Question 13.156: Collars A and B, of the same mass m, are moving toward each ...

Impulse-momentum principle (collars A and B):

\Sigma m \mathbf{v}_{1}+\Sigma \mathbf{Imp}_{1 \rightarrow 2}=\Sigma m \mathbf{v}_{2}

Horizontal components \overset{+}{→}: m_{A} \nu_{A}+m_{B} \nu_{B}=m_{A} \nu_{A}^{\prime}+m_{B} \nu_{B}^{\prime}

Using data, \quad m \nu+m(-ν)=m \nu_{A}^{\prime}+m \nu_{B}^{\prime}

or \quad\nu_{A}^{\prime}+\nu_{B}^{\prime}=0     (1) 

Apply coefficient of restitution.

\begin{aligned}& \nu_{B}^{\prime}-\nu_{A}^{\prime}=e\left(\nu_{A}-\nu_{B}\right) \\& \nu_{B}^{\prime}-\nu_{A}^{\prime}=e[\nu-(-\nu)] \\& \nu_{B}^{\prime}-\nu_{A}^{\prime}=2 e \nu & \text{(2)}\end{aligned}

Subtracting Eq. (1) from Eq. (2), 

\begin{aligned}-2 \nu_{A} & =2 e \nu \\\nu_{A} & =-e \nu && \mathbf{v}_{A}=e \nu\longleftarrow\end{aligned}

Adding Eqs. (1) and (2),

\begin{aligned}2 \nu_{B} & =2 e \nu \\\nu_{B} & =e \nu && \mathbf{v}_B=e\nu\longrightarrow \end{aligned}

Kinetic energies:

T_{1}=\frac{1}{2} m_{A} \nu_{A}^{2}+\frac{1}{2} m_{B} \nu_{B}^{2}=\frac{1}{2} m \nu^{2}+\frac{1}{2} m(-\nu)^{2}=m \nu^{2}

T_{2}=\frac{1}{2} m_{A}\left(\nu_{A}^{\prime}\right)^{2}+\frac{1}{2} m_{B}\left(\nu_{B}^{\prime}\right)^{2}=\frac{1}{2} m(e \nu)^{2}+\frac{1}{2} m(e \nu)^{2}=e^{2} m \nu^{2}

Energy loss: \quad\quad\quad\quad\quad\quad T_{1}-T_{2}=\left(1-e^{2}\right) m \nu^{2}\blacktriangleleft