1. The driver of an SUV (m = 1700 kg) isn’t paying attention and runs into the back end of a car (m = 950 kg) on level ground at a red light.

Question

On impact, both drivers lock their brakes. The SUV and car stick together and travel a distance of 8.2 m before they come to a stop. How fast was the SUV traveling just before the collision? The coefficient of friction between the tires and the road is 0.72. What is the KE before the collision and after the collision? Why might these not be the same?

Answer 1

Mass 1 = SUV

Mass 2 = Car

Apply conservation of momentum

V1 = Velocity of SUV before impact

V2 = Velocity of car before impact

V' = Velocity of both objects after impact (they travel together, coupled)

M1V1 + M2V2 = (M1+M2)V'

V1 = ((M1+M2)/M1)*V'

Now, apply conservation of energy after impact to the stop of both vehicles

Sum of work = Change in KE

Work done is by the friction force

Friction force

fk = uk * N = uk * (M1+M2)*g

Work of Friction = -fk * L = -uk * (M1+M2)*g*L

L = 8.2m

Now, put it all together

V' = After impact velocity

V'' = Velocity when they come to a stop

Sum of Work = Change in KE

-uk * (M1+M2)*g*L = (1/2)(M1+M2)(V''^2-V'^2)

Let = V'' = 0

-uk * (M1+M2)*g*L = (1/2)(M1+M2)(-V'^2)

V'^2 = [2*uk * (M1+M2)*g*L]/(M1+M2)

V'^2 = [2*uk * g*L] = 115.8

V' = 10.76 m/s

Now with this, plug into our conservation of momentum equation from the beginning

V1 = ((M1+M2)/M1)*V'

V1 = (1700+950)/1700 * 10.76 = 16.77 m/s

a) V1 = 16.77 m/s

b)

KE before collision = 1/2 * M1 * V1^2 = 239,047 J

KE after collision = 1/2 * (M1+M2) * V'^2 = 1/2* (1700+950)*10.76^2 = 153,405 J

c) There is a decrease of velocity, therefore the KE will decrease!