Forces on an elastic ball bouncing off a 45 degree slope

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How physics works when falling ELASTIC ball hits 45 degree slope?

At point of colision what forces are working? As do I believe- in time delta t- one force as moment was straght with y axis(prependicular to ground) and Normal force( or how do u call it) witch is prependicular to 45 degree slope? And then sum of vectors would be parallel to x axis? How?? When moment after colision would be also parallel to x axis? And speed before colision and after would be same? How to show this in equations hm?
 
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the only force (besides gravity) is the Normal, which is prependicular to 45 degree slope as you said. i am not sure what you ment by the force on the y axis, velocity? gravity? anyway the sum of the vectors doesn't have to be parallel to the x-axis in order for the ball's velocity to be parralel to the x axis, the forces basically "move" the velocity vector (which has nothing to do with the force vector) in 90 degrees (if the gravity is ignored) the speed should remain the same if energy is coserved.

hope i helped.
 
Okay, I figured out it, thanks
 
edoarad said:
the only force (besides gravity) is the Normal, which is prependicular to 45 degree slope as you said. ... the forces basically "move" the velocity vector (which has nothing to do with the force vector) in 90 degrees (if the gravity is ignored) the speed should remain the same if energy is coserved.

hope i helped.

Hi I am having a small fight with a work college and have searched high and low on google and this is about as close as I've gotten to an answer!

So, what your saying is if a ball hits a surface on an angle richochettes off at the opposite angle same velocity. the force (reaction seen by the surface) will be perpendicular to the surface?

And Hence if you were applying the Force from an impact of a rock on a bin wall (assuming no friction) that hit the wall on an angle you would apply the component of the force that acts perpendicular to the wall. Not in the direction that the rock was falling.
 

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