In a game of pool, the cue ball strikes another ball of the same mass and initia

Question

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at 4.75 m/s along a line making an angle of -18° (i.e. below the x-axis) with the cue ball's original direction of motion (the x-axis), and the second ball has a speed of 2.569 m/s. Find the angle between the direction of motion of the second ball and the original direction of motion of the cue ball. Find the original speed of the cue ball. Is kinetic energy (of the centers of mass, don't consider the rotation) conserved? (Y/N)

Explanation / Answer

apply the law on conservaton of momentum along the horizontal direction

as m1v1 + m2*0 = m1vf cos theta + m2V2f cos theta

V1i = (4.75 * cos 18 ) + (2.569 * cos theta)

Now along vertical path,

- m1v1f sin theta + m2v2f cos 18 = 0+ 0 = 0

sin phi = (4.75* sin 18/(2.569)

phi = 34.81 deg with +x axis

intial Speed of cue ball = Vi2 = (4.75 * cos 18 ) + (2.569 cos 34.81)

Vi2 = 6.626 m/s
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using the law of conservation of ENErgy , we get

KEi = 0.5 mv1^2 + 0.5 m2v22

KE i = (0.5 * m * (6.626^2 ) + 0

KE i = 21.95 *m Joules

KEf = 0.5 * m * (2.569^2 + 4.75^2)

KE f= 14.58 m J

so as KEF and initial are not same

KE is not conserved

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