Question

A pool player makes a shot on a pool table. The 0.17 kg cue ball strikes the 0.15 kg #8-ball, but it is not sunken on the table. Instead, the 8-ball bounces off a rail, and collides into the cue ball a second time. When they collide in a perfectly elastic collision, the cue ball is travelling 1.2 m/s [N10°E) and the 8-ball is travelling 0.89 m/s [W20°N]. Determine the final velocities of each ball.

Answer 1

cue ball final: 0.85 m/s at N65°W

8-ball final: 1.27 m/s at N12°E

Step-by-step explanation:

In a perfectly elastic collision, both momentum and energy are conserved. The momentum is the sum of products of mass and velocity. The energy is half the product of mass and the square of velocity. The difference of "particle" velocities changes sign as a consequence of the collision.

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equations for final velocity

If we let the masses be m1 and m2, the corresponding initial velocities be v1 and v2, and the final velocities be u1 and u2, we have ...

m1·u1 +m2·u2 = m1·v1 +m2·v2 . . . . . . . conservation of momentum

u1 -u2 = v2 -v1 . . . . . . . . . . . . . . . relative velocity sign changes

This pair of linear equations can be solved for u1 and u2 in any of the usual ways to give ...

u1 = v1(m1 -m2)/(m1 +m2) + v2(2·m2)/(m1 +m2)

u2 = v2(m2 -m1)/(m1 +m2) +v1(2·m1)/(m1 +m2)

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application to this problem

Using angle measures as clockwise from north, we are given ...

cue ball: m1 = 0.17 kg, v1 = 1.2∠10° m/s

eight ball: m2 = 0.15 kg, v2 = 0.89∠-70° m/s

A suitable calculator can find the final velocities using the above equations.

The first attachment shows the calculations using a TI-84 work-alike calculator.

cue ball final: 0.85 m/s at N65°W

8-ball final: 1.27 m/s at N12°E

The second attachment shows a vector diagram of the velocities.