# (Elastic collision / shadow problem) I suspect my professor is wrong.

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1. Dec 13, 2013

### firefight33

1. The problem statement, all variables and given/known data
Suppose that two elastic balls moving through three dimensional space collide and rebound, conserving both kinetic energy and momentum. The balls cast shadows, and these shadows moving over a flat two-dimensional surface come together, collide and rebound. Now pretend the shadows have mass. Suppose that the shadow of a call has a mass proportional to the mass of the ball. Then, the colliding shadows

A) conserve kinetic energy
B) conserve momentum
C) conserve both kinetic energy and momentum
D) conserve neither kinetic energy nor momentum

The attempt at a solution
This was a problem on a test in my physics class at a community college. I think my professor is fairly bad and doesn't know what he's talking about most of the time. This problem is also the only problem that I haven't been able to find online, so I assume he created it.

I originally picked D, and found out later that my professor marked this as wrong.

My reasoning for picking D was that I thought of a counter example to C. I imagined two spheres traveling towards each other completely horizontally (w/o gravity and parallel to the shadow plane) with one slightly higher than the other. When they both collide, both of the spheres will have some vertical velocity from the impact (which the shadows would not pick up). So in this collision between two elastic balls, the shadows are picking up both velocities perfectly before impact since the motion of the balls is parallel to the shadow plane, but after the impact some of this velocity is lost from both balls in the shadow plane even though in 3D it isn't.

The answer is of course not A or B, since both must be right if one were right. So, my professor implying that the answer is C is causing me some trouble. Is my counter example not really a counter example? If so where did I go wrong? Or am I right?

2. Dec 13, 2013

### voko

You argument about losing part of the velocity is correct. Yet it is not a counterexample to both A and B, because it is not correct that "both must be right if one were right".

3. Dec 13, 2013

### firefight33

But how can either A or B be correct is speed is lost due to distortion? (If you're implying that)

There can be a case of ratio of the speeds not changing since we can make this counter example to be perfectly symmetrical. (Which I think would rule out KE being the same, while momentum isn't)

And the sum of the speeds would decrease which would rule out momentum being the same.

Last edited: Dec 13, 2013
4. Dec 13, 2013

### voko

Your final sentence is incorrect. Non-conservation of speed does not imply non-conservation of momentum. Consider, for example, two identical balls going at an identical speed toward each other. The total momentum is zero. So the balls could lose all of their speed, yet total momentum would still be conserved.

5. Dec 13, 2013

### ehild

The velocities also "cast shadow" onto the plane: You can assume light rays perpendicular to the plane and the original velocities having some angle alpha with the plane . How the shadow-velocities are related to the original ones?

ehild

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6. Dec 13, 2013

### firefight33

Alright, two balls are moving in the exact same direction with one slightly higher than the other parallel to the shadow plane. One is moving faster than the other and they collide in an elastic collision. Both balls lose translation speed in the shadow plane. Momentum is conserved in 3D, and there's no difference in the directional components to the momentum so the magnitude of the speed is the only factor here. In the shadow plane, speed is lost. And in this case, momentum should be also.

This whole thread is going to be extremely embarrassing for me:(

7. Dec 13, 2013

### voko

No, that is not correct.

Yes. And because it is a vector, you can decompose it into three mutually orthogonal components. Each component must also be conserved. Let one of those be perpendicular to the shadow plane. The other two will be the shadow plane. So?

8. Dec 13, 2013

### firefight33

I'm confused about "Each component must also be conserved". Why are all three components still conserved when the two balls collide in such a way that alters their motion perpendicular to the shadow plane?

9. Dec 13, 2013

### voko

"Conserved" means "stays constant". When a vector is "conserved", it stays constant. When a vector stays constant, each of its components stays constant. I hope this is clear.

Now, when we says "momentum is conserved", we really mean "total momentum is conserved". The total momentum of two balls is the sum of each ball's momentum. It is the sum that is conserved, not each ball's momentum.

In an example given earlier, total momentum was zero when the balls were either moving or stationary; obviously each ball's individual momentum was not conserved, yet the total momentum was.

10. Dec 13, 2013

### D H

Staff Emeritus
That each component is conserved is what conservation of momentum is all about. Conservation of momentum is a vector equation: $m_1 \vec v_1 + m_2 \vec v_2 = m_1 \vec u_1 + m_2 \vec u_2$, where $\vec v_i$ and $\vec u_i$ denotes the velocities of object i before and after the collision. This vector equation is actually three scalar equations. The x component of velocity is conserved ($m_1 v_{1_x} m_2 v_{2_x} = m_1 u_{1_x} + m_2 u_{2_x}$), and similarly for the y and z components of velocity. Each component is conserved.

The shadow balls don't have a normal component. Call the axis normal to the plane the z axis. Momentum is conserved in all axes, so the shadow balls will still exhibit conservation of momentum. This means the answer must be either (B) or (C). Your example of two real balls colliding does indeed demonstrate that energy is not a conserved quantity. This rules out answer (C) (as well as (A)).