Car 1 has a mass of m1 = 65 multiplied by 10^3 kg and moves at a velocity of v01

Question

Car 1 has a mass of m1 = 65 multiplied by 10^3 kg and moves at a velocity of v01 = +0.82 m/s. Car 2, with a mass of m2 = 92 multiplied by 10^3 kg and a velocity of v02 = +1.1 m/s, overtakes car 1 and couples to it. Neglect the effects of friction in your answer.
(a) Determine the velocity of their center of mass before the collision
? m/s

(b) Determine the velocity of their center of mass after the collision
? m/s
(c) Should your answer in part (b) be less than, greater than, or equal to the common velocity vf of the two coupled cars after the collision?


Justify your answer.

Explanation / Answer

The mass of car1, m1 = 65000 kg Velocity of car1, v1 = 0.82 m/s The mass of car 2, m2 = 92000 kg The velocity of car 2, v2 = 1.1 m/s a) The velocity of centre of mass before collision, Mv = (m1v1 + m2v2) / (m1 + m2) = 0.98 m/s b) Since there are no external forces are acts on the cars the intial momentum of the cars before collision is equal to the final momentum of the cars after collision. So the velocity of centre of mass of cars after collision = 0.98 m/s c) From the answers of part a and part b it is confirm that the two results are equal.

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