Mark Scheme (Results)
January 2011
GCE
GCE Mechanics M3 (6679) Paper 1
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January 2011
Publications Code UA026583
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© Edexcel Ltd 2011
General Instructions for Marking
1. The total number of marks for the paper is 75.
2. The Edexcel Mathematics mark schemes use the following types of marks:
•
M marks: method marks are awarded for ‘knowing a method and attempting to apply it’,
unless otherwise indicated.
•
A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been
earned.
•
B marks are unconditional accuracy marks (independent of M marks)
•
Marks should not be subdivided.
3. Abbreviations
These are some of the traditional marking abbreviations that will appear in the mark schemes.
•
bod – benefit of doubt
•
ft – follow through
•
the symbol
•
cao – correct answer only
•
cso - correct solution only. There must be no errors in this part of the question to obtain
this mark
•
isw – ignore subsequent working
•
awrt – answers which round to
•
SC: special case
•
oe – or equivalent (and appropriate)
•
dep – dependent
•
indep – independent
•
dp decimal places
•
sf significant figures
•
¿ The answer is printed on the paper
•
will be used for correct ft
The second mark is dependent on gaining the first mark
January 2011
Mechanics M3 6679
Mark Scheme
Question
Number
1.
Scheme
v
dv
= 7 − 2x
dx
1
2
v 2 = 7 x − x 2 ( +c )
Marks
M1
M1A1
A1
x = 0 v = 6 ⇒ c = 18
v = 0 x 2 − 7 x − 18 = 0
( x + 2 )( x − 9 ) = 0
M1
A1
∴x = 9
[6]
GCE Mechanics M3 (6679) January 2011
1
Question
Number
Scheme
Marks
2.
(a)
Mass ratio
4m
3
r
8
Dist from O
B1
B1
(4 + k)m
km
1
− r
2
0
Moments about O:
3
1
r × 4m = r × km
8
2
M1
k =3
A1
(4)
(b)
30o
O
G
λ mg
7mg
tan 30 =
OG =
B1
OG
r
λ
(7 + λ )
M1
× 2r
1
λ
1
=
× 2r ×
√ 3 (7 + λ )
r
7 + λ = 2 √ 3λ
7
λ=
2 √ 3 −1
(
)
A1
A1
(o.e.) or 2.84
(4)
[8]
GCE Mechanics M3 (6679) January 2011
2
Question
Number
Scheme
Marks
3.
(a)
2
M1
2
Vol = π ∫ y 2 dx = π ∫ e 2 x dx
1
1
M1 A1
2
1
= π ⎡⎣e 2x ⎤⎦
1
2
1
= π ⎡⎣e 4 − e 2 ⎤⎦
2
(b)
C of M
∫
2
1
∫
=
2
1
A1
(4)
π y 2 x dx
vol
M1 A1
2
21
⎡1
⎤
e x dx = ⎢ xe 2 x ⎥ − ∫ e 2 x dx
⎣2
⎦1 1 2
2x
2
M1
2
⎡1
⎤ ⎡1 ⎤
= ⎢ xe 2 x ⎥ − ⎢ e 2 x ⎥
⎣2
⎦1 ⎣ 4 ⎦ 1
1
1
1 ⎞
⎛1
= × 2e 4 − × 1e 2 − ⎜ e 4 − e 2 ⎟
2
2
4 ⎠
⎝4
1 ⎞
⎛3
= ⎜ e4 − e2 ⎟
4 ⎠
⎝4
1 ⎞
⎛3
π ⎜ e4 − e2 ⎟
4
4 ⎠
= 1.656...
C of M = ⎝
1
4
2
π (e − e )
2
A1
M1 A1
= 1.66
(3 sf)
GCE Mechanics M3 (6679) January 2011
(6)
[10]
3
Question
Number
Scheme
Marks
4.
(a)
⎛ πt ⎞
x = 5sin ⎜ ⎟
⎝ 3⎠
x = 5 ×
π
⎛ πt ⎞
cos ⎜ ⎟
3
⎝ 3⎠
M1A1
⎛π ⎞
⎛ πt ⎞
x = −5 × ⎜ ⎟ sin ⎜ ⎟
⎝3⎠
⎝ 3⎠
2
x=−
(b)
π2
9
period =
A1
(∴ S.H.M.)
x
(3)
2π
π
=6
B1
3
B1
amplitude = 5
(c)
(d)
(2)
⎛ πt ⎞
cos ⎜ ⎟
3
⎝ 3⎠
5π
max . v =
3
... = 5 ×
π
At A x = 2
sin
π
3
tA =
M1
or vmax = aω
A1
(2)
M1
⎛ πt ⎞
2 = 5sin ⎜ ⎟
⎝ 3⎠
t = 0.4
3
π
A1
× sin −1 0.4
At B x = 3
tB =
time A → B =
3
π
3
π
× sin −1 0.6
× sin −1 0.6 −
3
π
A1
× sin −1 0.4
= 0.2215... = 0.22 s accept awrt 0.22
A1
(4)
[11]
GCE Mechanics M3 (6679) January 2011
4
Question
Number
Scheme
Marks
5.
A
Ta
Tb
B
mg
(a)
B1
l
√2
R ( ↑ ) Ta cos 45 = Tb cos 45 + mg
r=
M1A1
Ta − Tb = mg √ 2
R ( →)
(1)
Ta cos 45 + Tb cos 45 = mrω
M1A1
2
1
1
l
+ Tb ×
=m
ω2
√2
√2
√2
Ta + Tb = mlω 2
(2)
Ta − Tb = mg √ 2
(1)
Ta ×
(
1
= m ( lω
2
)
+ g √ 2)
M1
2Ta = m lω 2 + g √ 2
Ta
2
A1
Tb = mlω 2 − Ta
=
(b)
(
1
m lω 2 − g √ 2
2
Tb > 0
ω2 >
g√2
l
A1
)
(
(8)
M1
)
1
m lω 2 − g √ 2 > 0
2
A1
*
GCE Mechanics M3 (6679) January 2011
(2)
[10]
5
Question
Number
Scheme
6.
A
Marks
B
C
(a)
3
4
l
θ
Ta
P
Tb
3mg
5
length AP = length BP = l
4
1
kmg ( 4 l ) 1
Ta = Tb =
= kmg
4
l
R (↑)
Ta cos θ + Tb cos θ = 3mg
1
3 1
3
kmg × + kmg × = 3mg
4
5 4
5
3
kmg = 3mg
10
k = 10
*
B1
( or T = ...)
( or 2T cos θ = 3mg )
3
⎛ 1
⎞
⎜ or kmg × = 3mg ⎟
5
⎝ 2
⎠
M1A1
M1A1
A1
A1
(7)
(b)
12
5
initial extn =
l
B1
13
8
l −l = l
5
5
λ x2
2
10mg ⎛ 8l ⎞ 128mgl
E.P.E. lost = 2 ×
= 2×
⎜ ⎟ =
2l
2l ⎝ 5 ⎠
5
12l 36mgl
P.E. gained = 3mg ×
=
5
5
1
36mgl 128mgl
× 3mv 2 +
=
2
5
5
256 gl 72 gl
v2 =
−
15
15
⎛ 184 ⎞
v = √⎜
gl ⎟
⎝ 15 ⎠
M1A1
M1A1
A1
(6)
[13]
GCE Mechanics M3 (6679) January 2011
6
Question
Number
Scheme
7.
Marks
u
v
mg
(a)
mgl ( cos α − cos θ ) =
M1A1=A1
1 2 1 2
mv − mu
2
2
v 2 = u 2 + 2 gl ( cos α − cos θ )
A1
*
(4)
(b)
⎛3
⎞
v 2 = 2 gl ⎜ − cos θ ⎟ + u 2
⎝5
⎠
⎛3 ⎞
At top θ = 360° v 2 = 2 gl ⎜ − 1⎟ + u 2
⎝5 ⎠
2
v 2 > 0 − 2 gl × + u 2 > 0
5
4
gl
u2 >
5
gl
*
u>2
5
cos α =
3
5
M1A1
M1
A1
(4)
GCE Mechanics M3 (6679) January 2011
7
Question
Number
(c)
Scheme
Marks
Equation of motion along radius at lowest point:
T1 − mg =
θ = 180
M1A1
mv 2
l
M1
⎛3 ⎞
v 2 = 2 gl ⎜ + 1⎟ + u 2
⎝5 ⎠
16 gl
+ u2
5
m ⎛ 16 gl
⎞
T1 = ⎜
+ u 2 ⎟ + mg
l ⎝ 5
⎠
2
21mg mu
=
+
5
l
v2 =
A1
At highest point:
T2 + mg =
M1
mv 2
l
2
⎛ 2 ⎞ mu
T2 = 2mg ⎜ − ⎟ +
− mg
l
⎝ 5⎠
mu 2 9mg
T2 =
−
l
5
T1 = 5T2
M1
⎛ mu 2 9mg ⎞
21mg mu 2
+
= 5⎜
−
⎟
5
l
5 ⎠
⎝ l
66mg
4mu 2
=
5
l
33
gl
u2 =
*
10
M1
θ = 360
GCE Mechanics M3 (6679) January 2011
A1
A1
(9)
[17]
8
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