Question 9.118: In the two-sphere arrangement of Fig. 9-20, assume that sphe......

We refer to the discussion in the textbook (Sample Problem – “Elastic collision, two pendulums,” which uses the same notation that we use here) for some important details in the reasoning. We choose rightward in Fig. 9-20 as our +x direction. We use the notation v when we refer to velocities and v when we refer to speeds (which are necessarily positive). Since the algebra is fairly involved, we find it convenient to introduce the notation \vec{v} when we refer to velocities and v when we refer to speeds (which are necessarily positive). Since the algebra is fairly involved, we find it convenient to introduce the notation Δm = m_2 – m_1 (which, we note for later reference, is a positive-valued quantity).

(a) Since \vec{v}_{1 i}=+\sqrt{2 g h_1} where h_1 = 9.0 cm, we have

\vec{v}_{1 f}=\frac{m_1-m_2}{m_1+m_2}v_{1 i}=-\frac{\Delta m}{m_1+m_2}\sqrt{2 g h_1}

which is to say that the speed of sphere 1 immediately after the collision is

v_{1 f}=c \Delta m / b m_1+m_2 g h \sqrt{2 g h_1}

and that \vec{v}_{1 f} points in the –x direction. This leads (by energy conservation m_1 g h_{1 f}=\frac{1}{2}m_1 v_{1 f}^2 ) to

h_{1 f}=\frac{v_{1 f}^2}{2 g}=\left\lceil\frac{\Delta m}{m_1+m_2}|^2 h_1\right. \text{.}

With m_1 = 50 g and m_2 = 85 g, this becomes h_{1f} ≈ 0.60 cm .

(b) Equation 9-68 gives

v_{2 f}=\frac{2 m_1}{m_1+m_2}v_{1 i}\text{.}       (9-68)

v_{2 f}=\frac{2 m_1}{m_1+m_2}v_{1 i}=\frac{2 m_1}{m_1+m_2}\sqrt{2 g h_1}

which leads (by energy conservation m_2 g h_{2 f}=\frac{1}{2}m_2 v_{2 f}^2 ) to

h_{2 f}=\frac{v_{2 f}^2}{2 g}=\left\lceil\frac{2 m_1}{m_1+m_2}|^2 h_1\right. \text{.}

With m_1 = 50 g and m_2 = 85 g, this becomes h_{2f} ≈ 4.9  cm.

(c) Fortunately, they hit again at the lowest point (as long as their amplitude of swing was “small,” this is further discussed in Chapter 16). At the risk of using cumbersome notation, we refer to the next set of heights as h_{1ff} and h_{2ff}. At the lowest point (before this second collision) sphere 1 has velocity +\sqrt{2 g h_{1 f}} (rightward in Fig. 9-20) and sphere 2 has velocity -\sqrt{2 g h_{1 f}} (that is, it points in the –x direction). Thus, the velocity of sphere 1 immediately after the second collision is, using Eq. 9-75,

v_{1 f}=\frac{m_1-m_2}{m_1+m_2}v_{1 i}+\frac{2 m_2}{m_1+m_2}v_{2 i}            (9-75)