Collision with diagonal surface

Ok, I'm having a bit of a problem with this.

A ball is dropping in free fall (not spinning/rotating or anything).
After a given time it collides with a diagonal surface (a total inelastic collision), it then begins to roll down from it.

How do I find the velocity that it begins to roll down the diagonal surface with? Do I just regard the ball's velocity as a vector and find the component that is parallel to the surface?

It's 0.
This may sound counterintuitive, but that's only because there's no such thing as a totally inelastic collision.

>>2
I thought that was only true for a head on collision?

you need to know the angle of collision...

>>4
It falls down in a straight line.

Need: -How long was it falling for?
      -Was it being accelerated or falling at constant velocity?
      -Angle of inclination of the diagonal surface with respect  
       to the horizontal, or the angle the vector of the free fall
       velocity with respect to the diagonal.

Then you would just use superposition to find the resultant velocity vector

>>6
I'm just looking for a general term, nothing specific.
I'm using this to construct a simulator for a school project, so there's no specific values.

The velocity down the ramp is going to be equal to sqrt(2gh) sin(theta) where h is the height the ball is dropped from with zero initial velocity and theta is the angle of incline of the ramp.

>>8
The gh * sin(theta) I get, but why do you need to multiply by 2, and why on earth do you need a square root?

Just before it hits the ramp mgh = .5mv^2

Therefore the velocity when it hits the ramp is v = sqrt(2gh)

Then you just multiply it by sin(theta) to get the component down the ramp.

>>10
Oh I see.
So you're actually splitting the kinetic energy up, and letting the energy in the right angled direction getting destroyed.

Ok, thanks for the help.

if it rolls down you'll have to consider the moment of inertia, because some of the initial energy goes into the rotation so the velocity down the ramp will be lower then the velocity of a sliding ball.

I believe you may need to know the coefficient of friction becuase it does not instantly start rolling, it slides as w (omega) is increased, and some energy is lost.

>>11
That's a dangerous way to think about it. It's easy to see if you transform frames, to a frame where the ball is not falling straight but perpendicular to the slope. Naturally, it will stop completely (as you said earlier). But as this frame is moving parallel to the slope, in the frame where it is falling straight, it will continue to move with a speed parallel to the slope.

>>14
I'm not entirely sure I understand what you mean.

The way I'm doing it currently is splitting the kinetic energy it has build up from falling up into components, ignoring the right angled one, and using the amount of kinetic energy left in the parallel component to calculate the speed it would then have.
I'm still not sure at all if it is the correct method, but it sounds 'sort of' right. I mean, it falls from above and then just 'sticks' the surface, the more of an incline the surface has, the more energy is preserved because it hits at a higher angle.

>>13
Yes, that is true, but I have to set some limits for this project, it's on a limited time schedule.
Of course I discuss all the limitations and errors of the program's calculations in my rapport.

>>10 has the a pretty good answer, but it looks like he assumes that the impulse upon landing is perfectly perpendicular to the plane (ie no frictional impulse) this corresponds to 0 deformation of either the ball or the slope during the collision, this, of course, means no energy loss (aka perfectly elastic collision)  This also means that the ball picks up no rotational velocity in the collision and the ball begins by skidding down the slope.
But we specified an INelastic collision so let's consider the more likely scenario that the collision's frictional factor is enough that the ball doesn't skid at all.
r=radius
m=mass
I=moment of inertia
F=impulse due to friction
t=torque=F*r
w=final rotational velocity=t/I=v/(2*pi*r)
v0=speed before collision
v=final speed=v0*sin(theta)-F/m
This gives:  2*pi*F*r^2/I=v0*sin(theta)-F/m
F=v0*sin(theta)/(2*pi*r^2/I+1/m)
So our actual starting velocity is: v=v0*sin(theta)*(1-1/(1+2*pi*r^2*m/I))
If the ball is uniform then I=2*m*r^2/5 and v simplifies to:
v=v0*sin(theta)*(1-1/(1+5*pi))=v0*sin(theta)*(5*pi/(1+5*pi))
In practice you would have a slightly smaller value for v since r temporarily shrinks when the ball collides.

>>16
Wow, thanks a lot!