Chapter 7, Problem 68GP

To determine

To show: The speed of the car A before applying the brakes is exceeding $90\text{km}/\text{h}$ .

Explanation of Solution

Given:

The mass of the car A is $m_1=1,900\text{ kg}$ .

The mass of the car B is $m_2=1,100\text{ kg}$ .

The initial velocity of car B is $u_2=0$ .

The distance traveled by the car A before the collision is $x_{\text{A}}=15\text{ m}$ .

The distance traveled by the car A after the collision is $x_{\text{A}}{}^{\prime }=18\text{ m}$ .

The distance traveled by the car B after the collision is $x_{\text{B}}{}^{\prime }=30\text{ m}$ .

The coefficient of kinetic friction is ${\mu }_{\text{k}}=0.6$ .

Formula used:

The expression for the distance can be sliding with respect to the change in kinetic energy is,

  $-{\mu }_{\text{k}}g\Delta x=\frac{1}{2}\left(v_{\text{f}}^{\text{2}}-v_{\text{i}}^{\text{2}}\right)$

Here, ${\mu }_{\text{k}}$ is the coefficient of kinetic friction, g is the acceleration due to gravity, $\Delta x$ is the slid distance, $v_{\text{f}}$ is the final velocity, and $v_{\text{i}}$ is the initial velocity.

The expression for law of conservation of momentum is,

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Here, $m_1$ is the mass of the car B, $m_2$ is the mass of car B, $u_1$ is the initial velocity of car B, $u_2$ is the initial velocity of the car B, $v_1$ is the final velocity of car B, and $v_2$ is the final velocity of the car B.

Calculation:

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^{\text{2}}$ .

Consider the final velocity of car A and car B in this collision is $v_{\text{fA}}=v_{\text{fB}}=0$ .

The velocity of car A after collision is,

  $\begin{array}{l}-{\mu }_{\text{k}}g{x}_{\text{A}}{}^{\prime }=\frac{1}{2}\left(v_{\text{f}}^{\text{2}}-v_{\text{i}}^{\text{2}}\right)\\ -\left(0.60\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(18\text{ m}\right)=\frac{1}{2}\left(0-v_{\text{1}}^{\text{2}}\right)\\ -211.68=-v_{\text{1}}^{\text{2}}\\ v_{\text{1}}=14.55\text{m}/\text{s}\end{array}$

The velocity of car B after collision is,

  $\begin{array}{l}-{\mu }_{\text{k}}g{x}_{\text{B}}{}^{\prime }=\frac{1}{2}\left(v_{\text{f}}^{\text{2}}-v_{\text{i}}^{\text{2}}\right)\\ -\left(0.60\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\text{30 m}\right)=\frac{1}{2}\left(0-v_{\text{2}}^{\text{2}}\right)\\ -352.8=-v_{\text{2}}^{\text{2}}\\ v_{\text{2}}=18.78\text{m}/\text{s}\end{array}$

The velocity of car B after collision is,

  $\begin{array}{l}{m}_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\\ \left(1,900\text{ kg}\right)u_1+\left(1,100\text{ kg}\right)\left(0\right)=\left(1,900\text{ kg}\right)\left(14.55\text{m}/\text{s}\right)+\left(1,100\text{ kg}\right)\left(18.78\text{m}/\text{s}\right)\\ 1,900u_1=48,303\\ u_1=25.42\text{m}/\text{s}\end{array}$

Consider the final velocity of car A when the break were first applied is $v_{\text{f}}=u_1$ .

The velocity of car A the break were first applied using the relation is,

  $\begin{array}{l}-{\mu }_{\text{k}}g{x}_{\text{A}}=\frac{1}{2}\left(u_{\text{1}}^{\text{2}}-v_{\text{i}}^{\text{2}}\right)\\ -\left(0.60\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(15\text{ m}\right)=\frac{1}{2}\left({\left(25.42\text{m}/\text{s}\right)}^2-v_{\text{i}}^{\text{2}}\right)\\ -176.4=646.18-v_{\text{i}}^{\text{2}}\\ v_{\text{i}}^{\text{2}}=822.58\end{array}$

  $\begin{array}{l}{v}_{\text{i}}=28.68\text{m}/\text{s}\times \frac{1\text{ km}}{1,000\text{ m}}\times \frac{60\text{ s}}{1\text{ min}}\times \frac{60\text{ min}}{1\text{ h}}\\ v_{\text{i}}=103.25\text{km}/\text{h}>90\text{km}/\text{h}\end{array}$

Conclusion:

Thus, the speed of the car A before applying the brakes exceeds $\text{90 }\text{km}/\text{h}$ .