# B Can the fish move the ball?

#### metastable

no net momentum
I don't think we are trying to find the net momentum of the whole system which will of course always be 0 if the whole system starts at rest relative to the ground...

I thought we are instead looking for any net momentum of just the glass relative to the ground or walls of the space station at any point during the exercise (which would need to be offset by other momentum somewhere else to maintain net 0 momentum).

Obviously if the whole system starts at rest to the ground (such as a pair of skaters pushing off each other), one mass can acquire net momentum to the ground, while the system as a whole maintains 0 net momentum to the ground (the one skater's net momentum is offset by the other skaters net momentum in the opposite direction for a total of 0 net momentum at all times).

#### A.T.

I don't think we are trying to find the net momentum of the whole system which will of course always be 0 ...
If the net momentum of the whole system will always be 0, then the CoM of the whole system will always remain at rest. So whether the ball can move, boils down to whether the CoM can move relative to the ball. If it can, you can use the hamster method anyway.

#### metastable

whether the ball can move, boils down to whether the CoM can move relative to the ball.
Well in the skaters pushing off analogy, if they are the same weight the center of mass doesn't move at all, but both skaters do...

#### jbriggs444

Homework Helper
Well in the skaters pushing off analogy, if they are the same weight the center of mass doesn't move at all, but both skaters do...
If one skater surrounds the other completely then that push-off is going to be difficult. You cannot get much relative momentum if your centers of mass are constrained to coincide.

#### metastable

One skater could be a ring shape around the other skater, and if they both have the same mass, and the skater on the inside pushes on the skater on the outside, they will both move in opposite directions with the same velocity for a short time...

#### jbriggs444

Homework Helper
One skater could be a ring shape around the other skater, and if they both have the same mass, and the skater on the inside pushes on the skater on the outside, they will both move in opposite directions with the same velocity for a short time...
You persist in violating the assumptions of the problem.

#### A.T.

One skater could be a ring shape around the other skater....
You have replaced "ball" with "ring shape". The physics is still the one pointed out post #152:

Whether the ring shape can move, boils down to whether the net CoM can move relative to the ring shape.

#### metastable

So here we have a sealed rectangular aquarium resting on the ground, containing a floating, sealed spherical aquarium, containing a fish with an extendable arm for propulsion. The rectangular aquarium has its own fish with an extendable arm. If both fish push for the same duration with the same force, will the floating spherical aquarium move in the water?

#### A.T.

So here we have a sealed rectangular aquarium resting on the ground, containing a floating, sealed spherical aquarium, ...
When we answer this, will you put another aquarium in an aquarium in an aquarium... ?

Do you understand the general argument in post #152?

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#### metastable

If the glass is initially at rest to the ground, both fish have the same mass and displace the same amount of water, both arms exert the same amount of mechanical work in joules against the glass in opposite directions at the same point, both fish require the same amount of energy to move the same distance in the same time and also move the same amount of water the same distance within their respective tanks in the same amount of time. Since fish displacement relative to ground plus internal water displacement relative to ground within both tanks is equal, there is no surplus energy or momentum to move the spherical tank sideways through the water... or am I mistaken here?

#### jbriggs444

Homework Helper
If the glass is initially at rest to the ground, both fish have the same mass and displace the same amount of water, both arms exert the same amount of mechanical work in joules against the glass in opposite directions at the same point, both fish require the same amount of energy to move the same distance in the same time and also move the same amount of water the same distance within their respective tanks in the same amount of time. Since fish displacement relative to ground plus internal water displacement relative to ground within both tanks is equal, there is no surplus energy or momentum to move the spherical tank sideways through the water... or am I mistaken here?
Yes, you are.

#### metastable

If the inner tank has the same mass as the outer tank and both are initially floating at rest in a blue walled space station as shown below, and we release a spring (containers are filled with water and have water density glass of negligible mass), I don't understand why they won't simply go in opposite directions to conserve momentum, as shown below:

#### jbriggs444

Homework Helper
If the inner tank has the same mass as the outer tank and both are initially floating at rest in a blue walled space station as shown below, and we release a spring (containers are filled with water and have water density glass of negligible mass), I don't understand why they won't simply go in opposite directions to conserve momentum, as shown below:
They will. The outside fish produces an external force on the inside ball, just as if the inside fish were not there. That is not your mistake.

#### metastable

They will.
So then the ball is observed to move (see where it crosses over the grey line).

#### jbriggs444

Homework Helper
So then the ball is observed to move (see where it crosses over the grey line).
No. I do not. As I understand the drawing, the ball is pushed leftward and the surrounding water flows rightward past some imaginary line.

Edit: Sorry, I'd assumed you were sticking to the ball in an aquarium scenario. But you've changed to some kind of concentric balls in a space station scenario. In that scenario, the outer ball does not move. It obviously does not move since the only forces in play are internal and that means that the center of gravity is fixed.

#### metastable

you've changed to some kind of concentric balls in a space station scenario. In that scenario, the outer ball does not move.
It obviously does not move since the only forces in play are internal and that means that the center of gravity is fixed.
It isn’t necessary to change the center of gravity when forces are internal to change the external position in space. I take (2) 1kg blocks w/ a spring between them. One of the blocks is attached to the inner edge of a box of insignificant mass that encloses both blocks. the other block has a tiny weight to counterbalance the insignificant mass exterior box. when the spring between the (2) 1kg blocks is released the outer box of insignificant mass moves to one side. The position of the external enclosure shifts even though the forces are internal and the center of mass is fixed. Momentum and energy would be conserved — position isn’t necessarily.

#### jbriggs444

Homework Helper
It isn’t necessary to change the center of gravity when forces are internal to change the external position in space.
Stop changing the scenario. We are talking about rigid spheres containing uniform density incompressible fluids. Do try to keep track.

#### A.T.

The position of the external enclosure shifts even though the forces are internal and the center of mass is fixed.
We have been through this several times already:
1) If the CoM can move relative to the ball: Translate it in front of the contact (hamster method, works continuously against resistance)
So whether the ball can move, boils down to whether the CoM can move relative to the ball. If it can, you can use the hamster method anyway.

#### jbriggs444

Homework Helper
[A picture showing an un-covered containers subject to external forces]
Nothing to see here.

#### metastable

It would be quite interesting if someone did an experiment with a spring and covered containers...

#### sophiecentaur

Gold Member
This thread is typical of PF - but there is some excuse in this case, perhaps. Fact is that the initial assumptions were not laid down so people are bringing in more and more subtle effects. But the bottom line (i think) is that, if the fish moves through the water in one direction, water must move in the opposite direction. This will have an effect on the envelope due to friction so the envelope will move in the opposite sense to the fish. Once the fish stops swimming, fish and ball will come to a halt with the angles of the ball and fish, relative to some initial reference, different from how it started off.
If the fish can directly move the wall then it just gets more complicated and any loss of Energy will not affect the main principle because it'.
I like the wobbling ball factor!! But, if the fish has the same density as water, why would the ball be displaced- unless the fish actually hits the side?

#### jbriggs444

Homework Helper
But, if the fish has the same density as water, why would the ball be displaced- unless the fish actually hits the side?
You are falling into the same trap as @metastable. The ball does not move, even if the fish hits the side.

The gotcha is that if the fish comes to an abrupt stop then so does the water. Any momentum transferred from fish to wall is matched by an opposite momentum transferred from water to wall.

Subject to background assumptions of rigid wall, incompressible fluid, etc.

#### metastable

The gotcha is that if the fish comes to an abrupt stop then so does the water.
How can you prove when the fish hits the wall the water stops? Some glitter in the water should be able to test that proposal.

#### jbriggs444

Homework Helper
How can you prove when the fish hits the wall the water stops? Some glitter in the water should be able to test that proposal.

"Can the fish move the ball?"

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