Question

Car Accident at S. High and Market According to police records a totally inelastic collision between two cars occurred at 2:37 PM today at the intersection of S. High and Market St. The automobile driven by Jim was proceeding north on S. High while the automobile driven by Dan was moving west on Market (wrong direction). Following the impact the combined wreckage was observed to slide a distance of 3.55 meters in a direction 37.0° west of north. A representative of the wrecking company estimated that Dan and Jim’s automobiles had masses of 1600 kg and 1200 kg, respectively. In an exclusive interview following the accident Officer Teale of the Police Department noted that on a typical sunny afternoon the coefficient of sliding friction between automobile wreckage and the street surface is 0.600. Later a hospital spokesperson stated, “there were no serious injuries in the accident”. Determine the velocities of the two cars just prior to the collision.

Answer #1

work done by the frictional force = change in kinetic energy

mu_k*(m1+m2)*g = (0.5*(m1+m2)*v^2

0.6*9.8 = 0.5*v^2

v = 3.42 m/sec is the velocity of wreckage immediately after the
collision

Using law of conservation of momentum

(m1*u1)+(m2*u2) = (m1+m2)*v

along east -west

(1600*u1)+0 = (1600+1200)*v*sin(37) = (1600+1200)*3.42*sin(37) =
5763

1600*u1 = 5763

u1 = 5763/1600

u1 = 3.6 m/sec is the initial velocity of Dan

along North-south

(1200*u2) = (1600+1200)*3.42*cos(37)

u2 = 6.37 m/sec is the initial velocity of Jim

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