A 980-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.6 m before stopping. The police officer, estimating the coefficient of kinetic friction between tires and road to be 0.80, calculates the speed of the sports car at impact. What was that speed?

The linear momentum of a body is defined as the product of its mass and velocity. It is given as \(p = mv\).

The principle of conservation of linear momentum states that if two bodies collide with each other, total linear momentum before and after collision remains the same if no external force acts on the system.

\(\begin{array}{c}{p_{{\rm{before}}}} = {p_{{\rm{after}}}}\\{\left( {mv} \right)_{{\rm{before}}}} = {\left( {mv} \right)_{{\rm{after}}}}\end{array}\)

The mass of sports car is\(m = 980\;{\rm{kg}}\).

The mass of the SUV is\(M = 2300\;{\rm{kg}}\).

The distance covered by the two cars before stopping is\(x = 2.6\;{\rm{m}}\).

The coefficient of kinetic friction between tires and road is \({\mu _{\rm{k}}} = 0.80\).

Let the direction of motion of the sports car be along the positive x-direction. Suppose A represents the sports car and B represents the SUV. Initially, the SUV is at rest, so its initial velocity is \({v_{\rm{B}}} = 0\). The two cars move as one unit, so their final velocities are the same, \({v_{\rm{A}}}^\prime = {v_{\rm{B}}}^\prime = v'\).

The pseudo force is equal to the frictional force, and the normal force is equal to the combined weight of the cars. Therefore,

\(\begin{array}{c}ma = {\mu _{\rm{k}}}N\\ma = {\mu _{\rm{k}}}mg\\a = \left( {0.80} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\\ = 7.84\;{\rm{m/}}{{\rm{s}}^2}\end{array}\)

Using the following kinematic equation of motion, the final speed of both the cars as they slide together is got.

\(\begin{array}{c}{{v'}^2} - {u^2} = 2ax\\{{v'}^2} - 0 = 2ax\\v' = \sqrt {2ax} \end{array}\)

Substituting the numerical values in the above expression.

\(\begin{array}{c}v' = \sqrt {2\left( {7.84\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {2.6\;{\rm{m}}} \right)} \\ = \sqrt {40.768\;{{\rm{m}}^2}{\rm{/}}{{\rm{s}}^2}} \\ \approx 6.38{\rm{m/s}}\end{array}\)

\(\begin{array}{c}{p_{{\rm{before}}}} = {p_{{\rm{after}}}}\\{m_{\rm{A}}}{v_{\rm{A}}} + {m_{\rm{B}}}{v_{\rm{B}}} = \left( {{m_{\rm{A}}} + {m_{\rm{B}}}} \right)v'\\{v_{\rm{A}}} = \frac{{{m_{\rm{A}}} + {m_{\rm{B}}}}}{{{m_{\rm{A}}}}}v'\end{array}\) … (i)

Substitute the value of\(v' = 6.38{\rm{m/s}}\)in equation (i) to determine the initial speed of car A (sports car).

\(\begin{array}{c}{v_{\rm{A}}} = \left( {\frac{{980\;{\rm{kg}} + 2300\;{\rm{kg}}}}{{980\;{\rm{kg}}}}} \right)\left( {6.38\;{\rm{m/s}}} \right)\\ = 21.35\;{\rm{m/s}}\end{array}\)