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## Homework Statement

At a Route 3 highway on-ramp, a car of mass 1.50e3 kg is stopped at a stop sign, waiting for a break in traffic before merging with the cars on the highway. A pickup of mass 2.00e3 kg comes up from behind and hits the stopped car. Assuming the collision is elastic, how fast was the pickup

## Homework Equations

M1V1x + M2V2iX=M1Vfx + M2V2fx

V2ix-V1ix = -(V2fx-V1fx)

**3. The solution in the book**

"From conservation of momentum: M2V2i=M1Vf + M2V2f (1) because the intital velocity of mass 1 is 0 m/s

The collision is elastic, so the relative velocity after the collision is equal and opposite to the relative velocity before the collision: V2i = -(V2f-V1f) (2)

We watn to solve these two equations for V2i, so we can eliminate V2f. Multiplying eq. (2) through by M2 and rearanging yields: M2V2i = M2V1f - M2V2f (3)

Adding eqs. (1) and (3) gives: 2*M2V2i = (M1+M2)V1f (4)

Finally we solve eq. (4) for V2i: V2i = M1+M1/2M2 * V1f = 1500 kg + 2000 kg/4000 kg * 20.0 m/s = 17.5 m/s "

Could someone please explain to me how the textbook solved it this way? Or how it can be solved in a simpler way? Thank you very much!