Solving Pool Table Problem: Cue Ball Momentum Change

[seraphimhouse]

Homework Statement

Figure 9-48 gives an overhead view of the path taken by a 0.160 kg cue ball as it bounces from a rail of a pool table. The ball's initial speed is 1.47 m/s, and the angle θ1 is 62.9°. The bounce reverses the y component of the ball's velocity but does not alter the x component. What are (a) angle θ2 and (b) the magnitude of the change in the ball's linear momentum? (The fact that the ball rolls is irrelevant to the problem.)

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c09/fig09_47.gif

Homework Equations

p = mv where p and v are vectors.

The Attempt at a Solution

before the moment of impact, I found the Vx and Vy of the ball traveling through the pythagorean theorem and I know that the velocity along the x-axis is constant. I'm just having a hard time trying to comprehend this problem.

I tried this equation:

J = Pf - Pi = M (Vf-Vi) where J, P and V are vectors

Jx = 0 because the Vx is equal on before and after the collision.

But still found myself stuck.

 

[Delphi51]

Having found Vx and Vy initially, all you have to do is put a minus sign on the Vy and you have the components after the collision. It should be easy to use a tangent calc to get the angle. The change in velocity is the new Vy minus the old Vy. That will not be zero because of the sign change. The change in momentum is the mass times the change in velocity.

 

[kerimek]

I would approach this problem by first breaking down the given information and identifying what is known and unknown. From the given information, we know the mass of the cue ball (0.160 kg), its initial speed (1.47 m/s), and the initial angle θ1 (62.9°). We are also given the fact that the bounce does not alter the x component of the ball's velocity.

Next, I would use the conservation of momentum principle, which states that the total momentum of a system remains constant in the absence of external forces. In this case, the only external force acting on the cue ball is the bounce from the rail. Therefore, the total momentum of the cue ball before and after the bounce will be equal.

Using the equation p = mv, we can calculate the initial momentum of the cue ball as:

pi = (0.160 kg)(1.47 m/s) = 0.2352 kg m/s

Since the x component of the velocity remains constant, the final momentum in the x direction will also be 0.2352 kg m/s. However, the y component of the velocity will be reversed after the bounce, so the final momentum in the y direction will be -0.2352 kg m/s.

Now, we can use vector addition to find the final momentum of the cue ball after the bounce. This can be done by adding the x and y components of the final momentum together using the Pythagorean theorem:

pf = √(px^2 + py^2) = √(0.2352^2 + (-0.2352)^2) = 0.3324 kg m/s

Next, we can use the fact that momentum is conserved to find the angle θ2. This can be done by using the equation tan θ = py/px, where py and px are the y and x components of the final momentum, respectively. Therefore, we can solve for θ2 as:

θ2 = tan^-1(-0.2352/0.2352) = -45°

Finally, to find the magnitude of the change in the ball's linear momentum, we can simply subtract the initial momentum from the final momentum:

Δp = pf - pi = 0.3324 - 0.2352 = 0.0972 kg m/s

In conclusion, the angle θ2 is -45