Question 7.SP.39: Two billiard balls meet centrally with speeds of 2 and -4 m/......
Two billiard balls meet centrally with speeds of 2 and -4 m/s. What will be their final speeds after impact if the coefficient of restitution is assumed to be 0.8?
Let subscript 1 refer to the 2 \mathrm{~m} / \mathrm{s} ball and subscript 2 to the -4 \mathrm{~m} / \mathrm{s} ball. The masses are assumed equal \left(m_{1}=m_{2}=m\right) . Hence, u_{1}=2 \mathrm{~m} / \mathrm{s} and u_{2}=-4 \mathrm{~m} / \mathrm{s} . The problem is to determine v_{1} and v_{2} . Work directly from the following fundamental equations rather than apply the results of Problem 7.36:
\begin{array}{c}e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \quad 0.8=\frac{v_{2}-v_{1}}{2-(-4)} \quad \therefore v_{2}-v_{1}=4.8 \mathrm{~m} / \mathrm{s}\qquad \qquad &(1)\\ \\m_{1} u_{1}+m_{2} u_{1}=m_{1} v_{1}+m_{2} v_{2} \quad m(2)+m(-4)=m v_{1}+m v_{2} \quad \therefore v_{1}+v_{2}=-2 \mathrm{~m} / \mathrm{s}\qquad \qquad &(2)\end{array}
Solve the two equations simultaneously to get v_{1}=\underline{-3.8 \mathrm{~m} / \mathrm{s}} and v_{2}=\underline{1.8 \mathrm{~m} / \mathrm{s}} .