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Physics, 28.02.2020 20:56 nay191

Consider the following two-caraccident: Two cars of equal mass mcollide at an intersection. Driver E was traveling eastward, anddriver N, northward. After the collision, the two cars remainjoined together and slide, with locked wheels, before coming torest. Police on the scene measure the length d of the skid marks to be 9 meters. The coefficient offriction mubetween the locked wheels and the road is equal to 0.9.
Each driver claims that hisspeed 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 couldnot have been greater than 12 meters per second. Take the followingsteps to decide whether driver N's statement is consistent with thethird driver's contention.
Part A Let the speeds of drivers E and N prior tothe collision be denoted by v_e and v_n, respectively. Find v^2, the square of the speed of the two-car system theinstant 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 thecollision?
Express your answer in terms ofv_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 interms of the mass mof a single car, the magnitude of the acceleration due to gravityg, the coefficient of sliding friction mu, and the distance dthrough which the two-car system slides before coming torest.
Part D Using the information given in the problemintroduction and assuming that the third driver is telling thetruth, determine whether driver N has reported his speed correctly. Specifically, if driver E had been traveling with a speed ofexactly 12 meters per second before the collision, what must driverN'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.

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