# Collision in moving water-angle of rebound

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## Main Question or Discussion Point

Hi all!
I was having a conversation with a (rather young) friend about the following imaginary situation- suppose you are in a helicopter, at rest relative to the ground. So you are basically seeing a 'top' view. Say you have a a wall fixed to the ground, and it is kept in the middle of a moving river, parallel to the flow which has a horizontal speed $v_m$(relative to the ground). Suppose it is moving rightward. Note that the wall doesn't move relative to you. Now, there is a boat which is moving through the river, such that it's velocity relative to the river is purely vertical (i.e. its speed relative to the ground is is $(v_y)^2+(v_m)^2$), and so relative to the helicopter, it follows a 'diagonal' path, moving towards the wall. The property of the wall is such that if there was no water ( and the boat could somehow still move), the boat would rebound exactly at the same speed, making the same angle with the wall as before (so something like an elastic collision).

Now, what happens in the presence of the river? For you inside the helicopter, does it still rebound at the same angle with which it hits the wall?
My argument is yes, it does. Because, if you were 'in the water', you would see the ball go straight up to the moving mirror and come back down the same way, so no change in angle. Since the helicopter is also an inertial frame, and all of them are equivalent, there should still be no change in angle after collision.
His argument is that after the collision, the boat STARTS moving out with the same speed as before, but this new velocity is immediately affected by the river velocity, and so it rebounds at a different angle (i.e. makes a greater angle with the normal through the wall; it has been 'carried' further to the right.)
Now, I do not want to use conservation of momentum to back up my argument. Is there a simple physical argument I could use to point out the mistake in his reasoning? I tried to show him the component of the velocity in direction of flow remains unchanged, and the component directed towards the wall gets 'reversed'. But he still thinks that the 'reversed' component will be further deviated to the right by the river. I tried explaining that the horizontal component is doing just that, but perhaps there is a better, more precise argument to explain this. Any help would be appreciated.

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A.T.
Now, what happens in the presence of the river?
Fluid dynamics makes everything complicated. You might have different simplifying assumptions in mind.

Fluid dynamics makes everything complicated. You might have different simplifying assumptions in mind.
Basically, consider any 'ideal' medium moving with a certain speed. Would the rebound angle still be the same as when there was no motion of the medium? The specific example we had in mind was the 'ether' ( the one which SR got rid of). So I tried to draw an analogy there.

Nidum
Gold Member
You've buried a problem with a relatively simple configuration in a sea of rambling words .

Just draw picture .

You've buried a problem with a relatively simple configuration in a sea of rambling words .

Just draw picture .
I wanted to, but I do not know how to attach one!

A.T.
...'ideal' medium...
You and your friend have to agree on what that is, and what forces it exerts on the boat under which conditions.

Nidum
Gold Member
Ideally draw using paint or similar software but otherwise scan a paper sketch . Save as a .png file and use the UPLOAD button right of the reply box to upload .

You can also paste images directly into the reply box .

anorlunda
Mentor
I wanted to, but I do not know how to attach one!
Use the UPLOAD button that sits to the right of POST REPLY and PREVIEW.

Ideally draw using paint or similar software but otherwise scan a paper sketch . Save as a .png file and use the UPLOAD button right of the reply box to upload .

You can also paste images directly into the reply box .

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