An automobile has a mass of 2050 kg and a velocity of +18 m/s. It makes a rear-e

Question

An automobile has a mass of 2050 kg and a velocity of +18 m/s. It makes a rear-end collision with a stationary car whose mass is 1850 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?
? m/s

(b) Find the impulse (magnitude and direction) that acts on the skidding cars from just after the collision until they come to a halt. (Indicate direction by the sign of the impulse.)
? N

(c) If the coefficient of kinetic friction between the wheels of the cars and the pavement is µk = 0.71, determine how far the cars skid before coming to rest.
? m

Explanation / Answer

Given Mass of an automobile , m1 = 2050 kg Initial velocity of m1 is , v1 = 18 m/s Mass of the stationary car , m2 = 1850 kg a) By law of conservation of momentum     m1 v1 + 0 = (m1 + m2 ) V     V = (2050 kg) (18 m/s) / (2050 kg + 1850kg )      V = 9.46 m/s    is the velocity of the two cars after collision b) Impulse J = - (m1+m2) V                    = - (2050 kg + 1850kg ) * 9.46 m/s                     = 3.68 *10^4 kg m/s c) Acceleration of the two cars is           a = - g = - 0.71 *9.8 m/s^2 = -6.95 m/s^2 The distance travelled by the cars is         Vf^2 - V^2 = 2 as          0 - (9.46 m/s)^2 = 2(-6.95 m/s^2) *s                      s = 6.43 m

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