Mark Scheme (Final) Summer 2007
GCE
GCE Mathematics (6679/01)
Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH
June 2007 6679 Mechanics M3 Mark Scheme
Question Number
1.
Scheme
(a)
A
2 0
Marks
2 x x dx 2
M1 A1
...
2 x3 x 3 ...
A1
2
2 x3 8 4 A x 4 3 0 3 3 (b)
x 1
cso
(by symmetry)
2 4 1 1 y y 2 dx 2 x x 2 dx 3 2 2 1 4 x 2 4 x 3 x 4 dx 2 1 4 x3 x5 4 x 2 3 5
A1
(4)
B1 M1 A1 A1
2
4 1 4 x3 x5 8 4 y x 3 2 3 5 0 15 8 3 2 accept exact equivalents y 15 4 5
A1
(5) [9]
Question Number
2.
Scheme
(a) Base Mass ratios h 2
y
Cylinder 2 h2 h 2
0
Container 3 h2
(b) Liquid Mass ratios M h y 2
B1 B1
h 2
1 y h 3
Container M h 3
Ratio of 1 : 2 : 3
y
3 h 2 y 2 h 2
Leading to
Marks
M1 A1 cso A1
Total 2M
y
h h 2M y M M 2 3 5 y h 12
Ratio of 1 : 1 : 2
(5)
B1 B1
M1 A1 A1
(5) [10]
Question Number
3.
Scheme
Marks
(a) At surface k mg k mgR 2 2 R
(b)
mx
N2L
cso
(2)
mgR 2 x2
dv gR 2 d 1 2 gR 2 or v 2 v 2 dx x dx 2 x 1 1 2 1 2 or v gR 2 dx v dv gR x2 dx 2 x2 1 2 gR 2 v C 2 x gR x 2 R, v 0 C 2 2 2gR v2 gR x 2gR 2 v2 gR At x R , R v gR
M1 A1
M1 M1 A1 M1 A1
M1 A1
(7) [9]
Question Number
Scheme
Marks
A
4.
l T
r
P mg
T cos mg
M1 A1
T sin
mv 2 r
M1 A1
tan
r 2 l r
tan
v2 rg
Eliminating T
M1
r v2 2 2 l r rg
Eliminating
M1
cso
A1
2
gr 2 v 2 l 2 r 2
or equivalent
M1 A1
(9) [9]
Question Number
5.
Scheme
(a)
Marks
x 2 x 1 2 0.04
5
2 5
T
awrt 1.3
(b) v2 2 a 2 x2 0.22 52 a2 0.042
2 a 25
M1 A1
ft their
A1
(3)
M1 A1ft
accept exact equivalents or awrt 0.057 A1
(3)
(c) Using x a cos t 1 a a cos t 2 5t
t
ft their
M1 A1ft
3
15
T 4t
A1 4 15
awrt 0.84
M1 A1 (5) [11]
Alternative to (c) Using x a sin t 1 a a sin t 2 5t t
ft their
M1 A1ft
6
30
T T 4t
A1 4 15
awrt 0.84
M1 A1 (5)
Question Number
Scheme
Marks
6. V
W
W mg
O
U (a)
1 m U 2 v 2 mga 1 cos 2
Energy
T Leaves circle when T 0
Leading to
mg cos
M1 A1=A1
mv 2 a
U 2 2 ga 2 ga cos g cos a 2 U ag 2 3cos
M1 A1
Eliminating v
M1
cso
A1
(b) Using conservation of energy from the lowest point of the surface 1 m U 2 W 2 mga 2 W 2 U 2 2ag 1 3 Using cos , W 2 ag 2 2ag 3 3 cso ag 3
Alternatives for (b) are given on the next page.
(7)
M1 A1=A1
M1 A1
(5) [12]
Question Number
6.
Scheme
Marks
Alternative to part (b) using conservation of energy from the point where P loses contact with surface. ga 2 V ag cos 3
1 m W 2 V 2 mga cos 2 1 2 1 1 m W ag mga 2 3 3 2 W ag 3
Energy
Leading to
M1 A1 A1 cso
M1 A1
(5)
Alternative to part (b) using projectile motion from the point where P loses contact with surface.
ga 3 2 2 2 Wy V sin 2 ga cos V 2 ag cos
1 1 8 3 1 ag 1 2 ga ag 9 3 3 3 Vx V cos
8 3 1 W 2 Wy2 Vx2 ag ag ag 3 9 3
M1 A1 A1 cso
M1 A1
(5)
Question Number
7.
Scheme
(a)
A
Marks
1.5l
B
2l
T
T P mg
AP 1.5l 2l 2
2
2.5l
M1 A1
4 5 2.5l 1.5l 2 Hooke’s Law T 1.5l 3 5mg 2T cos mg T 8 2 4 2 5mg 2 mg 3 5 8 3 15mg 16 A 1.5l B cos
(b)
3.9l
B1 M1 A1 M1 A1 M1 cso
A1
(9)
h
P
h 3.9l 1.5l 2
2
3.6l
M1 A1
1 2 15mg 2.4l mv mg h 2 2 16 2 1.5l v0 2
Energy Leading to
ft their h cso
M1 A1ft = A1 A1
(6) [15]