Mark scheme - Mark scheme - 6678 Mechanics M2 June 2005 by wvm21293

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									                                                     June 2005
                                                 6678 Mechanics M2
                                                   Mark Scheme

Question                                                  Scheme                                         Marks
Number

                                             P
           1 (a)         Driving force =                                                 B1
                                             v
                              21000
                                     600  v  35 m s–1                                 M1 A1
                                v
                                                                                                   (3)
                             P                  1
                   (b)          600  1200.g.                                           M1 A1
                             v                 14
                              ( = 1440 N)
                         21000              21000
                                1440  v         14.6 or 15 m s–1             M1 A1
                           v                1440
                                                                                                   (4)




           2 (a)   B     4               x       C

                                                      5             (x = 3)


                   A          4          D

                         M(AB): 7  3.5  5  5.5  4  2  20  x                     M1 A2,1,0


                          20x  24.5  27.5  8  60  x  3 cm                 dep      M1 A1
                                                                                                   (5)

           (b)               X     3.5           kM   M(XY):

                                                      M  (3.5  3)  kM  3.5           M1 A1 
                         M                             k  1.
                                                            7
                                                                                         A1
                                                                                                   (3)
                             Y




6678 Mechanics
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
          3 (a)              v = (18 – 12t2)i + 2ctj                                                        M1 A1 A1

                             t=   3
                                  2
                                      : v = –9i + 3cj                                                          M1

                               v = 15  92 + (3c)2 = 152                                             M1

                                                   (3c)2 = 144  c = 4                                        A1
                                                                                                                       (6)

                  (b)                                a = – 24ti + 8j                                   M1

                             t3:
                               2
                                                     a = –36i + 8j                                     M1
                                                                                                               A1 
                                                                                                                       (3)



          4 (a)
                                       12.6                             12.6t = x                             B1

            0.1                                                         0.1 = 4.9 t2                          B1

                                                                                  x2
                                              x               0 .1  4 . 9                                   M1
                                                                                12 .6 2

                                                              x = 1.8 m                                       A1
                                                                                                                       (4)

          (b)                 u                                         u cos.t  2.5                        M1 A1
                        )                                               u sin .t      1
                                                                                          2
                                                                                              gt   2
                                                                                                               M1 A1
                                      2.5
                                                               24
                                                             u.   t  2.5
                                                               25
                                                               7         2.5.25
                                                             u.  4.9.
                                                               25          24u
                                                                  4.9  2.5  25 2
                                                             u2 
                                                                       7  24
                                                              u  6.75 or 6.8 m s–1                           M1 A1
                                                                                                                       (6)




6678 Mechanics
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
          5 (a)                 2u               2u

                                 3m                  2m


                                       v             2u


          CLM:               6mu  4mu  3mv  4mu                                                     M1 A1
                             v      2u
                                       3
                                                                                                          A1
          NLI:               2u  v  e.4u                                                             M1 A1
                              4eu  8 u  e  3 .
                                       3
                                               2                                               M1 A1
                                                                                                                (7)
            (b)                                  2u              0

                                                     2m              5m

                                                      x              y
                             5my  2mx  4mu                                                           M1 A1
                             y  x  5 .2u  6 u
                                     3
                                             5
                                                                                                          A1
          Solve:             x 7u
                                 2                                                                     M1 A1
                             2
                             7
                               u    2 u so B does not overtake A
                                     3
                                                                                               M1
                                       So no more collisions                                           A1 cso
                                                                                                                (7)

          6 (a)      Y
                     X          0.5

                            P                    M(A):
                                                     P  0.5 sin 60  30g  1.5                M1 A2
                                           30g
                                                     P  90g.   2
                                                                 3
                                                                      1020 N (1000N)          A1
                                                                                                       (4)
                   (b)                                       X  P cos60  P  1
                                                                               2
                                                                                                       M1 A1
                                                                (  509 N (510N) )

                                                             Y  P cos30  30g                        M1 A1
                                                              ( Y = –588 N)
                            resultant =     ( X  Y )  (5092  5882 )  778 N
                                                 2        2
                                                                                                       M1 A1
                                                                                     or 780N                    (6)

                   (c) In equilibrium all forces act through a point                        M1
                       P and weight meet at mid-point;
                      hence reaction also acts through mid-point so reaction horizontal A1 cso
                                                                                                                (2)
                   OR M(mid-point): Y x 1.5 = 0  Y = 0                                         M1

                         Hence reaction is horizontal                                           A1




6678 Mechanics
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
          7 (a) PE lost = 3  g  8 sin 30  3  g  8  0.5 = 117.6 J  118J                 M1 A1
                                                                             or 120J          (2)

            (b)      KE gained =   1
                                   2
                                             37.5 J
                                      3  52                                                 M1 A1
                  Work-energy:     F  8  117.6  37.5  80.1                               M1 A1
                                    F  10.0125  10 N               `                       A1
                                                                                                         (5)
            (c)            R  3g cos30 (= 25.46 N)                                           B1
                                            10
                           F  R              0.393 or 0.39                              M1 A1
                                          25.46
                                                                                                         (3)
            (d)          Work done by friction = 80.1 as before                        M1
                  Work-energy: 1  3  v 2  1  3  2 2  117.6  80.1
                               2              2
                                                                                            M1 A2,1,0
                                             v  5.39 or 5.4 m s–1                    A1
                                                                                                         (5)




6678 Mechanics
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
6678 Mechanics
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics

								
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