Question 14.5: In a game of billiards, ball A is given an initial velocity ......
In a game of billiards, ball A is given an initial velocity \text v _0 of magnitude v _0 = 10 ft/s along line DA parallel to the axis of the table. It hits ball B and then ball C, which are both at rest. Knowing that A and C hit the sides of the table squarely at points A^{\prime} ~\text { and } ~C^{\prime}, respectively, that B hits the side obliquely at B^{\prime}, and assuming frictionless surfaces and perfectly elastic impacts, determine the velocities \text v _A,\text v _B ,~\text { and }~\text v _C with which the balls hit the sides of the table. (Remark: In this sample problem and in several of the problems which follow, the billiard balls are assumed to be particles moving freely in a horizontal plane, rather than the rolling and sliding spheres they actually are.)
Conservation of Momentum. Since there is no external force, the initial momentum m\text v _0 is equipollent to the system of momenta after the two collisions (and before any of the balls hits the side of the table). Referring to the adjoining sketch, we write
\overset{+}{\text y}~x~components:~~~~~~~~ m(10~ ft / s )=m\left(v_B\right)_x+m v_C (1)
{+}{\text x}~y~components:~~~~~~~~ 0=m v_A-m\left(v_B\right)_y (2)
+l moments about O: -(2 ~ft ) m(10~ ft / s )=(8 ~ft ) m v_A \\ \\ -(7~ ft ) m\left(v_B\right)_y-(3 ~ft ) m v_C (3)
Solving the three equations for v_A,\left(v_B\right)_x, ~\text { and }~\left(v_B\right)_y ~\text { in terms of }~ v_C,
v_A=\left(v_B\right)_y=3 v_C-20 \quad\left(v_B\right)_x=10-v_C (4)
Conservation of Energy. Since the surfaces are frictionless and the impacts are perfectly elastic, the initial kinetic energy \frac{1}{2} m v_0^2 is equal to the final kinetic energy of the system:
\frac{1}{2} m v_0^2=\frac{1}{2} m_A v_A^2+\frac{1}{2} m_B v_B^2+\frac{1}{2} m_C v_C^2
v_A^2+\left(v_B\right)_x^2+\left(v_B\right)_y^2+v_C^2=(10~ ft / s )^2 (5)
Substituting for v_A,\left(v_B\right)_x,~ \text { and }~\left(v_B\right)_y from (4) into (5), we have
2\left(3 v_C-20\right)^2+\left(10-v_C\right)^2+v_C^2=100 \\ \\ 20 v_C^2-260 v_C+800=0Solving for v_C, we find v_C = 5 ft/s and v_C = 8 ft/s. Since only the second root yields a positive value for v_A after substitution into Eqs. (4), we conclude that v_C = 8 ft/s and
v_A=\left(v_B\right)_y=3(8)-20=4 ~ft / s \quad\quad\quad\left(v_B\right)_x=10-8=2 ~ft / s\text v _A=4~ ft / s~\text x \quad\quad \text v _B=4.47 ~ft / s \quad c ~~63.4^{\circ} \quad\quad \text v _C=8~ ft / s~\text y
