Three steel spheres of equal weight are suspended from the ceiling by cords of equal length which are spaced at a distance slightly greater than the diameter of the spheres. After being pulled back and released, sphere A hits sphere B, which then hits sphere C. Denoting by e the coefficient of restitution between the spheres and by \mathbf{ v }_{ 0 } the velocity of A just before it hits B, determine (a) the velocities of A and B immediately after the first collision, (b) the velocities of Band C immediately after the second collision. (c) Assuming now that n spheres are suspended from the ceiling and that the first sphere is pulled back and released as described above, determine the velocity of the last sphere after it is hit for the first time. (d) Use the result of Part c to obtain the velocity of the last sphere when n = 5 and e = 0.9.
Question 13.161: Three steel spheres of equal weight are suspended from the c...
(a) First collision (between A and B ):
The total momentum is conserved:
\begin{aligned}m \nu_{A}+m \nu_{B} & =m \nu_{A}^{\prime}+m \nu_{B}^{\prime} \\\nu_{0} & =\nu_{A}^{\prime}+\nu_{B}^{\prime}\quad\text{(1)}\end{aligned}
Relative velocities:
\begin{gathered}\left(\nu_{A}-\nu_{B}\right) e=\left(\nu_{B}^{\prime}-\nu_{A}^{\prime}\right) \\\nu_{0} e=\nu_{B}^{\prime}-\nu_{A}^{\prime}\quad\quad\text{(2)}\end{gathered}
Solving Equations (1) and (2) simultaneously,
\begin{gathered}\nu_{A}^{\prime}=\frac{\nu_{0}(1-e)}{2} \blacktriangleleft\\\nu_{B}^{\prime}=\frac{\nu_{0}(1+e)}{2}\blacktriangleleft\end{gathered}
(b) Second collision (between B and C ):
The total momentum is conserved.
m \nu_{B}^{\prime}+m \nu_{C}=m \nu_{B}^{\prime \prime}+m \nu_{C}^{\prime}
Using the result from (a) for \nu_{B}^{\prime}
\frac{\nu_{0}(1+e)}{2}+0=\nu_{B}^{\prime \prime}+\nu_{C}^{\prime} (3)
Relative velocities:
\left(\nu_{B}^{\prime}-0\right) e=\nu_{C}^{\prime}-\nu_{B}^{\prime \prime}
Substituting again for \nu_{B}^{\prime} from (a)
\nu_{0} \frac{(1+e)}{2}(e)=\nu_{C}^{\prime}-\nu_{B}^{\prime \prime} (4)
Solving equations (3) and (4) simultaneously,
\nu_{C}^{\prime}=\frac{1}{2}\left[\frac{\nu_{0}(1+e)}{2}+\nu_{0}(1+e) \frac{(e)}{2}\right] \\ \begin{aligned}\nu_{C}^{\prime} & =\frac{\nu_{0}(1+e)^{2}}{4} \blacktriangleleft\\\nu_{B}^{\prime \prime} & =\frac{\nu_{0}\left(1-e^{2}\right)}{4}\blacktriangleleft\end{aligned}
(c) For n spheres
n balls
(n-1) th collision,
we note from the answer to part (b) with n = 3
\nu_{n}^{\prime}=\nu_{3}^{\prime}=\nu_{C}^{\prime}=\frac{\nu_{0}(1+e)^{2}}{4}
or \quad \nu_{3}^{\prime}=\frac{\nu_{0}(1+e)^{(3-1)}}{2^{(3-1)}}
Thus, for n balls
\nu_{n}^{\prime}=\frac{ν_{0}(1+e)^{(n-1)}}{2^{(n-1)}}\blacktriangleleft
(d) For n = 5, e = 0.90,
from the answer to part (c) with n = 5
\begin{aligned}\nu_{B}^{\prime} & =\frac{\nu_{0}(1+0.9)^{(5-1)}}{2^{(5-1)}} \\& =\frac{\nu_{0}(1.9)^{4}}{(2)^{4}}\end{aligned}
\nu_{B}^{\prime}=0.815 \nu_{0}\blacktriangleleft
