# AP Problem - Elastic Collision at an Angle

1. Dec 8, 2007

### meganw

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

SEE BELOW: 4th Reply has a Diagram

2. Relevant equations

Conservation of Momentum: m1(v1i) +m2(v2i) = m1(v1f) + m2(v2f)
Vf^2=Vi^2 + 2a(delta y)
Conservation of Kinetic Energy (Elastic Collision): .5m1(v1i^2)+.5m2(v2i) = .5m1(v1f^2)+.5m2(v2f^2)

3. The attempt at a solution

See 4th post for newest question:

Last edited: Dec 8, 2007
2. Dec 8, 2007

### Staff: Mentor

Since the surface is frictionless, only the component of velocity normal to the surface is affected by the collision.

3. Dec 8, 2007

### meganw

What? Sorry I don't understand what you're saying....does that mean the angle doesn't change?

(By the way thank you for being so amazingly helpful!!!! :) )

4. Dec 8, 2007

### Staff: Mentor

Yes, the angle the velocity makes with the surface will not change.

5. Dec 8, 2007

### meganw

This is the problem and diagram:

http://img255.imageshack.us/img255/1966/55320227sg8.png [Broken]

I have done a-c, and these are my answers that I got. I know they're correct because I checked them with the solutions:

L= 4$$\sqrt{}2$$ (h)
Delta Y = Delta x = L/$$\sqrt{}2$$

For d I used the kinematic equation
Vf^2=Vi^2 + 2a(delta y)
I got delta y=4h
and vf = $$\sqrt{}8gh$$

but the ap board says the answer is (conservation of energy):

mgh + mgL/$$\sqrt{}2$$=.5mv^2

v = $$\sqrt{}10gh$$

But why is my answer for part d wrong?

note: $$\sqrt{}2$$
(this symbol is the square root symbol)

Last edited by a moderator: May 3, 2017
6. Dec 8, 2007

gahhhh

7. Dec 8, 2007

### Dick

You didn't add the Vi^2 (=2gh) to the 2a(delta y) (=8gh).