Chapter 13, Problem 114A

(a)

To determine

To Sketch: The situation, assigning coordinate axes and identifying “before” and “after”.

Explanation of Solution

Introduction:

The situation can be sketched by applying the conservation of momentum and the notation of vectors.

According the conservation of momentum,

Total initial momentum = total final momentum

To sketch the problem showing the states before and after collision, take the y-axis as north and the x-axis as East.

Before the collision, a compact car is shown by a vector which is heading towards the south while the full - sized car is represented by a vector which is heading towards the east.

After the collision, both cars would move as a wreck with a common velocity in southeast direction which is represented by a single vector.

  

Conclusion:

Hence, the sketch has been drawn.

(b)

To determine

To Find: The direction and speed of the wreck.

Answer to Problem 114A

  $\begin{array}{c}{v}_f=9.\text{4 m/s}\\ \text{θ}={34}^{\circ }\end{array}$

Explanation of Solution

Given:

Mas of compact car, $m_1=875\text{ kg}$

Mass of full-sized car, $m_2=1584\text{ m/s}$

Initial speed of compact car, $v_{1i}=15\text{ m/s}$

Initial speed of full − sized car, $v_{2i}=12\text{ m/s}$

Formula Used:

Velocity of an object is

  $v_f=\frac{p_f}{m_1+m_2}$

Here, $p_f$ is the momentum, $m_1$ and $m_2$ are the mass and $v_f$ is the velocity of the car.

Momentum of the car is given by

  $p_{i,x}=m_2{v}_{2i}$

Here, $p_{i,x}$ is the momentum, ${\text{m}}_2$ is the mass of the car and $v_{2,i}$ is the velocity of the car.

Calculation:

The x-component of the initial momentum of the system will be the sum of the x-components of the momentum of each car. Since the car moving south has no x-component,

  $p_{i,x}=m_2{v}_{2i}$

Substitute $1584\text{kg}$ for $m_2$ , $12\text{m/s}$ for $v_{2i}$ in equation (1)

  $\begin{array}{c}{p}_{i,x}=\left(1584\text{kg}\right)\left(12\text{m/s}\right)\\ =1.9\times {10}^4\text{kg}\text{m/s}\end{array}$........................... (2)

The y-component of the initial momentum is $m_1{v}_{1i}$ since only the car moving south has the y-component of momentum.

Substitute $875\text{kg}$ for $m_1$ , $15\text{m/s}$ for $v_{1i}$ in equation (1)

  $\begin{array}{c}{p}_{i,y}=\left(876\text{kg}\right)\left(15\text{m/s}\right)\\ =1.3\times {10}^4\text{kg}\text{m/s}\end{array}$........................... (3)

By the law of conservation of momentum, it can be written as,

  $p_{f,x}=p_{i,x}$

And,

  $p_{f,y}=p_{i,y}$

Using the vector diagram, it can be written as,

  $p_f=\sqrt{{\left(p_{fx}\right)}^2+{\left(p_{f,y}\right)}^2}$........................... (4)

Substitute $1.9\times {10}^4\text{kg}\cdot \text{m/s}$ for $p_{f,x}$ , $1.3\times {10}^4\text{kg}\cdot \text{m/s}$ for $p_{f,y}$ in equation (4)

  $\begin{array}{c}{p}_f=\sqrt{{\left(1.9\times {10}^4\text{kg}\cdot \text{m/s}\right)}^2+{\left(1.3\times {10}^4\text{kg}\cdot \text{m/s}\right)}^2}\\ =2.3\times {10}^4\text{kg}\cdot \text{m/s}\end{array}$

The direction of momentum is given by

  $\theta \text{=}{\text{tan}}^{-1}\left(\frac{p_{f,y}}{p_{f,x}}\right)$........................... (5)

Substitute $\text{1}{\text{.3×10}}^{\text{4}}\text{kg}\text{.m/s}$ for $p_{f,y}$ , $\text{1}{\text{.9×10}}^{\text{4}}\text{kg}\text{.m/s}$ for $p_{f,x}$ in equation (5)

  $\begin{array}{c}\text{θ}={\tan }^{-1}\left(\frac{\text{1}{\text{.3×10}}^{\text{4}}\text{kg}\text{.m/s}}{\text{1}{\text{.9×10}}^{\text{4}}\text{kg}\text{.m/s}}\right)\\ ={34}^{\circ }\end{array}$

Now, to find the velocity

  $v_f=\frac{p_f}{m_1+m_2}$........................... (6)

Substitute $875\text{kg}$ for $m_1$ and $1584\text{kg}$ for $m_2$ and $2.3\times {10}^4\text{kg}\text{.m/s}$ for $p_f$ in equation (6)

  $\begin{array}{c}{v}_f=\frac{2.3\times {10}^4\text{kg}\cdot \text{m/s}}{875\text{kg}+1584\text{kg}}\\ v_f=9.\text{4 m/s}\end{array}$

Thus, the wreck at an angle of ${34}^{\circ }$ moves with a speed of $9.\text{4 m/s}$ .

Conclusion:

Hence, the wreck is at an angle of ${34}^{\circ }$ with a speed of $9.\text{4 m/s}$ .

(c)

To determine

To Find: The distance travelled by the wreck.

Answer to Problem 114A

  $s=8.18\text{ m}$

Explanation of Solution

Given:

Coefficient of kinetics friction, ${\mu }_k=0.55$

Initial velocity of wreck, $v_i\text{'}=9.4\text{ m/s}$

Final velocity of wreck, $v_f\text{'}=0$

Formula Used:

Newton’s second law:

  $F=ma$

Where,

  $F=$ Force.

  $a=$ Acceleration of the wreck.

Friction force:

  $f={\mu }_k.mg$

Where,

  $f=$ Friction force on the wreck.

  $m=$ Mass of the wreck.

Calculation:

Let $a$ be the constant acceleration of the wreck.

The wreck will come to a stop when the fractional force equivalent to the net force acting on it.

Therefore,

  $\begin{array}{c}{\mu }_{k}mg=ma\\ a={\mu }_{k}g\end{array}$

Plugging in the given values:

  $\begin{array}{c}a=\left(0.55\right)\left(\text{9}\text{.8}{\text{m/s}}^{\text{2}}\right)\\ =5.4{\text{m/s}}^{\text{2}}\end{array}$

Since, the wreck comes to a stop, Therefore, acceleration would be negative

The distance covered can be determined by using the formula

  ${\left(v_f\text{'}\right)}^2={\left(v_i\text{'}\right)}^2-2as$

Plugging in the given values:

  $\begin{array}{c}{0}^2={\left(9.4\text{ m/s}\right)}^2-2\times \left(5.4{\text{ m/s}}^2\right)\times s\\ \therefore \\ s=\frac{{\left(9.4\right)}^2}{2\times 5.4}\text{ m}\\ s=8.18\text{ m}\end{array}$

Conclusion:

Hence, the wreck skids $8.18\text{m}$ far after the impact.