My friend and I both have different answers, and we would liketo know who or if

Question

My friend and I both have different answers, and we would liketo know who or if we both are even correct. Thank you!!
A truck mass of 9M is moving at a speed v=15km/h when it collideshead on with a parkd car of mass M. Sprint mounted bumpers ensurethat the collision is essentially elastic. Find the finalvelocities of each vehicle. Repeat for the case of the car movingat a speedy v=15km/h and colliding with the stationary truck. Whatfraction of the incident vehicle's kinetic energy is transferred ineach case. My friend and I both have different answers, and we would liketo know who or if we both are even correct. Thank you!!
A truck mass of 9M is moving at a speed v=15km/h when it collideshead on with a parkd car of mass M. Sprint mounted bumpers ensurethat the collision is essentially elastic. Find the finalvelocities of each vehicle. Repeat for the case of the car movingat a speedy v=15km/h and colliding with the stationary truck. Whatfraction of the incident vehicle's kinetic energy is transferred ineach case.

Explanation / Answer

In this case you can approach the question in two ways.In an elastic collision, the coefficient of restitution e =1.    e = velocity of separation / velocity ofapproach.    Suppose you take the direction of incidentvehicle as positive.    Velocity of approach = +u.    Final velocities be +v and +v' for incidentand parked vehicles respectively.    So, Velocity of separation = (v' - v)    So, v' - v = u    Also, by conservation of linear momentum,    9Mu = 9Mv + Mv'    => 9u = 9v + v' => v + u = 9u - 9v => 10v = 8u => v = 0.8u => v' = 1.8 u Now, in the second case of collision, Mu = Mv + 9Mv' v' = v+u => u = v + 9v' => v' - u = u - 9v' => v' = 2u/10 = 0.2u => v = -0.8u So, you get the final velocities in each case. Now, going for the fractions of kinetic energytransferred, f1 = (0.5*M*v'*v')/(0.5*9*M*u*u) = v'*v'/(9*u*u) =(1.8*1.8)/(9*1*1) = 0.36 f2 = (0.5*9M*v'*v')/(0.5*M*u*u) = (9*0.2*0.2)/(1*1) =0.36 So, it is same in both cases. The second method of approach is by considering the fact thattotal kinetic energy is conserved in an elastic collision. 0.5*9M*u*u = 0.5*9M*v*v + 0.5*M*v'*v' => 9*u*u =9*v*v + v'*v' and 0.5*M*u*u = 0.5*M*v*v + 0.5*9M*v'*v' => u*u = v*v+ 9*v'*v' for the two collisions. Again considering conservation of linear momentum, 9M*u = 9M*v + M*v' => 9u = 9v + v' M*u = M*v + 9M*v' => u = v + 9v' Solving these two sets of equations we get the sameanswers.    u+v = v'    u+v = v' So only one of you can be correct.

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