You're an accident investigator at a scene where a drunk driver in a \(1600-\mathrm{kg}\) car has plowed into a \(1300-\mathrm{kg}\) parked car with its brake set. You measure skid marks showing that the combined wreckage moved \(25 \mathrm{m}\) before stopping, and you determine a frictional coefficient of \(0.77 .\) What do you report for the drunk driver's speed just before the collision?

After the collision, both cars stick together, so their total mass \(M_{total}= m_{drunk} + m_{parked} = 1600 kg + 1300 kg = 2900 kg\).

According to the principle of linear momentum, before collision, total initial momentum is equal to the final momentum. Since the parked car was not moving before collision, total initial momentum only includes momentum of the drunk driver's car, which is \(m_{drunk} \cdot v_{drunk} = 1600 kg \cdot v_{drunk}\). After the collision, both cars are moving together, so final momentum is \(M_{total} \cdot v_{total } = 2900 kg \cdot v_{total}\). Therefore, the momentum equation is \(1600 kg \cdot v_{drunk} = 2900 kg \cdot v_{total}\).

The work done by friction eventually stopped the cars, which equals to the lost kinetic energy. According to the work-energy principle, work done by friction is defined as the friction force multiplied by the distance traveled, which is \(m_{total} \cdot g \cdot \mu \cdot d\), where \(g\) is the acceleration due to gravity (approximately \(9.8 m/s^2\)), \(\mu\) is the friction coefficient and d is the distance. The lost kinetic energy is \(0.5 \cdot M_{total} \cdot v_{total}^2\). So the energy equation is \(m_{total} \cdot g \cdot \mu \cdot d = 0.5 \cdot M_{total} \cdot v_{total}^2\). Substituting the given numbers in, we find \(2900 kg \cdot 9.8 m/s^2 \cdot 0.77 \cdot 25m = 0.5 \cdot 2900 kg \cdot v_{total}^2\).

Solve the energy equation from step 3 to find \(v_{total}\). Substitute \(v_{total}\) into the momentum equation in step 2, hence find \(v_{drunk}\).