What is the speed of the 8-ball after the collision?

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The discussion focuses on calculating the speed of the 8-ball after an elastic collision with the cue ball, which has an initial speed of 1.3 m/s. The 8-ball moves at an angle of 33.2 degrees post-collision, while both balls have equal mass. Key principles include the conservation of momentum and energy, which are essential for solving the problem. The mention of potential energy (Mgh) is deemed irrelevant in this context, as the collision occurs on a frictionless surface.

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  1. In a pool game, the cue ball, which has an initial speed of 1.3 m/s, makes an elastic collision with the 8-ball, which is initially at rest. After the collision, the 8-ball moves at an angle theta = 33.2o with respect to the original direction of the cue ball, as shown in the figure. Assume the pool table is frictionless and the masses of the cue ball and the 8-ball are equal. What is the speed of the 8-ball after the collision?

    prob11a.gif
2. Mgh + W = 1/2mv^2
p=mv
3. I was easily able to get that phi was 56.8 deg because of the 90 deg angle. Now I am lost at how to find the velocity of the 8 ball without the velocity of the cue ball?
 
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Hello White and welcome to PF :-)

Your collection of relevant equations needs revision and extension. Mgh and W are unlikely to play a role in this drama...
 
Conservation of momentum and energy are most important. But "Mgh" is irrelevant here. It is the potential energy of a mass M at height h, but there is no "height" here.
 

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