posted by Shaila on .
A car with mass of 1875kg is traveling along a country road when the driver sees a deer dart out onto the road. The driver slams on the brakes manages to stop before hitting the deer. The driver of a second car with 2135kg is driving too close and does not see the deer. When the driver realizes that the car ahead is stopping, he hits the brakes but is unable to stop. The cars lock together and skid another 4.58m. All of the motion is along a straight line. If the coefficient of friction between the dry concrete and rubber tires is 0.750, what was the speed of the second car when it hit the stopped car?
First solve for the speed of the cars locked together, V', using the fact that the kinetic energy lost equals work done against friction during the skid.
(1/2)(M+m)V'^2 = (M+m)(Uk)*g*X
V'^2 = 2*0.750*9.8*4.58 = 63.3 m/s^2
Then apply conservation of momentum to determine the speed of the faster car that caused the collision, V.
M V = (M+m)*V'
Solve for V