# 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
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
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
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