Question

A 2000 kg car slides on a patch of ice (no friction before the collision) into the back of a 1500 kg car. After the two cars stick together, they slide 10 meters into an intersection before coming to a stop. If the coefficient of friction between the cars’ tires and the road is 0.8, how fast was the first car traveling right before the collision? (vo1=22.1 m/s)

Answer 1

Here,

let the speed after collision is u1

as coefficient of friction, u = 0.80

for the acceleration, a = -u * g

a = -0.80 * 9.8 = -7.84 m/s^2

Using third equation of motion

v^2 - u1^2 = 2 * a * d

0 - u1^2 = 2 * (-7.84) * 10

u1 = 12.52 m/s

Now, for the initial velocity of first car is vo1

Using conservation of momentum

2000 * v01 + 1500 * 0 = (1500 + 2000) * 12.52

v01 = 22.1 m/s

the first car traveling right before the collision with 22.1 m/s