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A pair of bumper cars in an amusement park ride collide elastically as one approaches the other directly from the rear. One has a mass of 450kg and the other 550kg, owing to differences in passenger mass. if the lighter one approaches at 4.5m/s and the other is moving at 3.7m/s, calculate a)their velocities after the collision and b) the change in momentum of each

ok this is what i know

m1 = 450 kg, v1 = 4.5 m/s

m2 = 550 kg, v2 = 3.7 m/s

i know i'm suppose to use these formulas:

m1v1+m2v2 = m1v1' + m2v2'

and

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

my teacher gave us this hw without really explaining anything and the example my book gives to me is very confusing

so i know m1v1+m2v2 = total momentum in this system, which is:

450kg(4.5m/s)+550kg(3.7m/s)=4060 Ns

and the total KE after the collision is:

.5(450kg)(4.5m/s)^2+.5(550kg)(3.7m/s)^2 = 8321J

then my book becomes confusing and i get lost

what do i do?

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# Homework Help: Collision - need big help

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