Mark Scheme (Results) January 2011
GCE
GCE Mechanics M3 (6679) Paper 1
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January 2011 Publications Code UA026583 All the material in this publication is copyright Š 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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