Question 13.160: Two identical cars A and B are at rest on a loading dock wit...

m_{A}=m_{B}=m_{C}=m

Collision between B and C:

The total momentum is conserved:

\begin{gathered}\overset{+}{\longleftarrow } m \nu_{B}^{\prime}+m \nu_{C}^{\prime}=m \nu_{B}+m \nu_{C} \\\nu_{B}^{\prime}+\nu_{C}^{\prime}=0+1.5\quad\quad\quad\quad\text{(1)}\end{gathered}

Relative velocities:

\begin{aligned}\left(\nu_{B}-\nu_{C}\right)\left(e_{B C}\right) & =\left(\nu_{C}^{\prime}-\nu_{B}^{\prime}\right) \\(-1.5)(0.8) & =\left(\nu_{C}^{\prime}-\nu_{B}^{\prime}\right) \\-1.2 & =\nu_{C}^{\prime}-\nu_{B}^{\prime}\quad\quad\quad \text{(2)}\end{aligned}

Solving (1) and (2) simultaneously,

\begin{aligned}& \nu_{B}^{\prime}=1.35 \mathrm{~m} / \mathrm{s} \\& \nu_{C}^{\prime}=0.15 \mathrm{~m} / \mathrm{s}\end{aligned}

\mathbf{v}_{C}^{\prime}=0.150 \mathrm{~m} / \mathrm{s}\longleftarrow \blacktriangleleft

Since \nu_{B}^{\prime}>\nu_{C}^{\prime}, car B collides with car A.

Collision between A and B :

\begin{aligned}m \nu_{A}^{\prime}+m \nu_{B}^{\prime \prime} & =m \nu_{A}+m \nu_{B}^{\prime} \\\nu_{A}^{\prime}+\nu_{B}^{\prime \prime} & =0+1.35\quad\quad\quad\quad\text{(3)}\end{aligned}

Relative velocities:

\begin{aligned}\left(\nu_{A}-\nu_{B}^{\prime}\right) e_{A B} & =\left(\nu_{B}^{\prime \prime}-\nu_{A}^{\prime}\right) \\(0-1.35)(0.5) & =\nu_{B}^{\prime \prime}-\nu_{A}^{\prime} \\\nu_{A}^{\prime}-\nu_{B}^{\prime \prime} & =0.675\quad\quad\quad\quad\text{(4)}\end{aligned}

Solving (3) and (4) simultaneously,

2 \nu_{A}^{\prime}=1.35+0.675

\begin{gathered}\mathbf{v}_{A}^{\prime}=1.013 \mathrm{~m} / \mathrm{s}\longleftarrow \blacktriangleleft \\\mathbf{v}_{B}^{\prime \prime}=0.338 \mathrm{~m} / \mathrm{s}\longleftarrow \blacktriangleleft\end{gathered}

Since \nu_{C}^{\prime}<\nu_{B}^{\prime \prime}<\nu_{A}^{\prime}, there are no further collisions.