Elastic Collision Problem, WOULD REALLY APPRECIATE HELP

AI Thread Summary
The discussion centers on understanding the solution to an elastic collision problem involving a stopped car and a pickup truck. The key equation used, V2ix-V1ix = -(V2fx-V1fx), highlights the relationship between the relative velocities before and after the collision. This relationship is based on the principle that, in elastic collisions, the relative velocities are equal in magnitude but opposite in direction. An alternative method to solve the problem involves using conservation of kinetic energy, although this approach complicates the calculations due to the introduction of squared velocities. Overall, the forum participants emphasize the effectiveness of the textbook's method in deriving the final velocities.
starburst
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Hi everyone. I would really appreciate if I could get some help with this problem elastic collision problem. I will provide the question and the solution as shown in my textbook. What I would like to know if specifically how they used the equation, V2ix-V1ix = -(V2fx-V1fx), and why they arranged it the way they did. Is there perhaps a different way to solve the problem? Thank you very much!

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!
 
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starburst said:
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!

It's really a very simple way to get the required two equations in two unknowns that are required to solve for the two final velocities. It relies on a (provable) characteristic of elastic collisions whereby the relative velocities of the two objects, before and after collision, are of equal magnitude but opposite sign.

Another way to solve the problem is to use conservation of kinetic energy as the second equation (KE is conserved in elastic collisions). But this introduces the squares of the velocities into the mix, which turn into square roots of expressions, and are a bit harder to work with.

If you want you can prove the relative velocity relationship by solving the general case elastic collision using the conservation of energy approach to yield expressions for the final velocities, and then use them to find the final relative velocity. You will then find that the relationship is true in general.
 
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