An automobile has a mass of 2150 kg and avelocity of +13 m/s. It makes a rear-en

Question

An automobile has a mass of 2150 kg and avelocity of +13 m/s. It makes a rear-endcollision with a stationary car whose mass is 1850 kg. The cars lock bumpers and skid off togetherwith the wheels locked. (a) What is the velocity of the two cars justafter the collision?
1(No Response) m/s

(b) Find the impulse (magnitude and direction) that acts on theskidding cars from just after the collision until they come to ahalt. (Indicate direction by the sign of the impulse.)
2(No Response) N

(c) If the coefficient of kinetic friction between the wheels ofthe cars and the pavement is k =0.72, determine how far the carsskid before coming to rest.
3(No Response) m
(a) What is the velocity of the two cars justafter the collision?
1(No Response) m/s

(b) Find the impulse (magnitude and direction) that acts on theskidding cars from just after the collision until they come to ahalt. (Indicate direction by the sign of the impulse.)
2(No Response) N

(c) If the coefficient of kinetic friction between the wheels ofthe cars and the pavement is k =0.72, determine how far the carsskid before coming to rest.
3(No Response) m

Explanation / Answer

mass of automobile m = 2150 kg initial velocity of automobile u = 13 m / s mass of car M = 1850 kg initial speed of car U = 0 (a). from law of conservation of momentum , mu + MU = ( m + M) v                                                           27950 + 0 = 4000 v                the velocity of the two cars just after the collision v =27950 / 4000                                                                             = 6.9875 m / s (b).  coefficient of kinetic friction between thewheels of the cars and the pavement isk = 0.72 final speed V = 0 acclerationa = - k * g = -7.056 m / s ^ 2 from the relation V ^ 2 - v ^ 2 = 2aS required distance S = [ V ^ 2 - v ^ 2 ] / 2a                              = 3.45 m

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