Question 14.SP.6: In a game of billiards, ball A is given an initial velocity ......
In a game of billiards, ball A is given an initial velocity v_{0} with a magnitude of 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. Balls A and C hit the sides of the table squarely at points A′ and C′, respectively, and B hits the side obliquely at B′. Assuming frictionless surfaces and perfectly elastic impacts, determine the velocities v_{A},~v_{B},~and~v_{C} with which the balls hit the sides of the table. (Remark: In this Sample Problem and in several of the problems that follow, we assume the billiard balls are particles moving freely in a horizontal plane, rather than the rolling and sliding spheres they actually are.)
STRATEGY: Because there are no externally applied forces, use the conservation of linear and angular momentum. Because you are told that the impacts are perfectly elastic, you can also use the conservation of energy (but note that in general, energy is lost in an impact).
MODELING and ANALYSIS: Choose the system to be all three billiard balls and model them as particles.
Conservation of Momentum. There is no external force, so the initial momentum mv_{0} is equipollent to the system of momenta after the two collisions (and before any of the balls hit the sides of the table). Referring to Fig. 1, you have
\underrightarrow{+}x components: \qquad \qquad m(10\ \mathrm{ft}/s)=m(\nu_{B})_{x}+m\,\nu_{c} (1)
+↑ y components: \qquad \qquad \qquad 0=m\nu_{A}-m(\nu_{B})_{y} (2)
+↺ moments about O: (-2\mathrm{~ft})m(10\mathrm{~ft}/s)=(8\mathrm{~ft})m\nu_{A}-(7\mathrm{~ft})m(\nu_{B})_{y}-(3\mathrm{~ft})m\nu_{C} (3)
Solving the three equations for \nu_{A},(\nu_{B})_{x},\mathrm{and}\left(\nu_{B}\right)_{y} in terms of v_{C} gives
\nu_{A}=(\nu_{B})_{y}=3\nu_{C}-20\qquad(\nu_{B})_{x}=10-\nu_{C} (4)
Conservation of Energy. The surfaces are frictionless and the impacts are perfectly elastic, so the initial kinetic energy \textstyle{\frac{1}{2}}m\nu_{0}^{2} is equal to the final kinetic energy of the system:
\textstyle{\frac{1}{2}}m\nu_{0}^{2}={\frac{1}{2}}m\nu_{A}^{2}+{\frac{1}{2}}m\nu_{B}^{2}+{\frac{1}{2}}m\nu_{C}^{2}
\nu_{A}^{2}+(\nu_{B})_{x}^{2}+(\nu_{B})_{y}^{2}+\nu_{C}^{2}=(10\,\,\mathrm{ft}/s)^{2} (5)
Substituting for \nu_{A},(\nu_{B})_{x},\mathrm{and}\;(\nu_{B})_{y} from Eqs. (4) into Eq. (5), you have
2(3\nu_{C}-20)^{2}+(10-\nu_{C})^{2}+\nu_{C}^{2}=100
20\nu_{C}^{2}-260\nu_{C}+800=0
Solving for v_C, you find \nu_{c}=5\ \mathrm{ft}/s\ \mathrm{and}\ \nu_{c}=8\ \mathrm{ft}/s. Because only the second root yields a positive value for v_A after substitution into Eqs. (4), then v_C = 8 ft/s and
\nu_{A}=(\nu_{B})_{y}=3(8)-20=4\;\mathrm{ft}/s\qquad\qquad(\nu_{B})_{x}=10-8=2\;\mathrm{ft}/s
\mathbf{v}_{A}=4\mathrm{~ft}/s\uparrow\qquad\mathbf{v}_{B}=4.47\mathrm{~ft}/s~ ⦪ \mathrm{~63}.4^{\circ}\qquad\mathbf{v}_{C}=8\mathrm{~ft}/s\to ◂
REFLECT and THINK: In a real situation, energy would not be conserved, and you would need to know the coefficient of restitution between the balls to solve this problem. We also neglected friction and the rotation of the balls in our analysis, which is often a poor assumption in pool or billiards. We discuss rigid-body impacts in Chap. 17.
