Mark Scheme (Results) January 2009
GCE
GCE Mathematics (6678/01)
Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH
January 2009 6678 Mechanics M2 Mark Scheme Question Number
Scheme T
R
1
F = ma parallel to the slope,
M1*
T − 1500g sinθ − 650 = 1500a
A1
Tractive force, 30000 = T x 15
M1*
650
a=
1500g
Marks
30000 15
− 1500(9.8)( 141 ) − 650 1500
0.2 (m s-2)
d*M1 A1 (5) [5]
2
(a)
R (↑) : R = 25 g + 75 g (= 100 g )
11 × 100 g 25 = 44g (=431)
B
S
F = µR ⇒ F =
C
B1 M1 A1 (3)
(b)
M(A): 25 g × 2 cos β + 75 g × 2.8 cos β
R 75g
2 5g
β A
F
= S × 4 sin β
M1 A2,1,0
R ( ↔) : F = S 176g sin β = 260g cos β
M1A1
β = 56(°)
A1 (6)
(c) So that Reece’s weight acts directly at the point C.
6678/01 GCE Mathematics January 2009
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B1 [10]
Question Number 3
Scheme
R (b) : R = 10 g
R
(a)
Marks
µR
B1
4 10g = 56 7 4 ∴WD against friction = 10g 50 7 F = µR ⇒ F =
70
( )
( )( )
10g
2800(J)
B1 M1 A1 (4)
(b)
70(50) − " 2800 " =
1 2
M1*
(10)v 2 − 21 (10)(2)2
700 = 5v − 20 , 5v = 720 ⇒ v = 144
A1ft d*M1
Hence, v = 12 (m s-1)
A1 cao
2
2
2
(4) Or
(b)
4 R = 10a 7 4 70 − × 10 g = 10a , (a = 1.4) 7
N2L (→ ): 70 −
( )
M1* A1ft
AB → : v 2 = (2)2 + 2(1.4)(50)
d*M1
-1
Hence, v = 12 (m s )
A1 cao (4) [8]
4
(a)
∫
v = 10t − 2t 2 , s = vdt
= 5t 2 −
M1 A1
2t 3 ( +C ) 3
t = 6 ⇒ s = 180 − 144 = 36 (m)
A1 (3)
(b)
s=
∫
v dt =
− 432t −1
−1
(+ K ) =
t = 6, s = “36” ⇒ 36 = ⇒ K = − 36 At t = 10, s =
432 +K t
( )
B1
432 +K 6
M1* A1 d*M1
432 − 36 = 7.2 (m) 10
A1 (5) [8]
6678/01 GCE Mathematics January 2009
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Question Number 5
Scheme
Marks
(a) MR xi (→) from AD yi (↓) from BD
108
18π
108 + 18π
4
6
x
6
− π8
y
B1
M1
AD (→): 108(4) + 18π (6) = (108 +18π ) x x=
B1
432 + 108π = 4.68731... = 4.69 (cm) (3 sf) AG 108 + 18π
A1 (4)
(b)
yi (↓)
-
6
from BD
8
π
B1 oe
M1
BD (↓): 108(6) + 18π (− π8 ) = (108 +18π ) y y =
y
A1ft
504 = 3.06292... = 3.06 (cm) (3 sf) 108 + 18π
A1 (4)
(c) vertical
12 − x
D
M1
B
y G
tan θ = θ = required angle
y 12 − 4.68731.. 3.06392.. = 12 − 4.68731..
θ = 22.72641... = 23 (nearest degree)
dM1 A1
A1 (4) [12]
6678/01 GCE Mathematics January 2009
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Question Number 6
(a)
Scheme
Marks M1 A1
Horizontal distance: 57.6 = p x 3 p = 19.2
(2)
1 2 at for vertical displacement. 2 1 − 0.9 = q × 3 − g × 32 2 9g − 0.9 = 3q − = 3q − 44.1 2 43.2 *AG* q= = 14.4 3
(b) Use s = ut +
(c) initial speed
p 2 + 14.4 2
M1 A1
A1 cso (3)
(with their p) -1
M1 A1 cao
= 576 = 24 (m s )
(2) (d)
tan α =
14.4 3 (= ) 4 p
(with their p)
B1 (1)
(e) When the ball is 4 m above ground:
1 2 at used 2 1 3.1 = 14.4t − gt 2 o.e ( 4.9t 2 − 14.4t + 3.1 = 0 ) 2 3.1 = ut +
⇒t=
14.4 ± (14.4)2 − 4(4.9)(3.1) 2(4.9)
14.4 ± 146.6 = 0.023389… or 2.70488… 9.8 duration = 2.70488... − 0.23389... t=
M1 A1
seen or implied
M1
awrt 0.23 and 2.7
A1
= 2.47 or 2.5 (seconds) or 6 (e)
M1 A1 (6)
M1A1M1 as above t=
14.4 ± 146.6 9.8
A1
146.6 o.e. 9.8 = 2.47 or 2.5 (seconds)
Duration 2 ×
M1 A1 (6)
(f) Eg. : Variable ‘g’, Air resistance, Speed of wind, Swing of ball,
The ball is not a particle.
B1 (1) [15]
6678/01 GCE Mathematics January 2009
5
Question Number 7
(a)
Scheme
2u
Before P
After
Marks
u
3m
2m
x
y
Correct use of NEL
M1*
y − x = e(2u + u ) o.e.
A1
Q
CLM (→) : 3m(2u ) + 2m( −u ) = 3m( x ) + 2m( y ) (⇒ 4u = 3 x + 2y ) Hence x = y – 3eu, 4u = 3(y-3eu) + 2y, ( u (9e + 4) = 5 y ) Hence, speed of Q = 51 (9e + 4) u AG
B1 d*M1 A1 cso (5)
(b)
1 (9e + 4)u − 3eu 5 1 2u Hence, speed P = (4 − 6e)u = (2 − 3e) o.e. 5 5 1 2u x= u= (2 − 3e) ⇒ 5u = 8u − 12eu, ⇒ 12e = 3 2 5 gives, e = 123 ⇒ e = 41 AG
x = y − 3eu =
M1# A1
& solve for e
d#M1 A1 (4)
Or
(b)
M1 A1
Using NEL correctly with given speeds of P and Q 3eu =
1 5
(9e + 4)u − u
3eu =
9 5
eu + 54 u − 21 u ,
1 2
6 5
e=
3 10
⇒ e=
3e − 95 e = 15 60
4 5
−
1 2
#
d#M1
& solve for e
⇒ e = 41 .
A1 (4)
(c) Time taken by Q from A to the wall =
Distance moved by P in this time =
d ⎧ 4d ⎫ =⎨ ⎬ y ⎩ 5u ⎭
M1†
u d u ⎛ 4d ⎞ 2 × (= ⎜ ⎟ = d ) 2 ⎝ 5u ⎠ 5 2 y
⎛d⎞
Distance of P from wall = d − x ⎜ ⎟ ; = d − 52 d = 35 d ⎝y⎠
A1 d†M1; A1 cso
AG
(4) or
(c)
1 1 9 1 5 u : ( + 4)u = u : u = 2:5 2 5 4 2 4 So if Q moves a distance d, P will move a distance 52 d Distance of P from wall = d − 52 d ; = 35 d AG
M1†
Ratio speed P:speed Q = x:y =
A1
cso
d†M1; A1 (4)
6678/01 GCE Mathematics January 2009
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Question Number
Scheme
(d) After collision with wall, speed Q = 51 y =
Time for P, TAB =
3d 5 1 2
Hence TAB = TWB ⇒
(d)
( )= 5u 4
1 4
3d 5 1 2
their y
B1ft
from their y
B1ft
u
−x x , Time for Q, TWB = 1 u u 4
−x x = 1 u u 4
gives, 2 (3d5 − x ) = 4x ⇒ or
1 5
Marks
3d 5
M1
− x = 2x , 3x =
After collision with wall, speed Q = 51 y =
1 5
A1 cao
⇒ x = 51 d
3d 5
( )= 5u 4
(4) 1 4
u
their y
speed P = x = u , speed P: new speed Q = u : u = 2 :1 from their y 1 2
Distance of B from wall = nd
2
or (d)
1 2
1 4
1 3d d × ;= 3 5 5
their
After collision with wall, speed Q = 51 y =
1 5
( )= 5u 4
1 2 +1
B1ft B1ft M1; A1 (4)
1 4
u
their y
B1ft
Combined speed of P and Q = u + u = u 1 2
Time from wall to 2nd collision =
1 4
3d 5 3u 4
=
3 4
3d 5
×
4 3u
=
4d 5u
Distance of B from wall = (their speed)x(their time) =
6678/01 GCE Mathematics January 2009
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from their y u 4d ; = 51 d × 4 5u
B1ft M1; A1 (4) [17]