Note the picture...
11. Two cars approach an intersection as shown. Car 1 weighs 4500
lbs and has a speed of 55.0 mi/hr. Car 2 weighs 3750 pounds with
a speed of 60.0 mi/hr. They collide in a completely inelastic collision at the intersection.
a) Calculate the direction the wreckage moves after the collision. Express this as an angle measured counterclockwise from the positive x-axis.
b) If the coefficient of kinetic friction is 0.55 for the tires on this road, and the wheels of the car are locked (not rolling), calculate the distance the wreckage slides from the collision point.
The Attempt at a Solution
mc1 = 140slug vca = 55i
mc2 = 116.63 s vc2 = -30i + 51.96j
m1vc1 + m2vc2 = (m1+m2)v'
140(55i) + 116.63(-30i + 51.96j) = (140 + 116.63)v'
v' = 16.37i + 23.61j...theta = 55.26 degreees
|v'| = 28.73 mph
.5mv2 = ugmd....5v2 = ugd
.5(28.73) = (.55)(9.8)d...d = 2.67
im pretty sure myt last equation is wrong considering it says they slid 2 miles.. ugmd must not be correct, can i get some help