Chapter 13.4, Problem 13.162P

At an amusement park, there are 200-kg bumper cars A, B, and C that have riders with masses of 40 kg, 60 kg, and 35 kg, respectively. Car A is moving to the right with a velocity v_A = 2 m/s and car C has a velocity v_B = 1.5 m/s to the left, but car B is initially at rest. The coefficient of restitution between each car is 0.8. Determine the final velocity of each car, after all impacts, assuming (a) cars A and C hit car B at the same time, (b) car A hits car B before car C does.

To determine

Find the final velocity of each car after all impact, assuming car A $\left({v^{\prime }}_A\right)$ and C $\left({v^{\prime }}_C\right)$ hit car B $\left({v^{\prime }}_B\right)$ at the same time.

Answer to Problem 13.162P

The final velocity of each car after all impact, assuming car A $\left({v^{\prime }}_A\right)$ and C $\left({v^{\prime }}_C\right)$ hit car B $\left({v^{\prime }}_B\right)$ at the same time are $\underset{\_}{1.288\text{m}/\text{s}(\leftarrow )}$, $\underset{\_}{1.512\text{m}/\text{s}(\to )}$, and $\underset{\_}{0.312\text{m}/\text{s}(\to )}$ respectively.

Explanation of Solution

Given information:

The mass of the bumper car (m) is $200\text{kg}$.

The mass of the rider A $\left({m^{\prime }}_A\right)$ is $40\text{kg}$.

The mass of the rider B $\left({m^{\prime }}_B\right)$ is $60\text{kg}$.

The mass of the rider C $\left({m^{\prime }}_C\right)$ is $35\text{kg}$.

The velocity of A $\left(v_A\right)$ is $\text{2m/s}$.

The velocity of C $\left(v_C\right)$ is $-\text{1}\text{.5}\text{m/s}$.

The coefficient of restitution between each car (e) is 0.8.

Calculation:

Calculate the total mass of car A along with rider $\left(m_A\right)$ using the relation:

$m_A=\text{mass of car}\left(m\right)+\text{mass of rider}A\left({m^{\prime }}_A\right)$

Substitute $200\text{kg}$ for m and $40\text{kg}$ for ${m^{\prime }}_A$.

$\begin{array}{c}{m}_A=\text{200}\text{kg}+\text{40}\text{kg}\\ =\text{240}\text{kg}\end{array}$

Calculate the total mass of the car B along with rider $\left(m_B\right)$ using the relation:

$m_B=\text{mass of car}\left(m\right)+\text{mass of rider}B\left({m^{\prime }}_B\right)$

Substitute $200\text{kg}$ for m and $60\text{kg}$ for ${m^{\prime }}_B$.

$\begin{array}{c}{m}_B=\text{200}\text{kg}+\text{60}\text{kg}\\ =\text{260}\text{kg}\end{array}$

Calculate the total mass of the car C along with rider $\left(m_C\right)$ using the relation:

$m_C=\text{mass of car}\left(m\right)+\text{mass of rider}C\left({m^{\prime }}_C\right)$

Substitute $200\text{kg}$ for m and $35\text{kg}$ for ${m^{\prime }}_C$.

$\begin{array}{c}{m}_C=\text{200}\text{kg}+\text{35}\text{kg}\\ =\text{235}\text{kg}\end{array}$

Assume the velocities towards the right to be positive and the velocities towards the left to be negative.

The velocity will be zero as the car B $\left(v_B\right)$ is at rest initially.

The expression for the principle of conservation of momentum to the cars A, B, and C when cars A and C hit the car B at the same time as follows;

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

Here, $v_A$ is the initial velocity of car A, ${v^{\prime }}_A$ is the final velocity of car A after the impact, $v_B$ is the initial velocity of car B, ${v^{\prime }}_B$ is the final velocity of car B after the impact, $v_C$ is the initial velocity of car C and ${v^{\prime }}_C$ is the final velocity of car C after the impact.

Substitute $\text{240}\text{kg}$ for $m_A$, $\text{260}\text{kg}$ for $m_B$, $\text{235}\text{kg}$ for $m_C$, $2\text{m}/\text{s}$ for $v_A$, $-1.5\text{m}/\text{s}$ for $v_C$ and 0 for $v_B$.

$\begin{array}{l}\left\{\begin{array}{l}\left(240\text{kg}\right)\left(2\text{m}/\text{s}\right)+0+\\ \left(235\text{kg}\right)\left(-1.5\text{m}/\text{s}\right)\end{array}\right\}=\left\{\begin{array}{l}\left(240\text{kg}\right){v^{\prime }}_A+\left(260\text{kg}\right){v^{\prime }}_B\\ +\left(235\text{kg}\right){v^{\prime }}_C\end{array}\right\}\\ 480-352.5=240{v^{\prime }}_A+260{v^{\prime }}_B+235{v^{\prime }}_C\\ 240{v^{\prime }}_A+260{v^{\prime }}_B+235{v^{\prime }}_C=127.5\end{array}$ (1)

Calculate the coefficient of restitution (e) of the impact between the cars A and B using the formula:

$e=\frac{{v^{\prime }}_B-{v^{\prime }}_A}{v_A-v_B}$

Substitute 0.8 for e, $2\text{m}/\text{s}$ for $v_A$, and 0 for $v_B$.

$\begin{array}{l}0.8=\frac{{v^{\prime }}_B-{v^{\prime }}_A}{\left(2\text{m}/\text{s}\right)-0}\\ {v^{\prime }}_B-{v^{\prime }}_A=\left(0.8\right)\left(2\right)\\ {v^{\prime }}_B-{v^{\prime }}_A=1.6\end{array}$ (2)

Calculate the coefficient of restitution $\left(e\right)$ of the impact between the cars B and C using the formula:

$e=\frac{{v^{\prime }}_C-{v^{\prime }}_B}{v_B-v_C}$

Substitute 0.8 for e, $-1.5\text{m}/\text{s}$ for $v_C$, and 0 for $v_B$.

$\begin{array}{l}0.8=\frac{{v^{\prime }}_C-{v^{\prime }}_B}{0-\left(-1.5\text{m}/\text{s}\right)}\\ {v^{\prime }}_C-{v^{\prime }}_B=\left(0.8\right)\left(1.5\right)\\ {v^{\prime }}_C-{v^{\prime }}_B=1.2\end{array}$ . (3)

Solve the equations (1) and (2) and (3) to obtain velocities.

Add the equations (2) and (3) to eliminate ${v^{\prime }}_B$.

$\begin{array}{l}{v^{\prime }}_B-{v^{\prime }}_A+{v^{\prime }}_C-{v^{\prime }}_B=1.2+1.6\\ {v^{\prime }}_C-{v^{\prime }}_A=2.8\end{array}$ (4)

Multiply the equation (2) with 260 and subtract it from the equation (1).

$\begin{array}{l}240{v^{\prime }}_A+260{v^{\prime }}_B+235{v^{\prime }}_C-\left({v^{\prime }}_B-{v^{\prime }}_A\right)260=127.5-\left(1.6\right)\left(260\right)\\ 235{v^{\prime }}_C+500{v^{\prime }}_A=127.5-416\\ 235{v^{\prime }}_C+500{v^{\prime }}_A=-288.5\end{array}$ (5)

Multiply the equations (4) with 500 and add it to the equation (5) to obtain the final velocity of the car C.

$\begin{array}{l}\left({v^{\prime }}_C-{v^{\prime }}_A\right)500+500{v^{\prime }}_A+235{v^{\prime }}_C=\left(2.8\right)500-288.5\\ 500{v^{\prime }}_C+235{v^{\prime }}_C=1,400-288.5\\ 735{v^{\prime }}_C=1,111.5\\ {v^{\prime }}_C=1.512\text{m}/\text{s}(\to )\end{array}$

Substitute $1.512\text{m}/\text{s}$ for ${v^{\prime }}_C$ in the equation (3) to obtain the velocity of car B.

$\begin{array}{l}{v^{\prime }}_C-{v^{\prime }}_B=1.2\\ 1.512\text{m}/\text{s}-{v^{\prime }}_B=1.2\\ {v^{\prime }}_B=0.312\text{m}/\text{s}(\to )\end{array}$

Substitute $0.312\text{m}/\text{s}$ for ${v^{\prime }}_B$ in the equation (2) to obtain the velocity of the car A.

$\begin{array}{l}{v^{\prime }}_B-{v^{\prime }}_A=1.6\\ 0.312\text{m}/\text{s}-{v^{\prime }}_A=1.6\\ {v^{\prime }}_A=0.312-1.6\\ {v^{\prime }}_A=-1.288\text{m}/\text{s}\\ {v^{\prime }}_A=1.288\text{m}/\text{s}(\leftarrow )\end{array}$

Therefore, the final velocity of each car after all impact, assuming car A $\left({v^{\prime }}_A\right)$ and C $\left({v^{\prime }}_C\right)$ hit car B $\left({v^{\prime }}_B\right)$ at the same time are $\underset{\_}{1.288\text{m}/\text{s}(\leftarrow )}$, $\underset{\_}{1.512\text{m}/\text{s}(\to )}$ and $\underset{\_}{0.312\text{m}/\text{s}(\to )}$ respectively.

To determine

Find the final velocity of each car after all impact, assuming car A $\left({v^{‴}}_A\right)$ hits car B $\left({v^{‴}}_B\right)$ before car C does.

Answer to Problem 13.162P

The final velocity of each car after all impact, assuming car A $\left({v^{‴}}_A\right)$ hits car B $\left({v^{‴}}_B\right)$ before car C does are $\underset{\_}{0.9563\text{m}/\text{s}(\leftarrow )}$ and $\underset{\_}{0.02955\text{m}/\text{s }(\leftarrow )}$ respectively.

Explanation of Solution

Given information:

The mass of the bumper car (m) is $200\text{kg}$.

The mass of the rider A $\left({m^{\prime }}_A\right)$ is $40\text{kg}$.

The mass of the rider B $\left({m^{\prime }}_B\right)$ is $60\text{kg}$.

The mass of the rider C $\left({m^{\prime }}_C\right)$ is $35\text{kg}$.

The velocity of A $\left(v_A\right)$ is $\text{2m/s}$.

The velocity of B $\left(v_B\right)$ is $\text{-1}\text{.5m/s}$.

The coefficient of restitution between each car (e) is 0.8.

Calculation:

Calculate the final velocities of the cars when car A hits car B before car C does.

The expression for the principle of conservation of momentum to the first impact between car A and car B as follows:

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

Substitute $\text{240}\text{kg}$ for $m_A$, $\text{260}\text{kg}$ for $m_B$, $2\text{m}/\text{s}$ for $v_A$, and 0 for $v_B$.

$\begin{array}{l}\left(240\text{kg}\right)\left(2\text{m}/\text{s}\right)+0=\left(240\text{kg}\right){v^{\prime }}_A+\left(260\text{kg}\right){v^{\prime }}_B\\ 240{v^{\prime }}_A+260{v^{\prime }}_B=480\end{array}$ (6)

Calculate the coefficient of restitution (e) of the first impact between the cars A and B using the formula:

$e=\frac{{v^{\prime }}_B-{v^{\prime }}_A}{v_A-v_B}$

Substitute 0.8 for e, $2\text{m}/\text{s}$ for $v_A$, and 0 for $v_B$.

$\begin{array}{l}0.8=\frac{{v^{\prime }}_B-{v^{\prime }}_A}{\left(2\text{m}/\text{s}\right)-0}\\ {v^{\prime }}_B-{v^{\prime }}_A=\left(0.8\right)\left(2\right)\\ {v^{\prime }}_B-{v^{\prime }}_A=1.6\end{array}$ (7)

Multiply the equations (7) with 240 and add it to the equation (6) to obtain the final velocity of car B.

$\begin{array}{l}\left({v^{\prime }}_B-{v^{\prime }}_A\right)240+240{v^{\prime }}_A+260{v^{\prime }}_B=\left(1.6\right)\left(240\right)+480\\ 500{v^{\prime }}_B=384+480\\ {v^{\prime }}_B=\frac{864}{500}\\ {v^{\prime }}_B=1.728\text{m}/\text{s}(\to )\end{array}$

Substitute $1.728\text{m}/\text{s}$ for ${v^{\prime }}_B$ in the equation (7).

$\begin{array}{l}1.728\text{m}/\text{s}-{v^{\prime }}_A=1.6\\ {v^{\prime }}_A=1.728-1.6\\ {v^{\prime }}_A=0.128\text{m}/\text{s}(\to )\end{array}$

The expression for the principle of conservation of momentum for the second impact between car B and car C as follows:

$m_B{v^{\prime }}_B+m_C{v}_C=m_B{v^{″}}_B+m_C{v^{″}}_C$

Here, the final velocity of the car B after the second impact is ${v^{″}}_B$.

Substitute $\text{260}\text{kg}$ for $m_B$, $\text{235}\text{kg}$ for $m_C$, $-1.5\text{m}/\text{s}$ for $v_C$, and $1.728\text{m}/\text{s}$ for ${v^{\prime }}_B$.

$\begin{array}{l}\left\{\begin{array}{l}\left(260\text{kg}\right)\left(1.728\text{m}/\text{s}\right)+\\ \left(235\text{kg}\right)\left(-1.5\text{m}/\text{s}\right)\end{array}\right\}=\left(260\text{kg}\right){v^{″}}_B+\left(235\text{kg}\right){v^{″}}_C\\ 449.28-352.5=260{v^{″}}_B+235{v^{″}}_C\\ 260{v^{″}}_B+235{v^{″}}_C=96.78\end{array}$ (8)

The expression for the coefficient of restitution $\left(e\right)$ of the second impact between the cars B and C as follows:

$e=\frac{{v^{″}}_C-{v^{″}}_B}{{v^{\prime }}_B-v_C}$

Substitute 0.8 for e, $-1.5\text{m}/\text{s}$ for $v_C$ and $1.728\text{m}/\text{s}$ for ${v^{\prime }}_B$.

$\begin{array}{l}0.8=\frac{{v^{″}}_C-{v^{″}}_B}{\left(1.728\text{m}/\text{s}\right)-\left(-1.5\text{m}/\text{s}\right)}\\ {v^{″}}_C-{v^{″}}_B=3.228\left(0.8\right)\\ {v^{″}}_C-{v^{″}}_B=2.5824\end{array}$ (9)

Multiply the equation (9) with 260 and add it to the equation (8) to obtain the final velocity of car C after the impact.

$\begin{array}{l}\left({v^{″}}_C-{v^{″}}_B\right)\left(260\right)+260{v^{\prime }}_B+235v_C=\left(2.5824\right)\left(260\right)+96.78\\ 260{v^{″}}_C+235{v^{″}}_C=671.424+96.78\\ 495{v^{″}}_C=768.204\\ {v^{″}}_C=1.5519\text{m}/\text{s}(\to )\end{array}$

Substitute $1.552\text{m}/\text{s}$ for ${v^{″}}_C$ in the equation (9).

$\begin{array}{l}{v^{″}}_C-{v^{″}}_B=2.5824\\ 1.552\text{m}/\text{s}-{v^{″}}_B=2.5824\\ {v^{″}}_B=1.552-2.5824\\ {v^{″}}_B=-1.0304\text{m}/\text{s}\\ {v^{″}}_B=1.0304\text{m}/\text{s}(\leftarrow )\end{array}$

Consider the car A and car B again impact with each other.

The expression for the principle of conservation of momentum to the third impact between the car A and car B as follows;

$m_A{v^{\prime }}_A+m_B{v^{″}}_B=m_A{v^{‴}}_A+m_B{v^{‴}}_B$

Here, ${v^{‴}}_A$ the final velocity of the car A after the third impact and ${v^{‴}}_B$ the final velocity of the car after the third impact.

Substitute $\text{240}\text{kg}$ for $m_A$, $\text{260}\text{kg}$ for $m_B$, $0.128\text{m}/\text{s}$ for ${v^{\prime }}_A$, and $-1.0304\text{m}/\text{s}$ for ${v^{″}}_B$.

$\begin{array}{l}\left\{\begin{array}{l}\left(240\text{kg}\right)\left(0.128\text{m}/\text{s}\right)+\\ \left(260\text{kg}\right)\left(-1.0304\text{m}/\text{s}\right)\end{array}\right\}=\left(240\text{kg}\right){v^{‴}}_A+\left(260\text{kg}\right){v^{‴}}_B\\ 240{v^{‴}}_A+260{v^{‴}}_B=30.72-267.904\\ 240{v^{‴}}_A+260{v^{‴}}_B=-237.184\end{array}$ (10)

Calculate the coefficient of restitution (e) of the third impact between the cars A and B using the formula:

$e=\frac{{v^{‴}}_B-{v^{‴}}_A}{{v^{\prime }}_A-{v^{″}}_B}$

Substitute 0.8 for e, $0.128\text{m}/\text{s}$ for ${v^{\prime }}_A$ and $-1.0304\text{m}/\text{s}$ for ${v^{″}}_B$.

$\begin{array}{l}0.8=\frac{{v^{‴}}_B-{v^{‴}}_A}{\left(0.128\text{m}/\text{s}\right)-\left(-1.0304\text{m}/\text{s}\right)}\\ {v^{‴}}_B-{v^{‴}}_A=\left(0.8\right)\left(1.1584\right)\\ {v^{‴}}_B-{v^{‴}}_A=0.92672\end{array}$ (11)

Multiply the equation (11) with 240 and add it to the equation (1) to obtain the final velocity of the car B.

$\begin{array}{l}\left({v^{‴}}_B-{v^{‴}}_A\right)\left(240\right)+240{v^{‴}}_A+260{v^{‴}}_B=\left(0.92672\right)\left(240\right)-237.184\\ 240{v^{‴}}_B+260{v^{‴}}_B=222.413-237.184\\ 500{v^{‴}}_B=-14.771\\ {v^{‴}}_B=-0.02955\text{m}/\text{s }\\ {v^{‴}}_B=0.02955\text{m}/\text{s }(\leftarrow )\end{array}$

Substitute $-0.02955\text{m}/\text{s }$ for ${v^{‴}}_B$ in the equation (11).

$\begin{array}{l}{v^{‴}}_B-{v^{‴}}_A=0.92672\\ -0.02955\text{m}/\text{s}-{v^{‴}}_A=0.92672\\ -{v^{‴}}_A=0.92672+0.02955\\ {v^{‴}}_A=-0.9563\text{m}/\text{s}\\ {v^{‴}}_A=0.9563\text{m}/\text{s}(\leftarrow )\end{array}$

Therefore, the final velocity of each car after all impact, assuming car A $\left({v^{‴}}_A\right)$ hits car B $\left({v^{‴}}_B\right)$ before car C does are $\underset{\_}{0.9563\text{m}/\text{s}(\leftarrow )}$ and $\underset{\_}{0.02955\text{m}/\text{s }(\leftarrow )}$ respectively.