Inelastic Collision and incident angle

Question

This problem is puzzling me.

Scenario 1: A certain sphere is having an innelastic collision with a plane of a very large mass. Let the angle between the sphere's velocity and the surface be 90 degrees. The escape angle will be the same as the incident angle, therefore the sphere velocity will be 90 degrees from it's origin.

Scenario 2: The same as above, but now the ball is static, and the surface is moving towards the sphere with the same speed.

I was expecting exactly the same result, since on my mind (and obviously wrong), the speeds were relative. The sphere in scenario 2 leaves the collision plane on a trajectory normal to the surface. Note that I mention speed, because I have a gut feeling that the problem I'm having is with summing up the vectors.

I'm having problems visualizing and coming up with an intuitive explanation for why.

Answer 1

The escape angle will be the same as incidence angle, therefore the sphere velocity vector will be 90 degrees from it's original.

If I understand correctly, this is where your mistake is. The incidence angle is 90 degrees, which means that the sphere comes in at an inclination of 90 degrees with respect to the surface (i.e. normal to the surface). The angle of reflection (or escape angle, as you call it) is the same as the angle of incidence, which means that the sphere leaves at an inclination of 90 degrees also with respect to the surface. I'm not sure where you got the idea that it would leave at an angle of 90 degrees with respect to its original trajectory, but that's not how angle of reflection is measured. It's always with respect to the plane of impact.