A 1200-kg pick-up truck traveling south at 15.0 m/s collides with a 750-kg car traveling east. The two vehicles stick together. A patrolman investigating the accident determines that the final position of the wreckage after the collision is 25.0 m, at an angle of 50.0° south of east, from the point of impact. He also determines that the coefficient of friction between the tires and the road at that location was 0.400. What was the speed of the car before the collision?

Question

Quizplus · Accepted Answer

We can use conservation of momentum in two dimensions to solve the problem. The total momentum in the x-direction before the collision is:

Px = (1200 kg)(0 m/s) + (750 kg)(v) = 750v

The total momentum in the y-direction before the collision is:

Py = (1200 kg)(-15.0 m/s) + (750 kg)(0 m/s) = -18000 kg*m/s

After the collision, the two vehicles stick together and move at an angle of 50.0° south of east. We can resolve this momentum vector into its x- and y-components:

Pf,x = (1200 kg + 750 kg)(vf,x) = 1950 vf,x

Pf,y = 0

We can use the Pythagorean theorem to find the magnitude of the final momentum vector:

|Pf| = sqrt[(Pf,x)^2 + (Pf,y)^2] = 1950 vf,x

We also know that the final position of the wreckage is 25.0 m at an angle of 50.0° south of east, which means that the displacement in the x-direction is:

xf = 25.0 m cos(50.0°) = 16.0 m

We can use the coefficient of friction and the displacement in the x-direction to find the work done by friction:

Wf = f * d * cos(180°) = -f * d = -0.400 * 16.0 m = -6.40 J

Since there are no external forces acting on the system in the x-direction, the work done by friction must be equal to the change in kinetic energy:

Wf = Kf,x - Ki,x

-6.40 J = (1/2)(1950 kg)(vf,x)^2 - (1/2)(750 kg)(v)^2

We can solve this equation for vf,x:

vf,x = sqrt[((0.400)(16.0 m)(750 kg)(v)^2 - 6.40 J)/(1950 kg)]

vf,x = sqrt[(120000 kg*m^2/s^2)(v^2) - 20.8 J)/(1950 kg)]

Now we can use the fact that the magnitude of the final momentum vector is equal to the magnitude of the initial momentum vector, since there are no external forces acting in the y-direction:

|Pf| = |Pi| = sqrt[(750v)^2 + (-18000 kg*m/s)^2] = 18525 kg*m/s

|Pf| = sqrt[(1950 kg)^2 (vf,x)^2] = (1950 kg)(vf,x)

Combining both equations gives:

18525 kg*m/s = (1950 kg)(vf,x) = 1950 sqrt[(120000 kg*m^2/s^2)(v^2) - 20.8 J)/(1950 kg)]

18525 = sqrt[(120000v^2 - 20.8)]

342219375 = 120000v^2 - 20.8

v^2 = 2851.864

v = 53.4 m/s

However, we should check if this is reasonable based on the initial speeds of the two vehicles. The pick-up truck was traveling south at 15.0 m/s, which means that the car must have been traveling at least 43.4 m/s east to give the resultant velocity of 23.4 m/s southeast. Therefore, the correct answer is:

v = 43.4 m/s = 15.6 m/s

The answer in the given format is:

Answer: D

A 1200-Kg Pick-Up Truck Traveling South at 15