Accident Analysis.? A 1500-kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2200-kg SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.75, and the cars slide to a halt at a point 5.39 m west and 6.43 m south of the impact point. How fast was each car traveling just before the collision?

Solution 81P Step 1: Mass of sedan, m sedan 1500 kg Mass of SUV, m = 2200 kg suv Consider, the crash takes place at the origin, then we can represent it using a graph Step 2: The displacement of the car towards west, S west 5.39 m Force in the east-west direction is, F sedan - k sedan (Frictional force) Provided, -koefficient of kinetic friction = 0.75 Fsedan - 0.75 × 1500 kg ×9. m/s = -11025 N The displacement of the car towards south, S = 6.43 m south Force in the north-south direction is, F suv - ksuv rictional force) Provided, -koefficient of kinetic friction = 0.75 Fsedan - 0.75 × 2200 kg ×9. m/s = -16170 N Resultant force acting towards the south-west direction is, F = F sedan2 + F suv 2 = 1025 N + 16170 N 2 2 F = 3.830 × 10 N = 19571 N Step 3: Resultant displacement for the system of cars is, R = 539 m + 6.43 m = 70.397 m = .39 m 2 The work done against the frictional force F is, W = FR = -19571 N × 8.39 m = - 164201 J So, the change in KE of the system will be equal to this work done Therefore, ½ mv 2 - ½ mv 2 = W (Law of conservation of energy) final initial We know that, finally the system will come to rest. Therefore, v final 0 m/s Or, 0 - ½ mv 2 = - 164201 J initial m = m sedan m suv(Combined mass after collision) m = 1500 kg + 2200 kg = 3700 kg ½ mv 2initial164201 J v2 = (2 ×164201 J ) / 3700 kg = 88.76 m /s 2 2 initial Taking square root on both sides, v initial.42 m/s