0 for appropriate solid surfaces.So,it is more likely to have water spread itself on the surface (suppose made of glass).
I do not think you have to worry about the contraction in volume.The system will reposition itself in such a way that the CM is unaltered.

8. Apr 12, 2007

### andrevdh

Left to its own water tend to form spherical droplets under the influence of surface tension. When it gets large it deforms under its own weight and we find that it wets the a surface it rests on. With gravity removed the water will therefore melt into a ball. Since the ball will have a larger diameter than L (a sphere that fits inside of a cube has a smaller volume than the cube) they will be pushed apart a bit from their original positions.

Last edited: Apr 12, 2007
9. Apr 12, 2007

### neelakash

But,it is not that the water ball is wetting the surface for its own weight.I am saying that the surface tension co-efficients are such that the angle of contact will be acute and the water will wet the surface rather than building a ball.

10. Apr 12, 2007

### Robert J. Grave

This is a tough one. Assuming that all of the melting energy is used and stored in the melted water. Surface tension will probably leave the water touching the tray since 1/6 th of it was already in contact with the tray at the start. Now it probably will take the shape of a drop of water hanging from the end of a faucet or some other surface only inverted of course. The only difference would be it would be spherical not oval. What do you think? -Robert.

11. Apr 12, 2007

### neelakash

I see what you are saying.But why should the surface tension "leave" the water touching the tray?What do you mean by leave?

Surface tension is a surface property.Here also,it should depend on the nature of the solid surface,water and a possible atmosphere.Whatever surface is formed,it will depend on the three co-efficients.

12. Apr 12, 2007

### Robert J. Grave

The ice is already touching the tray. As it melts the melting water in contact with the tray will cling to the tray. Remember the water will occupy a smaller volumn than the ice cube did. surface tension would tend to pull it away from the tray in to a drop or sphere shape. The result, I think, may be that it ends up shaped like an upside down drop of water but as sphere shaped. This is do to the two part affect of surface tension. At the tray and above the tray. As far as the trays material and atnosphere, since they are not given then the solution can't consider them. Like my first statement, this problem could be more complex if other factors were known. -Robert.

13. Apr 12, 2007

### neelakash

Sounds OK.

14. Apr 13, 2007

### andrevdh

The intermolecular forces (cohesion) between the water molecules are quite strong and gives rise to the surface tension. There also exist attractive forces (addhesion) between the water molecules and the atoms of the tray. Depending on which is stronger the contact region between the ball of water and the tray will exhibit either a small or large contact angle. The rest of the water further away from the tray will be pulled into a ball by the surface tension. So it is just the small contact region that gets distorted.

15. Apr 13, 2007

### neelakash

It is more sound.

16. Apr 13, 2007

### andrevdh

Since the com of the ball will cause the com of the system to shift a little towards the ball an external force is needed to accomplish this displacement of the system. No such force exists. This means that the system generates a set (Newton's third law) of internal forces between the tray and the ball to shift the components (ball and tray) to reposition the com to its original point before the ice melted.