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 = (b) What is the kinetic energy lost in the collision

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Explanation / Answer

We will apply conservation of momentum here

lets take east along +ve X-axis and north along +ve Y-axis

Please remember that momentum and velocity are both vector quantities.

initial velocity of car 1 = 5.62i

initial velocity of car 2 = -10cos60i - 10sin60j = -5i - 5*(31/2)j

initial momentum = m1v1 + m2v2 = 590*(5.62i) + 613( - 5i - 5*(31/2)j )

This initial momentum will be equal to final momentum due to conservation of momentum law

final momentum = m1v1 + m2v2 = 590*(-3.24i) + 613*v2 = 590*(5.62i) + 613( - 5i - 5*(31/2)j )

velocity of bumper car 2 after collision = v2 = 3.527i - 5*(31/2)j = 3.527i - 8.660j

kinetic energy lost = initial energy - final energy = (1/2)*590*(5.62^2) + (1/2)*613*100 - (1/2)*590*(3.24^2) - (1/2)*613*(v2)

using pythogoras theorem v2 = 3.5272 + 8.6602 = 87.435

energy lost = 9317.398 + 30650 - 3096.792 - 26798.827 = 10071.779J

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