- #1

Graador

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- Homework Statement
- If two cars of equal mass hit each other head on, car A travelling at 25 m/s towards the right and car B travelling at 20 m/s to the left, what will be the speed of each car post-collision if 75% of the energy is lost in bending the metal of the cars.

- Relevant Equations
- m1v1 + m2v2 = m1V1 + m2V2

e = Kf/Ki

Since the mass of both vehicles is the same, it's possible to calculate Ki which happens to be 512,5J and from there, multiply it by 0,25 since 75% of the energy is gone and I end up with 128,125J.

Now my problem is that for the velocity, I have: 25,0 + -20,0 = V1 + V2 which is 5,00 = V1 + V2

When I take Kf, I have: 128,125 = V1(sq)/2 + V2(sq)/2 so 256,25 = V1(sq) + V2(sq)

If I try and get the square root, I end up with 16,0 = V1 + V2 which doesn't fit with the first equation which said 5,00 = V1 + V2

After some trial and error, I found that the velocity of the first car is -8,5 m/s and the second 13,5 m/s which when added up, equal 5,00 and when their squares are added up and divided by two, equal 127,25 which is close enough. I just don't understand how to find both velocities from the actual equations themselves because all I find is a quadratic formula that gives me wrong answers.

Now my problem is that for the velocity, I have: 25,0 + -20,0 = V1 + V2 which is 5,00 = V1 + V2

When I take Kf, I have: 128,125 = V1(sq)/2 + V2(sq)/2 so 256,25 = V1(sq) + V2(sq)

If I try and get the square root, I end up with 16,0 = V1 + V2 which doesn't fit with the first equation which said 5,00 = V1 + V2

After some trial and error, I found that the velocity of the first car is -8,5 m/s and the second 13,5 m/s which when added up, equal 5,00 and when their squares are added up and divided by two, equal 127,25 which is close enough. I just don't understand how to find both velocities from the actual equations themselves because all I find is a quadratic formula that gives me wrong answers.

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