Question

Consider the following two-caraccident: Two cars of equal mass collide 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 of the skid marks to be 9 meters. The coefficient offriction between 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 of $v_e$ and $v_n$ .

Part B

What is the kinetic energy of the two-car system immediately after thecollision?

Express your answer in terms of $v_e$ , $v_n$ , and .

Part C

Write an expression for the work $W_fric$ done on the cars by friction.

Express your answer symbolically interms of the mass of a single car, the magnitude of the acceleration due to gravity,the coefficient of sliding friction , and the distance through 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 , the magnitude of the acceleration due to gravity, to be9.81 meters per second per second.

Answer 1

Answer 2

²PART A: Use vector components and cons ofmomentum

       Σp_e= Σp'_e                                    Σp_n= Σp'_n

mv_e  + 0 = m+mv'_x                     0 + mv_n = m+mv'_n     

      mv_e = 2mv'_e                        mv_n =2m v'_n     

         v_e=2v'_e                               v_n = 2v'_n

∴v'² = v'_e² +v'_n²

        = (1/2v_e)² + (1/2 v_n)²

          = 1/4 v_e² + 1/4v_n²

           =1/4 (v_e² + v_n²)

PART B:

K = 1/2 m' v' ²    But we need to becareful, because the mass of the 2-car system is 2m

    = m v^{' 2}

    = m / 4 (v_e² +v_n²)

PART C:

Friction is a non-conservative force ; the work done will benegative as it opposes motion and you can't recovery theenergy back.

W_fric = F_fricd     

         = -u m'g d      --> careful again, as m' is the mass of the two cars  

        = - 2 u mgd

PART D:    Now we can put all this togetherwith the work-energy theorum

W_fric = ΔKE      and   KE_f = 0 as the cars skid to a stop.

W_fric = - K = - 2 u mgd

∴              - m / 4 (v_e² + v_n²) =- 2 u mg d          Clean up the algebra, plug innumbers, get v_n

                                                                              Already the mass will cancel, provided we

                                                                              got the two expressions correct above.

                      1/4 (v_e² + v_n²) = 2 u g d

                      1/8 (v_e² + v_n²) = u gd

                            (v_e² + v_n²) = 8 ug d

                                       v_n² = 8 u g d -v_e²

                                             = 8*0.9*9.81 m/s² * 9m - (12m/s)²

                                             = 491.688

                                        v_n = 22 m/s

Looks like Driver N was lying