Physics problem help ---- Actual speed given weight and friction.?

Question

Your accident-reconstruction team has been hired by the local police to analyze the following accident. A careless driver rear-ended a car that was halted at a stop sign. Just before impact, the driver slammed on his brakes, locking the wheels. The driver of the struck car had his foot solidly on the brake pedal, locking his brakes. The mass of the struck car was 970 kg, and that of the initially moving vehicle was 1130 kg. On collision, the bumpers of the two cars meshed. Police determine from the skid marks that after the collision the two cars moved 0.68 m together. Tests revealed that the coefficient of kinetic friction between the tires and pavement was 0.92. The driver of the moving car claims that he was traveling at less than 15 km/h as he approached the intersection.

What was the moving car's actual velocity?

______ km/h

Answer 1

The strategy is to work backwards thru this problem

Step 1: calculate the work done 'W' on the cars after the collision by the friction force 'F':

The friction force is the weight of the cars times the coefficient of friction:

F = 0.92 * (970 + 1130) * 9.81 = 18950 N

The work done is 'F' times the distance 'x' that the force was applied

W = F x = 18950 N  * 0.68 m = 12890 J

Step 2: since the cars come to a stop, the work done is equal to the kinetic energy 'KE' they had just after the collision:

KE = 12890 J

Step 3: calculate the velocity 'v2' of the cars just after the collision:

KE = 1/2 m v2^2

v2 = sqrt(12890 * 2 / (1130+970)) = 3.5 m/s

Step 4: since momentum is conserved in a collision, calculate the velocity of the moving car bfore the collision:

m1 v1 = m2 v2

v1 = (1130+970)/1130 * 3.5 = 6.5 m/s

= 23.4 km/hr