Posted by **Emily** on Tuesday, July 10, 2012 at 3:07pm.

Car 1 has a mass of m1 = 65 103 kg and moves at a velocity of v01 = +0.76 m/s. Car 2, with a mass of m2 = 92 103 kg and a velocity of v02 = +1.5 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.

(b) Determine the velocity of their center of mass after the collision

(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

- Physics -
**Emily**, Tuesday, July 10, 2012 at 3:22pm
I figured out a and b, but I can't seem to figure out C.

- Physics -
**drwls**, Tuesday, July 10, 2012 at 3:41pm
The velocity of the center of mass should not change, because no outside forces act (other than weight, which remains balanced by the upward force of the road.) They tell you to neglect friction. Friction (such as braking during impact) could alter the CM velocity.

The numbers used for the masses of the cars and their velocity are preposterous. They are as massive as houses and not much faster than turtles.

- Physics -
**Emily**, Tuesday, July 10, 2012 at 3:48pm
For part c, it says regarding the common velocity Vf isn't that different than Vcm?

- Physics -
**bobpursley**, Tuesday, July 10, 2012 at 3:55pm
if the cars are hooked, the common velocity has to be the velocity of the center of mass.

- Physics -
**drwls**, Tuesday, July 10, 2012 at 3:56pm
If they have a common velocity, it must be the CM velocity.

The formula for the final velocity, assuming conservation of momentum, is the same as the formula for the initial CM velocity. If V1 and V2 are initial velocities,

Vcm = [V1*M1 + V2*M2]/(M1 + M2)

(M1+M2)Vf = M1*V1 + M2*V2

Vf = [V1*M1 + V2*M2]/(M1 + M2)

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