[luis00]

1) In a chain reaction car accident on the North-South tollway, a BMW with a mass of 2000 kg collides elastically with a Honda Civic (mass 1500 kg) at rest. The Honda in turn collides with a Hummer of mass 3000 kg in front of it, that's also at rest, and the Honda and the Hummer lock bumpers and slide forward with a velocity of 3 m/s. (The BMW that originated the chain of events is not involved in the second collision). What was the velocity of the Honda right after the first collision and before it hits the Hummer?

My attempt: V=MA/MA+MB

(2000kg)(0m/s)+(1500)(3m/s)+(3000kg)(3m/s) / 1500kg+3000kg
Is this correct?

2)With regard to the BMW-Honda-Hummer collision problem, what was the velocity of the BMW right before the first collision?

[ebaines]

1) Not quite. The momentum of the Honda prior to its collision with the Hummer is equal to the momentinm of the combined Honda + Hummer after the collision:

M_Honda x V1 = (M_Honda + M_Hummer) x V2

V1 = (1500 Kg + 3000 Kg) x 3m/s /(1500 Kg)

2. To solve for the BMW's initial velocity, now that you know the Honda's velocity after the first collision you can set up two equations - one for conservation of momentum and one for conservation of energy. This will let you solve for both the BMW's initial velocity and the BMW's final velocity.