An automobile has a mass of 1860 kg and a velocity of 20.1 m/s. It makes a rear-

Question

An automobile has a mass of 1860 kg and a velocity of 20.1 m/s. It makes a rear-end collision with a stationary car whose mass is 2180 kg. The cars lock bumpers and skid off together with the wheels locked. (a) What is the velocity of the two cars just after the collision? (b) Find the impulse that acts on the skidding cars from just after the collision until they come to a halt. (c) If the coefficient of kinetic friction between the wheels of the cars and the pavement is ?k = 0.695, determine how far the cars skid before coming to rest.

Explanation / Answer

(a)

Momentum before collision = 1860 x 20.1 = 37386kg/m/s.
After collision total mass is 1860 + 2180 = 4040kg.
Momentum before = momentum after

37386 = 4040 x v where v is the common velocity of the two vehicles.
v = 37386/4040 = 9.25 m/s

(b) Impulse = change in momentum, the final momentum of the cars = 0 since they come to rest.
Magnitude of impulse = 37386 - 0 = 37386kg/m/s.

(c) The kinetic energy lost is 1/2{mv^2} = work done on cars by frictional force.
Frictional force = 0.695 x mg = 0.695 x 4040 x 9.81 ( Remember frictional force = coefficient of friction x normal component of force)
Using work done = force x distance
1/2{4040 x 9.25^2} = 0.695 x4040 x 9.81 x d where d is distance travelled by cars.

d = 9.25^2/(2 x 0.695 x 9.81) = 6.27m .

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