Chapter 13.4, Problem 13.162P

To determine

(a)

The velocities of each car if A and C hits B at the same time.

Answer to Problem 13.162P

$\begin{array}{l}\leftarrow v_A{}^1=1.288m/s\\ \to v_B{}^1=0.312m/s\\ \to v_C{}^1=1.512m/s\end{array}$

Explanation of Solution

Given information:

Mass of each bumper car is equal to $200kg$.

Riders A, B and C have mass of $40kg,60kg,35kg.$

$\begin{array}{l}{v}_A=2m/s\\ v_C=1.5m/s\\ e=0.8\end{array}$

The total linear momentum of two particles is conserved. Therefore,

$m_A{v}_A+m_B{v}_B=m_A{v}_A{}^1+m_B{v}_B{}^1$

The co-efficient of restitution is defined as:

$v_B{}^1-v_A{}^1=e\left(v_A-v_B\right)$

Calculation:

The mass of each bumper car with riders:

$\begin{array}{l}{m}_A=240kg\\ m_B=260kg\\ m_C=235kg\end{array}$

Apply conservation of linear momentum for all 3 bumper cars:

$\to m_A{v}_A+m_B{v}_B+m_C{v}_C=m_A{v}_A{}^1+m_B{v}_B{}^1+m_C{v}_C{}^1$

Substitute,

$\begin{array}{l}240\left(2\right)+0+235\left(-1.5\right)=240v_A{}^1+260v_B{}^1+235v_C{}^1\\ 240v_A{}^1+260v_B{}^1+235v_C{}^1=127.5\to \left(1\right)\end{array}$

The co-efficient of restitution for A and B:

$v_B{}^1-v_A{}^1=e\left(v_A-v_B\right)$

Substitute,

$\begin{array}{l}{v}_B{}^1-v_A{}^1=0.8\left(2\right)\\ v_B{}^1-v_A{}^1=1.6\to \left(2\right)\end{array}$

The co-efficient of restitution for B and C,

$v_C{}^1-v_B{}^1=e\left(v_B-v_C\right)$

Substitute,

$\begin{array}{l}{v}_C{}^1-v_B{}^1=0.8\left(0-\left(-1.5\right)\right)\\ v_C{}^1-v_B{}^1=1.2\to \left(3\right)\end{array}$

By solving equations 1,2 and 3,

$\begin{array}{l}{v}_A{}^1=-1.288m/s\\ v_B{}^1=0.312m/s\\ v_C{}^1=1.512m/s\end{array}$

Conclusion:

The velocities of each car are calculated by solving equations 1, 2 and 3.

To determine

(b)

The velocities of each car if A hits B before C does.

Answer to Problem 13.162P

$\begin{array}{l}\leftarrow v_A{}^{111}=0.9563m/s\\ \leftarrow v_B{}^{111}=0.0296m/s\\ \to v_C{}^{11}=1.552m/s\end{array}$

Explanation of Solution

Given information:

Mass of each bumper car is equal to $200kg$

Riders A, B and C have mass of $40kg,60kg,35kg$

$\begin{array}{l}{v}_A=2m/s\\ v_C=1.5m/s\\ e=0.8\end{array}$

The total linear momentum of two particles is conserved. Therefore,

$m_A{v}_A+m_B{v}_B=m_A{v}_A{}^1+m_B{v}_B{}^1$

The co-efficient of restitution is defined as,

$v_B{}^1-v_A{}^1=e\left(v_A-v_B\right)$

Calculation:

The mass of each bumper car with riders,

$\begin{array}{l}{m}_A=240kg\\ m_B=260kg\\ m_C=235kg\end{array}$

Apply conservation of linear momentum for A and B,

$\to m_A{v}_A+m_B{v}_B=m_A{v}_A{}^1+m_B{v}_B{}^1$

Substitute,

$\begin{array}{l}240\left(2\right)+0=240v_A{}^1+260v_B{}^1\\ 240v_A{}^1+260v_B{}^1=480\to \left(1\right)\end{array}$

The co-efficient of restitution for A and B,

$v_B{}^1-v_A{}^1=e\left(v_A-v_B\right)$

Substitute,

$\begin{array}{l}{v}_B{}^1-v_A{}^1=0.8\left(2\right)\\ v_B{}^1-v_A{}^1=1.6\to \left(2\right)\end{array}$

Solve equations 1 and 2,

$\begin{array}{l}{v}_A{}^1=0.128m/s\\ v_B{}^1=1.728m/s\end{array}$

For the second impact,

Apply conservation of linear momentum for A and B,

$\to m_B{v}_B{}^1+m_C{v}_C=m_B{v}_B{}^{11}+m_C{v}_C{}^1$

Substitute,

$\begin{array}{l}260\left(1.728m/s\right)+235\left(-1.5\right)=260v_B{}^{11}+235v_C{}^{11}\\ 260v_B{}^{11}+235v_C{}^{11}=96.78\to \left(1\right)\end{array}$

The co-efficient of restitution for A and B,

$v_C{}^{11}-v_B{}^{11}=e\left(v_B{}^1-v_C\right)$

Substitute,

$\begin{array}{l}{v}_C{}^{11}-v_B{}^{11}=0.8\left(1.728-\left(-1.5\right)\right)\\ v_C{}^{11}-v_B{}^{11}=2.5824\to \left(2\right)\end{array}$

Solve above equations,

$\begin{array}{l}{v}_B{}^{11}=-1.0304m/s\\ v_C{}^{11}=1.552m/s\end{array}$

For the third impact,

Apply conservation of linear momentum for A and B,

$\to m_A{v}_A{}^1+m_B{v}_B{}^{11}=m_A{v}_A{}^{111}+m_B{v}_B{}^{111}$

Substitute,

$\begin{array}{l}240\left(0.128\right)+260\left(-1.0304\right)=240v_A{}^{111}+260v_B{}^{111}\\ 240v_A{}^{111}+260v_B{}^{111}=-237.184\to \left(3\right)\end{array}$

The co-efficient of restitution for A and B,

$\left(v_B{}^{111}-v_A{}^{111}\right)=e\left(v_A{}^1-v_B{}^{11}\right)$

Substitute,

$\begin{array}{l}{v}_B{}^{111}-v_A{}^{111}=0.8\left(0.128-\left(-1.0304\right)\right)\\ v_B{}^{111}-v_A{}^{111}=0.92672\to \left(4\right)\end{array}$

Solve equations 3 and 4,

$\begin{array}{l}{v}_A{}^{111}=-0.9563m/s\\ v_B{}^{111}=-0.02955m/s\end{array}$

Therefore, the final velocities are,

$\begin{array}{l}\leftarrow v_A{}^{111}=0.9563m/s\\ \leftarrow v_B{}^{111}=0.0296m/s\\ \to v_C{}^{11}=1.552m/s\end{array}$

Conclusion:

The velocities of all 3 cars after, car A hits B before C does,

$\begin{array}{l}\leftarrow v_A{}^{111}=0.9563m/s\\ \leftarrow v_B{}^{111}=0.0296m/s\\ \to v_C{}^{11}=1.552m/s\end{array}$