An automobile has a mass of 2050 kg and a velocity of +14 m/s. It makes a rear-e

Question

An automobile has a mass of 2050 kg and a velocity of +14 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?

(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.)

(c) If the coefficient of kinetic friction between the wheels of the cars and the pavement is ?k = 0.66, determine how far the cars skid before coming to rest.

Explanation / Answer

1) the ball hits the floor at 6.18m/s and rebounds at 3.43m/s. so impulse = 0.280(6.18+3.43) =2.69kgm/s down

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(2) 1150(36) = 2390v so v = 17.3m/s

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(3) mom. east = 138.8kgm/s and mom. south = 465.8kgm/s giving a resultant of 486kgm/s
at an angle of 73.4 degrees south of east and a velocity of 4.26m/s

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(4)(a) 2050(15) = (2050+1850)v so v = 7.89m/s

(b) impulse = mv = 30750kgm/s

(c) force of friction = 0.65(3900)(9.8) = 24843N = ma = 3900a and a = 6.37m/s^2
v^2 = 2aS so 7.89^2 = 2(6.37)S and s = 4.89m

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