How Do You Calculate the Change in Momentum of a Billiard Ball?

[ctlpsl]

I have this problem that I'm not sure I am on the right track. The problem is A .20 kg billiard ball traveling at a speed of 15m/s strikes the side rail of the pool table at an angle of 60 degrees. If the ball rebounds at the same speed and angle, what is the cahnge in its momentum.
I have the following formulas:
∆p=p2-p1 = ∆px + ∆py

∆px = p2x-p1x 2pcos θ

∆py = p2y – p1y = p2sin θ – p1 sin θ = 0

So, if I calculate p = mv I get .3kg*m/s
Do I plug that value in for p in the equations above, and then I'm not sure what to do with the ∆py equation.

 

[Hootenanny]

I'm not sure what you doing with this bit;

∆p=p2-p1 = ∆px + ∆py

Also, p = mv $\neq$0.3.

I suggest writing down the intial and final momentums taking note of their direction.

~H

 

[dx]

Notice that the vertical component of the momentum is the same since the speed of the ball after the collision is the same and the angle is the same. Therefore, the only change is in the horizontal component ( let's call it the x component ). What happens to the horizontl component of the momentum? The horizontal component of the velocity is reversed right? It has to be brought to zero first, and then back to its initial speed in the opposite direction. So the magnitude of the change in velocity is 2v. Therefore the change in momentum is
$$2P_x$$

 

[ctlpsl]

Problem worked

Here is the full description and how I worked the problem

 

[ctlpsl]

the final answere

I forgot to include the final answere
then I have
∆p = ∆px - ∆py = -5.714 kg*m/s – 2.08 kg*m/s = -7.794 kg*m/s

 

[Hootenanny]

As dx said before, there is no change in the x component (the diagram changed things slightly) of the momentum. This equation; ∆p = ∆px - ∆py is just plain wrong. I suggest you start by writing it down like this;

Inital momentum;
p_xi = ...
p_yi = ...

Final momentum;
p_xf = ...
p_yf = ...

Resolve you momentums into x and y components before you begin. For continuity make any motion in the left to right plane x, and in the bottom to top plane y.

~H