Mark Scheme (Results) January 2010
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 ? Edexcel Ltd 2010
Question Number
Q1. (a) (b) LM
Alternative to (b)
January 2010 6677 Mechanics M1
Mark Scheme
Scheme
I = 2?12 - 2? 3 = 18 ( N s)
2?12 - 8m = 2? 3 + 4m Solving to m = 1.5
I = m(4 - (-8)) = 18
Solving to m = 1.5
Marks
M1 A1 (2)
M1 A1 DM1 A1 (4)
[6]
M1 A1 DM1 A1 (4)
Q2. (a) s
8
First two line segments B1
Third line segment B1
8, 75 B1
(3)
O (b)
1 ?8? (T + 75) = 500
2 Solving to T = 50
75 t
M1 A2 (1,0) DM1 A1 (5)
[8]
GCE Mechanics M1 (6677) January 2010
Question Number
Q3.
(a)
A R( )
Scheme 30?
20 N
60 ?
B
T N
C mg
20 cos 30? = T cos 60? T = 20 3 , 34.6, 34.64,...
Marks
M1 A2 (1,0)
A1
(4)
(b)
R( ) mg = 20sin 30? + T sin 60?
m = 40 ( 4.1) ,4.08 g
Q4. (a)
X
Y
1.8 m
A
1.5 m W
1.5 m
20
M ( A)
W ?1.5 + 20? 3 = Y ?1.8 Y = 5 W + 100 ? 63
(b)
X + Y = W + 20 X = 1 W - 40
63
(c)
5W 6
+
100 3
=
8
1W 6
-
40 3
W = 280
Alternative to (b) M(C) X ?1.8 + 20 ?1.2 = W ? 0.3
X = 1 W - 40 63
M1 A2 (1,0)
A1
(4)
[8]
M1 A2 (1, 0)
cso
A1
(4)
or equivalent M1 A1
A1
(3)
M1 A1 ft
A1
(3)
[10]
M1 A1 A1
GCE Mechanics M1 (6677) January 2010
Question Number
Scheme
Marks
Q5. (a) (b)
s = ut + 1 at2 2.7 = 1 a ? 9
2
2
( ) a = 0.6 m s-2
M1 A1
A1
(3)
R
? R
R = 0.8g cos 30? ( 6.79)
B1
Use of F = ?R
B1
0.8g
0.8g sin 30? - ?R = 0.8? a
M1 A1
30?
(0.8g sin 30? - ?0.8g cos 30? = 0.8? 0.6)
? 0.51
accept 0.507 A1
(5)
(c)
R
X
? R
0.8g
30?
R cos 30? = ?R cos 60? + 0.8g
( R 12.8)
X = R sin 30? + ?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 ? 9.8sin 60? ?R + 0.8g cos 60? = X cos 30?
Solving for X,
X
=
?0.8g sin 60? + 0.8g cos 60? cos30? - ? sin 30?
X 12
accept 12.0
M1 A2 (1,0) M1 A1
DM1 A1 (7)
GCE Mechanics M1 (6677) January 2010
Question Number
Q6. (a) N2L A:
(b) N2L B:
Scheme
5mg - T = 5m ? 1 g 4
T = 15 mg ? 4
T - kmg = km ? 1 g 4
k =3
Marks
M1 A1
cso
A1
(3)
M1 A1
A1
(3)
(c) The tensions in the two parts of the string are the same
(d) Distance of A above ground Speed on reaching ground
s1
=
1 2
?
1 4
g
?1.22
=
0.18g
( 1.764)
v = 1 g ?1.2 = 0.3g ( 2.94)
4
For B under gravity
(0.3g )2 = 2gs2
s2
=
(0.3)2
2
g
(
0.441)
S = 2s1 + s2 = 3.969 4.0 (m)
B1
(1)
M1 A1 M1 A1
M1 A1
A1
(7)
[14]
GCE Mechanics M1 (6677) January 2010
Question Number
Scheme
Marks
Q7. (a)
v = 21i +10j - (9i - 6j) = 3i + 4j
4
( ) ( ) speed is 32 + 42 = 5 km h-1
M1 A1 M1 A1 (4)
(b)
tan = 3 ( 36.9?)
4
bearing is 37, 36.9, 36.87, ...
M1
A1
(2)
(c)
s = 9i - 6j + t (3i + 4j)
M1
= (3t + 9)i + (4t - 6)j ?
cso A1
(2)
(d) Position vector of S relative to L is
(3T + 9) i + (4T - 6) j - (18i + 6j) = (3T - 9) i + (4T -12) j
M1 A1
(3T - 9)2 + (4T -12)2 = 100
M1
25T 2 -150T +125 = 0
(T 2 - 6T + 5 = 0)
or equivalent DM1 A1
T = 1, 5
A1
(6)
[14]
GCE Mechanics M1 (6677) January 2010
GCE Mechanics M1 (6677) January 2010
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