Mark Scheme (Results) January 2010
GCE
Mechanics M1 (6677)
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January 2010 Publications Code UA023034 All the material in this publication is copyright Š Edexcel Ltd 2010
January 2010 6677 Mechanics M1 Mark Scheme Question Number Q1.
Scheme
I = 2 × 12 − 2 × 3 = 18 ( N s )
(a) (b)
LM
Alternative to (b)
Q2.
Marks M1 A1
2 × 12 − 8m = 2 × 3 + 4m Solving to m = 1.5
M1 A1 DM1 A1
I = m ( 4 − ( −8 ) ) = 18
M1 A1
Solving to m = 1.5
DM1 A1
(2)
(4) [6]
(4)
(a) First two line segments B1 Third line segment B1 8, 75 B1
s
(3)
8
O (b)
75 t
1 × 8 × (T + 75 ) = 500 2 Solving to T = 50
M1 A2 (1,0) DM1 A1
(5) [8]
GCE Mechanics M1 (6677) January 2010
Question Number
Scheme
Marks
Q3.
60°
30°
A
20 N
B TN
C mg (a)
R( → )
(b)
R( ↑ )
20 cos 30° = T cos 60° T = 20 3 , 34.6, 34.64,…
M1 A2 (1,0) A1 (4)
mg = 20 sin 30° + T sin 60° 40 m= (≈ 4.1) ,4.08 g
M1 A2 (1,0) A1
(4) [8]
Q4.
(a)
X
Y 1.8 m
A 1.5 m M ( A)
W
1.5 m
W × 1.5 + 20 × 3 = Y × 1.8
5 100 ¿ Y= W+ 6 3 (b)
↑
(c)
X + Y = W + 20 1 40 X= W− 6 3
5 100 40 ⎞ ⎛1 W+ = 8⎜ W − ⎟ 6 3 3 ⎠ ⎝6 W = 280
Alternative to (b) M(C) X × 1.8 + 20 × 1.2 = W × 0.3 1 40 X = W− 6 3
GCE Mechanics M1 (6677) January 2010
20
M1 A2 (1, 0)
cso or equivalent
A1
(4)
M1 A1 A1
(3)
M1 A1 ft A1
M1 A1 A1
(3) [10]
Question Number Q5.
Scheme
Marks
1 1 s = ut + at 2 ⇒ 2.7 = a × 9 2 2 −2 a = 0.6 ( m s )
(a)
M1 A1 A1
(3)
(b) R = 0.8 g cos 30° ( ≈ 6.79 )
µR
R
Use of F = µ R 0.8 g sin 30° − µ R = 0.8 × a ( 0.8 g sin 30° − µ 0.8 g cos 30° = 0.8 × 0.6 )
0.8g 30°
µ ≈ 0.51
(c)
accept 0.507
B1 B1 M1 A1 A1
(5)
R X
µR
0.8g
30°
R cos 30° = µ R cos 60° + 0.8 g ( R ≈ 12.8 ) → X = R sin 30° + µ R sin 60° Solving for X, X ≈ 12 accept 12.0 ↑
M1 A2 (1,0) M1 A1 DM1 A1
(7) [15]
Alternative to (c) R = X sin 30° + 0.8 × 9.8 sin 60° µR + 0.8 g cos 60° = X cos 30°
µ 0.8 g sin 60° + 0.8 g cos 60° cos 30° − µ sin 30° X ≈ 12 accept 12.0
M1 A2 (1,0) M1 A1
X =
Solving for X,
GCE Mechanics M1 (6677) January 2010
DM1 A1
(7)
Question Number Q6.
Scheme
(a) N2L A:
(b) N2L B:
Marks
1 5mg − T = 5m × g 4 15 T = mg ¿ 4
M1 A1
cso
1 T − kmg = km × g 4 k =3
A1
(3)
M1 A1 A1
(3)
(c) The tensions in the two parts of the string are the same
B1
(1)
1 1 (d) Distance of A above ground s1 = × g × 1.22 = 0.18 g ( ≈ 1.764 ) 2 4 1 Speed on reaching ground v = g ×1.2 = 0.3g ( ≈ 2.94 ) 4
M1 A1
For B under gravity
( 0.3g )
2
= 2 gs2
⇒ s2
( 0.3) = 2
S = 2 s1 + s 2 = 3.969 ≈ 4.0 (m)
GCE Mechanics M1 (6677) January 2010
2
g ( ≈ 0.441)
M1 A1
M1 A1
A1
(7) [14]
Question Number Q7.
Scheme
(a) v=
(b)
(c)
Marks
21i + 10 j − ( 9i − 6 j) = 3i + 4 j 4 speed is √ ( 32 + 42 ) = 5 ( km h −1 )
M1 A1 M1 A1
3 ( ⇒ θ ≈ 36.9° ) 4 bearing is 37, 36.9, 36.87, …
tan θ =
M1 A1
s = 9i − 6 j + t ( 3i + 4 j)
cso
(d) Position vector of S relative to L is ( 3T + 9 ) i + ( 4T − 6 ) j − (18i + 6 j) = ( 3T − 9 ) i + ( 4T − 12 ) j 2
2
= 100
25T 2 − 150T + 125 = 0 (T 2 − 6T + 5 = 0 ) T = 1, 5
GCE Mechanics M1 (6677) January 2010
(2)
M1
= (3t + 9)i + (4t − 6) j ¿
( 3T − 9 ) + ( 4T − 12 )
(4)
A1
(2)
M1 A1 M1
or equivalent
DM1 A1 A1
(6) [14]
GCE Mechanics M1 (6677) January 2010
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