Physics Question (Sorta)

[Vladamir]

Right, since me and tris couldn't come up with a reasonable explanation i need help

With regards to the angle of rebound, i.e i throw a ball towards the floor at X degrees, it bounces off the floor at Y degrees.

According to my lecture notes, the size of the rebound angle is dependant on the elasticity of the striking object (the ball) and the friction between the surfaces (the ball & floor).

So what i'm trying to work out is what the rebound angle of a perfectly elastic object is (i.e an object that loses no velocity when impacting with the floor).

Any ideas?

 

[tris-]

ok if it loses no velocity, there is no friction. would help if you give the formula too ffs.

 

[Nate]

Garrys mod has all the answers!

 

[Mas]

Google Answer :-

If you attempt to apply the idea that rebound angle equals incident angle in the real world, you must take into account the coefficient of restitution, the spin of the ball before the collision, and the frictional coefficients (both sliding and rolling) to model the situation. So in the real world the answer is that the rebound angle is almost never equal to the incident angle.

 

[Achilles]

Newtons law of restitution is: Speed of approach x e(coefficient of restitution)=speed of separation. So in your example e is 1, if you had the speed of approach you could resolve it into horizontal and vertical vectors because you know the angle of approach? So Vcosa for horizontal and Vsina for vertical? So (not so sure), say you take the vertical component and plug into above formula. So you get the vertical component of the speed of separation and do the same with horizontal one. So then you use tany= vertical component (opp)/ horizontal compenent (adj) and you should have your angle. Disclaimer, probably not right.

 

[Lamp]

The physics break down when you try to throw a glass bottle on a concrete floor to test the bounce.