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Consider the following problem:

A ball is dropped with an angle a between its direction and the normal of the floor and bounces up with an angle b between its direction and the normal of the floor. Does the momentum conservation principle apply?

I would say that it does since there do not exist any external forces on the system.

But when I apply the law I get this:

$-mv_0cosa+0 = mv_1cosb+0$

(I use the y-axis to calculate the speed)

which in turns gives this:

$v_1 = -mv_0cosa/cosb$

It just feels wrong. Can I really apply the law of conservation of momentum???

Any help is appreciated. Thanks :)

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  • $\begingroup$ Note that there is change in the velocity of the ball bouncing around, which means that there is an acceleration and hence a force acting on it. Since you don't include that force and the ground to your system you cannot apply conservation of momentum. $\endgroup$
    – Gonenc
    May 11, 2015 at 19:23

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You didn't cancel out the mass $m$ properly.

$$-mv_0\cos a+0=mv_1\cos b+0\\ -v_0\cos a=v_1\cos b\\ \frac{-v_0\cos a}{\cos b}=v_1$$

The incident angle is equal to the exit angle in such collision, $a=b$. The above reduces to:

$$v_1=\frac{-v_0\cos a}{\cos b}=\frac{-v_0\cos b}{\cos b}=-v_0$$

And here you see what you probably expected. In the perpendicular the ball leaves with the same speed it came with, just backwards.

If angles $a$ and $b$ are not equal, an external force has made it's ground-shaking effect! Then your expression will not be simple.

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