## A Case Of Momentum Conservation

My question is...

when a ball falls vertically on an inclined plane with a velocity [v][/0] and let it collide elastically with the incline plane...let the angle of inclination be 'β' ...

Now, here we conserve the momentum of the ball in the direction of commomn normal of the two objects....

So we write an equation
initial momentum of the ball (along common normal) = final momentum of the ball (along common normal)...

Assuming that the inclined plane is fixed...

THE REAL QUESTION (CONFUSION)..
i think that the net force on the system (ball + plane) is acting along the common normal ..as gravitational force is acting ..so how are we conserving the momentum along common normal and even along the line tangent to the inclined plane....

Even along that line Gravity is acting....

Sorry for the length of the question...I was too descriptive..:D..

 PhysOrg.com physics news on PhysOrg.com >> Iron-platinum alloys could be new-generation hard drives>> Lab sets a new record for creating heralded photons>> Breakthrough calls time on bootleg booze
 Sorry, I'm a little confused by your use of the term "common normal" in this problem. I have never really heard that term. Momentum is conserved. Momentum is a vector quantity, so the direction of momentum after must also equal the direction of momentum before, not just the magnitude. To properly conserve momentum as a vector quantity in this problem, we must realize that the ball imparts a change in momentum in the Earth. Let's consider the Earth at rest in some frame, with the slope firmly attached to it, the ball drops onto the slope and bounces away at a different direction. The momentum of the ball, therefore, has changed. That change in momentum is balanced out by the fact that this change in momentum imparts a change in momentum of the entire Earth. In our previously established reference frame, the Earth is now moving--very slowly, but moving--so the slope is "fixed" only in the sense that its velocity of the Earth is always approximately zero, but p = mv, and the Earth has a lot of mass. Not sure if that answers your question...it was a little tough for me to understand.
 Thnx...@Soothsayer.... I got it now....Since, Earth has very large mass as compared to the ball...its vel. is negligible...and that's why when we give the velocity back to the ball... So whenever we conserve momentum along y direction ...it is understood that (objects + earth) has been taken as a system...Am i correct here????... By saying common normal ..i meant that impulse imparted by the ball to the (incline + earth) and by (incline + earth) to the ball is along common normal...

Recognitions:

## A Case Of Momentum Conservation

 Quote by D_DaYwAlKeR So whenever we conserve momentum along y direction ...it is understood that (objects + earth) has been taken as a system...Am i correct here????...
Yes. Earth ends up being part of your system if you want to consider momentum conservation, but it's mass might as well be infinite compared to everything else, so it can absorb any amount of momentum without changing velocity measurably.

By the way, momentum conserved in every direction. But yes, since force is applied along normal, only component along that normal changes.

 Right, force is essentially defined as being the change in momentum with respect to time. Forces also come in equal and opposite pairs, so the force the slope imparts on the ball, the ball also imparts on the slope.