Two bumper cars at the county fair are sliding toward one another (see figure be

Question

Two bumper cars at the county fair are sliding toward one another (see figure below). Initially, bumper car 1 is traveling to the east at 5.62 m/s, and bumper car 2 is traveling 60.0° south of west at 10.00 m/s. After they collide, bumper car 1 is observed to be traveling to the west with a speed of 3.17 m/s. Friction is negligible between the cars and the ground.

(a) If the masses of bumper cars 1 and 2 are 586 kg and 619 kg respectively, what is the velocity of bumper car 2 immediately after the collision? (Express your answer in vector form. Enter your answer to at least three significant figures.) in m/s

(b) What is the kinetic energy lost in the collision? in J

Explanation / Answer

m1 = 586 kg, m2 =619 kg , u1x = 5.62 m/s, u1y =0

u2x = -10cos (60)i = -5 i

u2y = -10 sin(60)j = -8.66 j

v1x = -3.17 m/s , v1y =0

From conservation of momentum along x -direction

m1u1x+m2u2x = m1v1x+m2v2x

(586* 5.62) - (619*5) = -(586*3.17) +(619v2x)

3293.32 - 3095 = - 1857.62 +619 v2x

v2x = 2055.94/619 = 3.32 m/s

From conservation of momentum along Y -direction

m1u1y+m2u2y = m1v1y+m2v2y

(586* 0) - (619*8.66) = (586*0) +(619v2x)

v2y= -8.66 m/s

v2 = [(3.32)2+(-8.66)2]1/2

Velocity of bumper car 2 immediately after collision v2 = 9.28 m/s

(b) Ki = (1/2)m1u1^2+(1/2)m2u2^2

Ki =[(1/2)(586)(5.62*5.62)]+[(1/2)(619)(10*10)]

Ki =40204 J

Kf = (1/2)m1v1^2+(1/2)m2v2^2

Kf =[(1/2)(586)(3.17*3.17)]+[(1/2)(619)(9.28*9.28)]

Kf =29598 J

Kinetic eneergy lost =Ki -Kf= 40204 - 29598

= 10606 J

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