Chapter 7, Problem 35P

To determine

The speed of the sports car.

Answer to Problem 35P

The initial speed of the sports car before impact is $23 \text{m/s}$

Explanation of Solution

Given:

Mass of the sports car and SUV is $920 \text{kg}$ and $2300 \text{kg}$ . Initially, SUV is at rest. Skid distance is $2.8 \text{m}$ . The coefficient of kinetic friction between tires and the road is $0.80$ . After the collision, both sports car and SUV skids with the same velocity.

Formula used:

According to the conservation of momentum principle, the total momentum before and after the collision is always constant,

  $\begin{array}{l}{p}_i=p_f\\ {\displaystyle \sum m_i{u}_i}={\displaystyle \sum m_f{u}_f}\end{array}$

And, kinetic energy between both the cars is lost due to the negative work done by the friction. Thus, work done by the friction is

  $W=F\cdot s=Fs \cos \theta$

Calculation:

Let, the mass of sports car is $m_A$ and the mass of SUV is $m_B$ . Final velocity using the momentum conservation principle is

  $\begin{array}{l}{m}_A{u}_A+0=(m_A+m_B)v_A\\ u_A=\frac{(m_A+m_B)v_A}{m_A}\left(1\right)\end{array}$

Now, formulating final velocity using work done formula

  $W={(K{E}_f-K{E}_i)}_{after collision}=0-\frac{1}{2}\left(m_A+m_B\right)v_A{}^2$

Also, work done due to friction is

  $\begin{array}{l}{W}_{fr}=Fs \cos \theta =-\mu (m_A+m_B)gs \cos {180}^{\circ }\\ W_{fr}=-\mu (m_A+m_B)gs\end{array}$

Now, equating both the equation of work done to evaluate final velocity,

  $\begin{array}{l}-\frac{1}{2}\left(m_A+m_B\right)v_A{}^2=-\mu (m_A+m_B)gs\\ v_A=\sqrt{2\mu gs}\end{array}$

Now, inserting the value of final velocity in equation $1$ to calculate the initial velocity of sports car

  $\begin{array}{l}{u}_A=\frac{(m_A+m_B)\sqrt{2\mu gs}}{m_A}=\frac{(920+2300)\sqrt{2\times 0.8\times 9.8\times 2.8}}{920}\\ u_A=23.1\approx 23\text{m/s}\end{array}$

Conclusion:

The initial speed of the sports car before impact is $23 \text{m/s}$