A 1055 kg van, stopped at a traffic light, is hit directly in the rear by a 655 kg car traveling with a veloci

Question

A 1055 kg van, stopped at a traffic light, is hit directly in the rear by a 655 kg car traveling with a velocity of +2.25 m/s. Assume that the transmission of the van is in neutral, the brakes are not being applied, and the collision is elastic. What is the final velocity of each vehicle?

a) car
m/s
(b) van
m/s

Answer 1

conservation of momentum (mv): mv_final=mv_initial:
655kg*2.25m/s =1055kg*v_van_f +655kg*v_car_f (sign of velocity to be determined)
conservation of energy: Ek_car_initial =Ek_van_final +Ek_car_final
655kg*(2.25m/s)^2 =1055kg*v_van_f^2 +655kg*v_car_f^2. sub v_van or v_car from momentum eq-on into energy eq-on:

655kg*(2.25m/s)^2 =655kg*v_car_f^2 + 1055kg * (655kg*2.25m/s/1055kg -655kg/1055kg*v_car_f)^2, expand and solve quadratic equation to get
Vcar=2.25m/s or -0.526m/s. the first one is nonrealistic, since it implies no collision ,ie velocity didn't change.
so Vcar is -0.526m/s or opposite to where it was travelling before
Vvan is found from momentum eq-on: v_van = 655kg* 2.25m/s/1055kg - 655kg/1055kg*v_car_f=1.723m/s
I checked this by plugging into the momentum formula.
so
a) -0.526m/s (ie in opposite direction)
b) 1.723 m/s

[To Michael: the point of Yahoo Answers is that I answer the question and explain as good as I can/have time for. It's not up to me or you to judge educational value of my answer or to assume that this question is for homework.]

Answer 2

Way to do his/her homework for them.

That's a good way to teach laziness.