Question

A 920 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 3.3 m before stopping. the police officer, knowing that the coefficient of kinetic friction between tires and road is 0.36, calculates the speed of the sports car at impact. what was that speed?

Answer 1

16.9 m/s, or 60.8 km/h  
The combined mass of both cars is 920 kg + 2300 kg = 3220 kg.
 The normal force of those cars is 3220 kg * 9.8 m/s^2 = 31556 kg*m/s^2.
 The frictional force is 31556 kg*m/s^2 * 0.36 = 11360.16 kg*m/s^2
 The work done to stop the cars is 11360.16 kg*m/s^2 * 3.3 m = 37488.528
kg*m^2/s^2
 Since the work done to stop the cars has to match the energy the cars had, we can determine their combined velocity using the equation for kinetic energy which is:
 E = 0.5 M V^2
 Solve for V, then substitute the known values and calculate:
 E = 0.5 M V^2
 2E = M V^2
 2E/M = V^2
 sqrt(2E/M) = V
 sqrt(2*37488.528 kg*m^2/s^2 / 3220 kg) = V
 sqrt(74977.056 kg*m^2/s^2 / 3220 kg) = V
 sqrt(23.2848 m^2/s^2) = V
 4.825432623 m/s = V
 
 Now we can calculate the momentum of both cars, so
 4.825432623 m/s * 3220 kg = 15537.89305 kg*m/s
  Finally the momentum of both cars has to be the same as the momentum of
the sports car just prior to the collision. So
 15537.89305 kg*m/s / 920 kg = 16.88901418 m/s

 
 So the sports car was moving at 16.9 m/s at the time of the collision. Let's convert that to km/h. 
 16.9 m/s * 3600 s/h / 1000 m/km = 60.8 kph

Answer 2

Let the velocity of the sports car before impact be v. 
 The velocity of the locked system of car and SUV is 
 [920*v/(2300+920)] 
 the police officer applies the following equation to get v
 [920*v/(2300+920)] ^2 = 2*[{0.36*(2300+920)*9.8}/(2300+920)}]*3.3
 .... or
 0.0816*v^2 =  23.2848 or v = sq rt[ 23.2848/0.0816] =  16.89 m/s