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 3.24 m/s. Friction is negligible between the cars and the ground. (a) If the masses of bumper cars 1 and 2 are 590 kg and 613 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.) v2 = m/s (b) What is the kinetic energy lost in the collision?

Explanation / Answer

assuming east -west direction along X-axis and North-south direction along the Y-axis

V1i = initial velocity of bumper car 1 before collision = 5.62 i

V2i = initial velocity of bumper car 2 before collision = - 10 Cos60 i - 10 Sin60 j = - 5 i - 8.7 j

V1f = final velocity of bumper car 1 after collision = - 3.24 i

V2f = final velocity of bumper car 2 after collision = ?

m1 = mass of bumper car 1 = 590 kg

m2 = mass of bumper car 2 = 613 kg

using conservation of momentum

m1 v1i + m2 v2i = m1 v1f + m2 v2f

590 (5.62 i) + 613 ( - 5 i - 8.7 j) = 590 (- 3.24 i) + 613 V2f

613 V2f = 5227.4 i - 3065 i - 5333.1 j

V2f = 3.53 i - 8.7 j

b)

v1i = 5.62 m/s

v2i = sqrt((-5.2)2 + (-8.7)2) = 10.14 m/s

v1f = 3.24 m/s

v2f = sqrt((3.53)2 + (-8.7)2) = 9.4 m/s

initial total kinetic energy is given as

KEi = (0.5) (m v1i2 + m2 v2i2) = (0.5) (590 (5.62)2 + 613 (10.14)2)= 40831.6 J

final total kinetic energy is given as

KEf = (0.5) (m v1f2 + m2 v2f2) = (0.5) (590 (3.24)2 + 613 (9.4)2)= 30179.13 J

Loss = KEi - KEf = 40831.6 - 30179.13 = 10652.5 J

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