Chapter 9.2, Problem 28PP

To determine

To identify:Whether the car will exceed speed limit or not.

Answer to Problem 28PP

The velocity of the north-moving car is exceeding the $20.1\text{m}/\text{s}$ speed limit.

Explanation of Solution

Given:

The velocity of the car moving towards east is $v_{\text{1i}}=15.7\text{m}/\text{s}$ .

The final velocity of the system is $\text{14}\text{.5 }\text{m}/\text{s}$ .

The mass of the system towards east is $m_1=1,345\text{kg}$ .

The mass of the system towards north is $m_2=1,923\text{kg}$ .

The angle is $\theta =63.5°$

Formula used:

The expression for the momentum is

  $p=mv$

Here, $m$ is the mass of the car and $v$ is the velocity of the car.

Calculation:

Consider the y-axis as north and x-axis as east. The state of the car before and after collision and the vector diagram are shown below.

  

The x-component of the initial momentum of the system $1$ is $p_{1\text{i}}$ .

The y-component of the momentum of the system $2$ is $p_{\text{2i}}$ .

The final momentum of the system is $p_{\text{f}}$ .

The velocity of the car moving towards east is $v_{\text{1i}}=15.7\text{m}/\text{s}$ .

The velocity of the car moving towards north is $v_{\text{2i}}$ .

The x-component of the initial momentum is,

  $\begin{array}{l}{p}_{\text{1i}}=m_1{v}_{\text{1i}}\\ p_{\text{1i}}=\left(1,345\text{ kg}\right)\times \left(15.7\text{m}/\text{s}\right)\\ p_{\text{1i}}=2.11\times {10}^4\text{kg}\cdot \text{m}/\text{s}\end{array}$

The final momentum of the system after the car sticks together is,

  $\begin{array}{l}{p}_{\text{f}}=\left(m_1+m_2\right)v_f\\ p_{\text{f}}=\left(1,345\text{ kg}+1,923\text{ kg}\right)\times \left(14.5\text{m}/\text{s}\right)\\ p_{\text{f}}=4.74\times {10}^4\text{kg}\cdot \text{m}/\text{s}\end{array}$

The y-component of the final momentum of the system is,

  $\begin{array}{l}{p}_{\text{f,y}}=p_{\text{f}}\sin \theta \\ p_{\text{f,y}}=\left(4.74\times {10}^4\text{kg}\cdot \text{m}/\text{s}\right)\times \left(\sin 63.5°\right)\\ p_{\text{f,y}}=4.24\times {10}^4\text{kg}\cdot \text{m}/\text{s}\end{array}$

By the law of conservation of momentum, the y-component of the final momentum and initial momentum of the systemare equal.

The velocity of the north-moving car is,

  $\begin{array}{l}{p}_{\text{f,y}}=m_2{v}_{2\text{i}}\\ \left(4.24\times {10}^4\text{kg}\cdot \text{m}/\text{s}\right)=\left(1,923\text{ kg}\right)\times v_{2\text{i}}\\ v_{2\text{i}}=22.05\text{m}/\text{s}\end{array}$

Hence, the velocity of the north-moving car is exceeding the speed limit.

Conclusion:

Therefore, the velocity of the north-moving car is exceeding the $20.1\text{m}/\text{s}$ speed limit.