GCE Edexcel GCE Mechanics M2 (6678)

Document Sample
GCE Edexcel GCE Mechanics M2 (6678) Powered By Docstoc
					                                GCE
                                Edexcel GCE
                                Mechanics M2 (6678)




                        Summer 2005

                        Mark Scheme (Results)
              Mechanics M2 (6678)
Edexcel GCE
                                                     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


                         ⇒ 20 x = 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 α .t = 2.5                     M1 A1
                        )α                                              ↑ u sin α .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
                                                             u =
                                                              2
                                                                        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 = −2u3
                                                                                                              A1
          NLI:               2u − v = e.4u                                                                 M1 A1
                             ⇒ 4eu = 8 u ⇒ e = 2 .
                                       3       3
                                                                                                   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 = 30 g × 1.5                M1 A2
                                              30g
                                                        P = 90 g.   2
                                                                     3
                                                                         ≈ 1020 N (1000N)          A1
                                                                                                           (4)
                   (b)                              →            X = P cos 60 = P  1
                                                                                   2
                                                                                                           M1 A1
                                                                    ( ≈ 509 N (510N) )

                                                    ↑            Y + P cos 30 = 30 g                       M1 A1
                                                                 (⇒ Y = –588 N)
                             resultant =       ( X + Y ) = (509 2 + 588 2 ) ≈ 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
                                       × 3 × 5 2 = 37.5 J                                     M1 A1
                  Work-energy:     F × 8 = 117.6 − 37.5 = 80.1                               M1 A1√
                                   ⇒ F = 10.0125 ≈ 10 N                 `                     A1
                                                                                                         (5)
            (c)            R = 3g cos 30 (= 25.46 N)                                          B1
                                             10
                           F = µR ⇒ µ =           ≈ 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