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In: Physics

Two automobiles with equal mass approach an intersection. One vehicle is traveling eastward and the other...

Two automobiles with equal mass approach an intersection. One vehicle is traveling eastward and the other vehicle is traveling northward. The vehicles collide and stick together. After impact they travel on the road and then travel on a gravel covered shoulder until both cars stop. Only the northward bound vehicle left skid marks at an angle of 55 degrees north of east. A police officer posted at the intersection clocked the eastward vehicle at 29 mph before impact. The northward bound car skid on the road for 25 ft and continued to skid on the gravel covered shoulder. What is the total distance, in feet, the two cars traveled after the collision? Assume us=0.96 and uk=0.85 for the road and us=0.74 and uk=0.65 for the shoulder.

Solutions

Expert Solution

Momentum before impact = 29m  x   lb-mph ( east bound)

= 25m y lb-mph ( north bound)

Total momentum before impact = (29m x + 25m y ) kg-m/s

momentum is conserved before and after impact

after impact the vehicles stick together , total mass = 2m

The combined mass makes 55 deg NoE

2mv Cos(55) = 29m

speed of the combined mass after impact v = 25.28 mph = 37.08 ft/s

KE after impact = 1/2* 2m*v2 = mv2 = 1375 m lb-ft2/s2

Force of friction on the road = k*2mg = 1.7 mg

distance moved = 25 ft

work done = 25 *1.7 mg = 42.5 mg

Friction force on the gravel = 0.65 *2mg = 1.3 mg

d-distance moved on the gravel

work done on gravel = d* 1.3 mg

total work done = (42.5+1.3d)mg

= 1375m ( KE of the combined mass after impact )

The combined mass will move until all of its KE is used up against friction

d= 0.36 ft

Total distance moved by the combined cars = 25 + 0.36 = 25.36 ft

genius_generous answered 2 years ago
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