Consider the following two-car accident: Two cars of equal mass m collide at an intersection. Driver E was traveling eastward, and driver N, northward. After the collision, the two cars remain joined together and slide, with locked wheels, before coming to rest. Police on the scene measure the length d of the skid marks to be 9 meters. The coefficient of friction mu between the locked wheels and the road is equal to 0.9.

Each driver claims that his speed was less than 14 meters per second (50 mph). A third driver, who was traveling closely behind driver E prior to the collision, supports driver E's claim by asserting that driver E's speed could not have been greater than 12 meters per second. Take the following steps to decide whether driver N's statement is consistent with the third driver's contention.

Part A

Let the speeds of drivers E and N prior to the collision be denoted by v_e and v_n, respectively. Find v^2, the square of the speed of the two-car system the instant after the collision.
Express your answer terms ofv_e and v_n.

Part B

What is the kinetic energy K of the two-car system immediately after the collision?
Express your answer in terms of v_e, v_n, and m.

Part C

Write an expression for the work W_fric done on the cars by friction.
Express your answer symbolically in terms of the mass m of a single car, the magnitude of the acceleration due to gravity g, the coefficient of sliding friction mu, and the distance d through which the two-car system slides before coming to rest.

Part D

Using the information given in the problem introduction and assuming that the third driver is telling the truth, determine whether driver N has reported his speed correctly. Specifically, if driver E had been traveling with a speed of exactly 12 meters per second before the collision, what must driver N's speed have been before the collision?
Express your answer numerically, inmeters per second, to the nearest integer. Take g, the magnitude of the acceleration due to gravity, to be9.81 meters per second per second.