Conservation of KE, wedge striking ball

[alkaspeltzar]

TL;DR

Looking for clarity on how kinetic energy is conserved in an assume elastic collision between a moving wedge and stationary ball, looking at the work and energy transfer.

So assume we have a wedge traveling at a constant V horizontally, that is braced so it CANNOT move vertically. Ignore air and friction. See picture.

It hits a stationary tennis ball and due to the angle, there is a net force on the ball as shown.

The energy should come from the kinetic energy of the wedge. However, if i were to look at the Work of the the horizontal reaction force on the wedge over the distance(wedge slows and loses energy), wouldn't the ball have the same force horizontally, over the same distance PLUS the work done from vertical component? what am i missing as that would be more energy right?

Only thing i can think of it that both items deform, and thru their deformation, the energy is stored as potential and then back to kinetic energy. As such both the energy lost of the wedge is equal to that of the ball. Trying to break it down is going to cause confusion.

 

[BvU]

Only thing i can think of it that both items deform

Does that mean you think the situation and the outcome would be different for 'hard' ball & wedge ?

$\$

 

[jambaugh]

In the frame you are using, the wedge is moving and also experiences a reaction force, and thus work is being done on the wedge. Choose a frame moving with the wedge. In that frame, there is no work being done on the wedge. But, of course, the relative velocities all change, and the kinetic energies change. Also, note that in this frame the angle at which the ball bounces changes since the velocity vectors are not the same as the old frame.

Edit: I typed this in haste... checking some things...

 

[alkaspeltzar]

Does that mean you think the situation and the outcome would be different for 'hard' ball & wedge ?

$\$

you are right, it doesn't change. But that is not the question. My questions is the ball feel a equal horizontal force over the same distance as the horizontal force acting on the wedge. But there is a downward force on the ball, over some distance. Where does that work come from as wouldn't that be more than that lost from the wedge?

 

[alkaspeltzar]

In the frame you are using, the wedge is moving and also experiences a reaction force, and thus work is being done on the wedge. Choose a frame moving with the wedge. In that frame, there is no work being done on the wedge. But, of course, the relative velocities all change, and the kinetic energies change. Also, note that in this frame the angle at which the ball bounces changes since the velocity vectors are not the same as the old frame.

Edit: I typed this in haste... checking some things...

reference frames is not my question

My questions is the ball feels an equal horizontal force over the same distance as the horizontal force acting on the wedge. The ball should gain KE and the wedge lose KE. But there is a downward force on the ball, over some distance. Where does that work come from as wouldn't that be more than that lost from the wedge?

 

[jambaugh]

Follow up: Yep, in the ball's initial comoving frame you get a standard reflection, angle of incidence equal angle of reflection and kinetic energy is conserved. In the wedge frame, The ball is launched normal to the wedge at a velocity $2\cos(\theta)$ where $\theta$ is the angle from vertical.

I think that your example is demonstrating that the concept of energy is not something imbued within the objects but is relational between object and observer.

The other take-away is that you should be very careful with non-physical infinities. Treating the wedge as infinite mass and moving means you have an infinite energy reserve. The kinetic energy of the wedge after the collision is $\infty - T = \infty$ so it changes and doesn't change at the same time.

 

[jambaugh]

The answer to your second qualifier is that you can either treat the wedge as an infinite mass in which case it obtains a vertical motion of $0 - k/\infty = 0$ or you are fixing the wedge to a track-earth infinite mass system which then absorbs that reaction momentum with a momentum/infinity velocity change. Again, remove the implicit infinities and you should find everything sensible.

 

[alkaspeltzar]

The answer to your second qualifier is that you can either treat the wedge as an infinite mass in which case it obtains a vertical motion of $0 - k/\infty = 0$ or you are fixing the wedge to a track-earth infinite mass system which then absorbs that reaction momentum with a momentum/infinity velocity change. Again, remove the implicit infinities and you should find everything sensible.

You are too deep for me. I just don't see how the work/energy balances out between the wedge and ball. I see the forces, and that makes sense, but when i work it backwards, trying to see how the wedge slows and the ball speed ups, doesn't appear to add up.

 

[alkaspeltzar]

nevermind, i see it now. The forces applied to the wedge will cover more distance in the same time than the forces applied to the ball. As a result, there are forces on the ball that make it move away, but the total of the work comes out the same. Darn geometry