A 1200-kg pick up truck traveling south at 15.0 m/s collides with a750-kg car tr

Question

A 1200-kg pick up truck traveling south at 15.0 m/s collides with a750-kg car traveling east. The two vehicles stick together. Apatrolman investigating the accident determines that the finalposition of the wreckage after the collision is 25.0 m, at an angleof 50.0 degrees south of east, from the point of impact. He alsodetermines that the coefficient of friction between the tires andthe road at that location was 0.400. What is the speed of the carbefore the collision?

a) 19.6 m/s
b) 17.4 m/s
c) 23.4 m/s
d) 14.0 m/s

Explanation / Answer

Hmmm... the problem is overconstrained. That is, they havegiven too much information. That's ok... . The angle of the final momentum is found by .    tan = initial y momentum /initial x momentum . tan50 = 1200 * 15.0 / 750 *v       or        tan50 = 24 / v .    solve for v, youget      20.1 m/s       Hmmm... this is the problemwith too much information: there is more than one way to"solve" the problem and you usually get different answers. Theperson who wrote this problem made a mistake. <<sigh>>All right, we'll do it the other way... we'll ignore theangle. .       The final speed of thecollision can be found by using cons of energy: .           v = 2ugd = 2*0.4*9.8*25 =    14 m/s .           so thefinal momentum is    mv = (1200 + 750) *14 =   27300 .      initial momentumis      1200 * 15 = 18000 south     and .                                        750 * v    east      . so       273002   = 180002 +   (750v)2     solve forv, you get      27.4 m/s . Hmmm... yeah... there's something wrong with this problem. I'massuming your prof made this one up. You can tell him that it's"overconstrained". He'll know what that means, and hopefullyhe'll understand his mistake. . Basically there are two different ways to use the given infoto solve this problem. I did it both ways, got two differentanswers... and neither of them match the given choices.

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