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Mechanics edxcel 2007 mark scheme

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                Mark Scheme (Results)
                January 2007

                 GCE




             GCE Mathematics

             Mechanics M1 (6677)




Edexcel Limited. Registered in England and Wales No. 4496750
Registered Office: One90 High Holborn, London WC1V 7BH
                                      January 2007
                                    6677 Mechanics M1
                                      Mark Scheme

Question
                                            Scheme                                            Marks
Number

   1.      (a)           P sin 30° = 24                                                     M1 A1
                                       P = 48                                               A1    3

           (b)                      Q = P cos 30°                                           M1 A1
                                        ≈ 41.6                   accept 24 √ 3 , awrt 42    A1    3     6


   2.      (a)               Μ(C) 80 × x = 120 × 0.5                                        M1 A1
                                        x = 0.75                                   cso      A1    3

           (b)              Using reaction at C = 0                                         B1
                        Μ(D) 120 × 0.25 = W ×1.25                              ft their x   M1 A1
                                      W = 24 (N)                                            A1    4

           (c)    ¡             X = 24 + 120 = 144 (N)                        ft their W M1 A1ft
                                                                                               2
           (d) The weight of the rock acts precisely at B.                               B1    1        10




   3.      (a)               a=
                                  (15i − 4 j) − ( 3i + 2 j) = 3i − 1.5 j                    M1 A1 2
                                              4


           (b)         N2L        F = m a = 6i − 3 j                          ft their a    M1 A1

                          F = √ ( 6 2 + 32 ) ≈ 6.71        (N)   accept √ 45 , awrt 6.7     M1 A1 4




           (c)           v 6 = ( 3i + 2 j) + ( 3i − 1.5 j) 6                   ft their a   M1 A1ft

                             = 21i − 7 j   (m s ) −1                                        A1      1   9




                                                       1
Question
                                    Scheme                                          Marks
Number

   4.      (a)      CLM       0.3u = 0.3 × ( −2 ) + 0.6 × 5                       M1 A1
                                 u =8                                             M1 A1 4

           (b)                I = 0.6 × 5 = 3     ( N s)                          M1 A1 2

           (c)   v = u + at ⇒ 5 = a ×1.5         ( a = 10 )
                                                        3                         M1 A1
                                         10
                    N2L      R = 0.6 ×      =2                                    M1 A1 4     10
                                          3


   5.      (a)   v 2 = u 2 + 2as ⇒ 02 = 212 − 2 × 9.8 × h                         M1 A1
                                        h = 22.5       ( m)                       A1      3

           (b)   v 2 = u 2 + 2as ⇒ v 2 = 02 + 2 × 9.8 × 24        or equivalent M1 A1
                                         ( = 470.4 )
                                        v ≈ 22     (m s )−1
                                                                   accept 21.7    A1      3

                                                                                  M1 A2 (1, 0)
           (c)    v = u + at ⇒ − √ 470.4 = 21 − 9.8t              or equivalent

                                                                 − 1 each error

                                              t ≈ 4.4      (s)     accept 4.36    A1      4   10




                                             2
Question
                                       Scheme                                     Marks
Number

   6.      (a)                          R
                   µR                                  P
                                              20°


                                              30g


                              Use of F = µ R                                    B1
                             P cos 20° = µ R                                    M1 A1
                   ¡       R + P sin 20° = 30 g                                 M1 A1
                              P cos 20° = µ ( 30 g − P sin 20° )                M1
                                                0.4 × 30 g
                                      P=                                        M1
                                           cos 20° + 0.4sin 20°
                                         ≈ 110 ( N )                  accept 109 A1     8

           (b)     ¡       R + 150 sin 20° = 30 g                               M1 A1
                                       ( R ≈ 242.7 )
             N2L        150 cos 20° − µ R = 30a
                                                                                M1 A1

                                              150 cos 20° − 0.4 × 242.7         M1
                                         a≈
                                                         30
                                                                                A1      6   14
                                            = 1.5 (ms-2)       accept 1.46




                                               3
Question
                                           Scheme                                                 Marks
Number

   7.      (a)   N2L     Q                2 g − T = 2a                                       M1 A1
                 N2L     P       T − 3 g sin 30° = 3a                                        M1 A1 4

           (b)                  2 g − 3 g sin 30° = 5a                                       M1
                                                a = 0.98 (ms-2)                     cso      A1      2

           (c)                           T = 2( g − a)                      or equivalent    M1
                                           ≈ 18    ( N)                     accept 17.6      A1      2

           (d)   The (magnitudes of the) accelerations of P and Q are equal                  B1      1

           (e)         v 2 = u 2 + 2as ⇒ v 2 = 2 × 0.98 × 0.8          ( = 1.568)            M1
                                             v ≈ 1.3      (m s )
                                                              −1
                                                                              accept 1.25 A1         2

           (f)    N2L for P          −3 g sin 30° = 3a
                                                         1
                                                  a = (−) g                                  M1 A1
                                                         2
                                 1                                 1                         M1 A1
                         s = ut + at 2    ⇒ 0 = √ 1.568t − 4.9t 2            or equivalent
                                 2                        2

                                                  t = 0.51   (s)            accept 0.511     A1      5 16




           A maximum of one mark can be lost for giving too great accuracy.




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