Question 13.155: The coefficient of restitution between the two collars is kn...

Impulse-momentum principle (collars A and B ):

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

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

Using data, \quad (5)(1)+(3)(-1.5)=5 \nu_{A}^{\prime}+3 \nu_{B}^{\prime}

or \quad 5 \nu_{A}^{\prime}+3 \nu_{B}^{\prime}=0.5    (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}=0.70[1-(-0.5)] \\& \nu_{B}^{\prime}-\nu_{A}^{\prime}=1.75 \quad\quad \text{(2)}\end{aligned}

(a) Solving Eqs. (1) and (2) simultaneously for the velocities,

Kinetic energies: T_{1}=\frac{1}{2} m_{A} \nu_{A}^{2}+\frac{1}{2} m_{B} \nu_{B}^{2}=\frac{1}{2}(5)(1)^{2}+\frac{1}{2}(3)(-1.5)^{2}=5.875 \mathrm{~J}

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}(5)(-0.59375)^{2}+\frac{1}{2}(3)(1.15625)^{2}=2.8867 \mathrm{~J}

(b) Energy loss: \quad\quad\quad\quad T_{1}-T_{2}=2.99 \mathrm{~J}\blacktriangleleft