A 1200kg pick up truck traveling south at 15m/s collides witha 750kg car traveling east. The two vehicles stick together. Thefinal position of the wreckage after the collision is 25m, atan angle of 50 south of east, from the point of impact. If thecoefficient of friction between the tires and the road, from thelocation of the collision to the point at which the wreckage comesto rest is 0.4, what was the speed of the car just before thecollision?

A 1200kg pick up truck traveling south at 15m/s collides witha 750kg car traveling east. The two vehicles stick together. Thefinal position of the wreckage after the collision is 25m, atan angle of 50 south of east, from the point of impact. If thecoefficient of friction between the tires and the road, from thelocation of the collision to the point at which the wreckage comesto rest is 0.4, what was the speed of the car just before thecollision?

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asked 2021-03-26
A 1200kg pick up truck traveling south at 15m/s collides witha 750kg car traveling east. The two vehicles stick together. Thefinal position of the wreckage after the collision is 25m, atan angle of 50 south of east, from the point of impact. If thecoefficient of friction between the tires and the road, from thelocation of the collision to the point at which the wreckage comesto rest is 0.4, what was the speed of the car just before thecollision?

Answers (1)

2021-03-28
Final x position: 25m cos(50degrees) = 16.06969...m
Let \(\displaystyle{v}_{{{1}\xi}}\) be initial velocity, \(\displaystyle{v}_{{{x}{f}}}\) be velocity immediately after collision, and \(\displaystyle{v}_{{{x}{3}}}\) be resting velocity.
\(\displaystyle{{v}_{{{x}{3}}}^{{2}}}={\left({0}\frac{{m}}{{s}}\right)}^{{2}}={{v}_{{{x}{f}}}^{{2}}}+{2}\cdot{a}?{x}={{v}_{{{x}{f}}}^{{2}}}+{2}{\left({0.4}\cdot-{9.8}\frac{{m}}{{s}^{{2}}}\right)}{16.06969}\) m
\(\displaystyle{v}_{{{x}{f}}}={11.22}\) m/s
\(\displaystyle{m}_{{1}}{v}_{{{1}\xi}}+{m}_{{2}}{v}_{{{2}\xi}}={\left({m}{1}+{m}{2}\right)}{v}_{{{x}{f}}}\)
\(\displaystyle{\left({750}{k}{g}\right)}{v}_{{{1}\xi}}+{1200}{k}{g{{\left({0}\frac{{m}}{{s}}\right)}}}={\left({750}{k}{g}+{1200}{k}{g}\right)}{\left({11.2}\frac{{m}}{{s}}\right)}\)
\(\displaystyle{v}_{{{1}\xi}}={29.2}\frac{{m}}{{s}}\)
I may be wrong though
0

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