Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and that all wheels on both cars lock at the time of impact.

Question

A 2110 kg car traveling at 11.1 m/s collides with a 2780 kg car that is initially at rest at the stoplight. The cars stick together and move 2.80 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and that all wheels on both cars lock at the time of impact.

Answer 1

Suppose v is the speed of the cars after the impact

Then by conservation of momentum

2110 x 11.1 = (2780 + 2110) v so v = 4.79 m/s

v² = u² + 2as

v = 0, u = 4.79, s = 2.80 giving a = - 4.10 m/s/s

F=ma so F = 4890 x 4.10 = 20049 N (force due to kinetic friction)

This force = normal force x coefficient = 4890 x 9.8 x coefficient = 20049

giving coefficient = 0.4184