A 1890 kg car traveling at 13.3 m/s collides with a 2970 kg car that is initally at rest at a stoplight. The cars stick together and move 1.96 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 all wheels on both cars lock at the time of impact.
By Conservation of momentum
m1V1+m2V2=(m1+m2)V
1890*13.3+2970*0=(1890+2970)V
V=5.1722 m/s
From Kinematic equation
V2=Vo2+2ad
Since car is decelerating and coming to stop means V=0
0=5.17222+2*(-a)*1.96
a=26.75/3.92=6.824 m/s2
from newtons second law
ukmg=ma
=>uK=a/g =6.824/9.8 =0.696
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