Chapter 14.1, Problem 14.20P

Knowing that the coordinates of the utility pole are $x_p=46$ ft and $y_p=59$ ft, determine (a) the time elapsed from the first collision to the stop at P, (b) the speed of car A.

To determine

(a)

Time elapsed from first collision to the stop at $P$.

Answer to Problem 14.20P

The time elapsed from first collision to stop at pole is $2.61\text{ s}$.

Explanation of Solution

Given:

Speed of car B is $45\text{ mi/hr}$.

Speed of car C is $60\text{ mi/hr}$.

Weight if car A is $3000\text{ lb}$.

Weight of car B is $2600\text{ lb}$.

Weight of car C is $2400\text{lb}$.

Coordinates of utility pole are $\left(46\text{ft},59\text{ft}\right)$.

Concept used:

Draw the diagram for the system mentioned as shown below:

Assume origin as point O where, Car A and Car B collides.

Write the expression for the mass center of first collision.

$\left(m_A+m_B+m_C\right)\left(x_0\hat{i}+y_0\hat{j}\right)=m_A\left(x_A\hat{i}+y_A\hat{j}\right)+m_B\left(x_B\hat{i}+y_B\hat{j}\right)+m_C\left(x_C\hat{i}+y_C\hat{j}\right)$ ...... (1)

Multiply the whole equation (1) with $g$ on both sides.

$\left(W_A+W_B+W_C\right)\left(x_0\hat{i}+y_0\hat{j}\right)=W_A\left(x_A\hat{i}+y_A\hat{j}\right)+W_B\left(x_B\hat{i}+y_B\hat{j}\right)+W_C\left(x_C\hat{i}+y_C\hat{j}\right)$ ...... (2)

Here, $W_A$ is the weight of car A, $W_B$ is the weight of car B, $W_C$ is the weight of car C, $\left(x_0,y_0\right)$ is coordinate of center of mass, $\left(x_A,y_A\right)$ is the coordinate of car A, $\left(x_B,y_B\right)$ is the coordinate of car B and $\left(x_C,y_C\right)$ is the coordinate of car C.

Write the expression for conservation of momentum for the system.

$\left(m_A+m_B+m_C\right)\overrightarrow{v}=m_A{\overrightarrow{v}}_A+m_B{\overrightarrow{v}}_B+m_C{\overrightarrow{v}}_C$ ...... (3)

Multiply the equation (3) with $g$ on both sides.

$\left(W_A+W_B+W_C\right)\overrightarrow{v}=W_A{\overrightarrow{v}}_A+W_B{\overrightarrow{v}}_B+W_C{\overrightarrow{v}}_C$ ...... (4)

Here, ${\overrightarrow{v}}_A$ is the velocity of car A, ${\overrightarrow{v}}_B$ is the velocity of car B, ${\overrightarrow{v}}_C$ is the velocity of car C and $\overrightarrow{v}$ is the velocity of center of mass of system.

Write the expression for the position of pole $P$.

$x_P\hat{i}+y_p\hat{j}=x_0\hat{i}+y_0\hat{j}+\overrightarrow{v}t$ ...... (5)

Here, $\left(x_P,y_P\right)$ is the coordinate of pole $P$.

Calculation:

Substitute $3000\text{ lb}$ for $W_A$, $2600\text{ lb}$ for $W_B$, $2400\text{ lb}$ for $W_C$, $0$ for $x_A$, $0$ for $y_A$, $0$ for $x_B$, $0$ for $y_B$, $32\text{ ft}$ for $x_C$ and $10\text{ ft}$ for $y_C$ in equation (2).

$\begin{array}{c}\left(3000+2600+2400\right)\left(x_0\hat{i}+y_0\hat{j}\right)=\left[\begin{array}{l}3000\left(0\hat{i}+0\hat{j}\right)+2600\left(0\hat{i}+0\hat{j}\right)\\ +2400\left(32\hat{i}+10\hat{j}\right)\end{array}\right]\\ 8000x_0\hat{i}+8000y_0\hat{j}=76800\hat{i}+24000\hat{j}\end{array}$

Equate $\hat{i}$ components of above expression.

$x_0=9.6\text{ ft}$

Equate $\hat{j}$ components of the above expression.

$y_0=3\text{ ft}$

Substitute $3000\text{ lb}$ for $W_A$, $2600\text{ lb}$ for $W_B$, $2400\text{ lb}$ for $W_C$, $-\left(60\text{mi/hr}\right)\hat{i}$ for ${\overrightarrow{v}}_C$ and $\left(45\text{ mi/hr}\right)\hat{j}$ for ${\overrightarrow{v}}_B$ in equation (4).

$\begin{array}{c}\left(\begin{array}{l}3000+2600\\ +2400\end{array}\right)\overrightarrow{v}=\left[\begin{array}{l}3000{\overrightarrow{v}}_A+2600\left(45\text{ mi/hr}\left(\frac{1.467\text{ ft/s}}{1\text{mi/hr}}\right)\right)\hat{j}\\ +2400\left(-60\text{mi/hr}\left(\frac{1.467\text{ ft/s}}{1\text{mi/hr}}\right)\right)\hat{i}\end{array}\right]\\ 8000\overrightarrow{v}=3000{\overrightarrow{v}}_A+171639\hat{j}-211248\hat{i}\\ \overrightarrow{v}=0.375{\overrightarrow{v}}_A+21.455\hat{j}-26.406\hat{i}\end{array}$

Substitute $9.6\text{ ft}$ for $x_0$, $3\text{ ft}$ for $y_0$, $0.375{\overrightarrow{v}}_A+21.455\hat{j}-26.406\hat{i}$ for $\overrightarrow{v}$, $46\text{ ft}$ for $x_P$, $59\text{ ft}$ for $y_P$ in equation (5).

$\begin{array}{c}46\hat{i}+59\hat{j}=9.6\hat{i}+3.0\hat{j}+\left(0.375{\overrightarrow{v}}_A+21.455\hat{j}-26.406\hat{i}\right)t\\ =9.6\hat{i}+3.0\hat{j}+0.375t{\overrightarrow{v}}_A+21.455t\hat{j}-26.406t\hat{i}\end{array}$

Substitute $v_A\hat{i}$ for ${\overrightarrow{v}}_A$ in above expression.

$\begin{array}{l}46\hat{i}+59\hat{j}=9.6\hat{i}+3.0\hat{j}+0.375t{v}_A\hat{i}+21.455t\hat{j}-26.406t\hat{i}\\ 46\hat{i}+59\hat{j}=\left[9.6+\left(0.375v_A-26.406\right)t\right]\hat{i}+\left(3+21.455t\right)\hat{j}\end{array}$

Equate $\hat{j}$ components of above expression.

$\begin{array}{l}59=\left(3+21.455t\right)\\ 56=21.455t\end{array}$

Rearrange the above expression for $t$.

$t=2.61\text{ s}$

The time elapsed from first collision to stop at pole is $2.61\text{ s}$.

Conclusion:

Thus, the time elapsed from first collision to stop at pole is $2.61\text{ s}$.

To determine

(b)

Speed of car A.

Answer to Problem 14.20P

The velocity of car A is $73.3\text{}\text{ mi/hr}$.

Explanation of Solution

Calculation:

Substitute $9.6\text{ ft}$ for $x_0$, $3\text{ ft}$ for $y_0$, $0.375{\overrightarrow{v}}_A+21.455\hat{j}-26.406\hat{i}$ for $\overrightarrow{v}$, $46\text{ ft}$ for $x_P$, $59\text{ ft}$ for $y_P$ in equation (5).

$\begin{array}{c}46\hat{i}+59\hat{j}=9.6\hat{i}+3.0\hat{j}+\left(0.375{\overrightarrow{v}}_A+21.455\hat{j}-26.406\hat{i}\right)t\\ =9.6\hat{i}+3.0\hat{j}+0.375t{\overrightarrow{v}}_A+21.455t\hat{j}-26.406t\hat{i}\end{array}$

Substitute $v_A\hat{i}$ for ${\overrightarrow{v}}_A$ in above expression.

$\begin{array}{l}46\hat{i}+59\hat{j}=9.6\hat{i}+3.0\hat{j}+0.375t{v}_A\hat{i}+21.455t\hat{j}-26.406t\hat{i}\\ 46\hat{i}+59\hat{j}=\left[9.6+\left(0.375v_A-26.406\right)t\right]\hat{i}+\left(3+21.455t\right)\hat{j}\end{array}$

Equate the $\hat{i}$ components of both side in above expression.

$\begin{array}{c}46=\left[9.6+\left(0.375v_A-26.406\right)t\right]\\ 36.4=\left(0.375v_A-26.406\right)t\end{array}$

Substitute $2.61\text{ s}$ for $t$ in above expression.

$\begin{array}{l}36.4=\left(0.375v_A-26.406\right)\left(2.61\right)\\ 36.4=0.97875v_A-68.92\end{array}$

Simplify above expression for $v_A$.

$\begin{array}{c}{v}_A=107.6\text{ft/s}\left(\frac{0.681\text{ mi/hr}}{1\text{ft/s}}\right)\\ =73.3\text{ mi/hr}\end{array}$

The velocity of car A is $73.3\text{}\text{ mi/hr}$.

Conclusion:

Thus, the velocity of car A is $73.3\text{}\text{ mi/hr}$.