# Homework Help: Collision problem

1. Apr 27, 2007

### neelakash

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

A ball moving parallel to the y axis undergoes an elastic collision with a parabolic mirror y^2=2px.Prove that no matter where the prompt of impact lies it will arrive at the mirror's focus.

2. Relevant equations

3. The attempt at a solution

I believe this can be done in the same way we prove a result for reflection of light in a parabolic mirror.Please guide me if I am wrong.

2. Apr 27, 2007

### Dick

It can be done the same way. Assuming the mass of ball is negligible compared with the mirror.

3. Apr 27, 2007

### neelakash

Ok,thank you.

4. Apr 27, 2007

### neelakash

Sorry,I need to talk more.In that equiavalent optics problem,we used Fermat's law of least path.Here it is not available.So,how should we proceed?

5. Apr 28, 2007

### Dick

The 'least path' law still basically applies in the form of a least action principle. But what you really need is just angle of incidence equals angle of reflection. To see this you have to assume there are no frictional forces involved in the collision. This means the wall can only exert a normal force. So the parallel component of the momentum can't change. Conservation of energy now tells you that the normal component of the momentum must just reverse.

6. Apr 28, 2007

### neelakash

That's right.But I cannot see anything using only "law" of reflection.How does it mean that the reflected ball passes through focus(P)?One possibility is that the undeviated ball strikes the directrics and from the definition of parabola,e=1...so,SP=SN.That is (d/dt)SN=(d/dt)SPThis means magnitude of velocity remains unchanged iff the ball reflects along SP...Because,for other points the relation is not true.

I understand the way is hanging around this.But,I cannot develop logically...
by the way,I hope in the same way one can prove for two focii in an ellipse?

7. Apr 28, 2007

### HallsofIvy

This can be done by calculating the "slope of the tangent line" to the curve of the mirror (i.e. the derivative).

However, I have a problem with the statement. You give the equation of the parabolic mirror as y2= 2px, a parabola "opening" to the right but say the ball is coming "parallel to the y-axis", vertically. The ball is going to hit the back of the mirror and will bounce AWAY from the focus!

8. Apr 28, 2007

### Dick

Magnitude of velocity is always unchanged. You have to look at the angles.

9. Apr 28, 2007

### neelakash

HallsofIvy,you are right.Actually,I posted the question directly from a book...When I tried,I started with a the geometrical optics analogy in another...so,I overlooked it.

However,how does the differentiation help?
dy/dx=(2p)^(1/2)*(-1/2)[1/{(x)^(1/2)}]

dick,I could not get anything out of ONLY ANGLES.What about the way suggested?

10. Apr 29, 2007

### neelakash

I think I got a way.I have to show GEOMETRICALLY the reflected ray passes through (p/2,0)

11. Apr 30, 2007

### neelakash

*First you find the co-ordinates of the point of incidence.
*differentiate y^2 and 2px and equate them and find the slope of the normal in terms of co-ordinates of the point of incidence.
*Now you have incident slope 0,normal's slope known,and reflected ray's slope m,say.
*use standard co-ordinate formula that i=r.This gives the value of m.
*The reflected ray's equation can be found.and it is seen that the line passes through (p/2,0)