Mark Scheme (Results) January 2009
GCE
GCE Mathematics (6679/01)
Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH
January 2009 6679 Mechanics M3 Mark Scheme Question Number
1
Scheme
⎛ 15 ⎞ 3a = − ⎜ 9 + ⎟ 2 ⎜ ⎟ t + 1 ( ) ⎝ ⎠ 15 3v = −9t + ( + A) t +1 v = 0, t = 4 ⇒ 0 = −36 + 3 + A ⇒ 5 v = −3t + + 11 t +1 t = 0 ⇒ v = 16
N2L
Marks
B1 M1 A1ft
A = 33
M1 A1
M1 A1
(7) [7]
2
θ T 4 3
mg
mg (a)
4 mg 3 T cos θ = mg
( ←)
T sin θ =
(↑)
M1 A1 A1
2
Leading to
(b)
HL
2 ⎛4 ⎞ T 2 = ⎜ mg ⎟ + ( mg ) ⎝3 ⎠ 5 T = mg 3
T=
λx
a 5 e= a 9
λ x2
⇒
5 3mge mg = 3 a
M1 A1
ft their T
(5)
M1 A1ft
2
3mg ⎛ 5 ⎞ 25 = × ⎜ a ⎟ = mga E= 2a 2a ⎝ 9 ⎠ 54
M1 A1
(4) [9]
6679/01 GCE Mathematics January 2009
2
Question Number
Scheme
ω=
3
Marks
80 × 2π ⎛ 8π ⎞ rad s −1 ⎜ = ≈ 8.377... ⎟ 60 ⎝ 3 ⎠ 16π ≈ 0.67 ms −1 as equivalent Accept v = 75 ( ↑ ) R = mg
( ← ) µ mg = mrω 2
For least value of µ
B1
B1 M1 A1=A1
2
µ=
0.08 ⎛ 8π ⎞ × ⎜ ⎟ ≈ 0.57 9.8 ⎝ 3 ⎠
accept 0.573
M1 A1
(7) [7]
4
(a)
T=
25 2π = 2 ω
a =8
⇒ ω=
B1
4π ( ≈ 0.502 ...) 25
M1 A1
2
⎛ 4π ⎞ 2 2 v = ω (a − x ) ⇒ v = ⎜ ⎟ (8 − 3 ) ⎝ 25 ⎠ 4π −1 v= √ 55 ≈ 3.7 ( m h ) 25 2
(b)
2
2
2
2
⎛ 4π ⎞ x = a cos ωt ⇒ 3 = 8cos ⎜ t⎟ ⎝ 25 ⎠ t ≈ 2.3602 … time is 12 22
6679/01 GCE Mathematics January 2009
3
ft their a, ω awrt 3.7
ft their a, ω
M1 A1ft M1 A1
(7)
M1 A1ft M1 A1
(4) [11]
Question Number 5
(a)
Scheme Let x be the distance from the initial position of B to C GPE lost = EPE gained
6mgx 2 mgx sin 30° = 2a a Leading to x = 6 7a AC = 6 (b)
Marks
M1 A1=A1 M1 A1
(5)
The greatest speed is attained when the acceleration of B is zero, that is where the forces on B are equal.
(
)
T = mg sin 30° = e=
6mge a
a 12
M1 A1
2
CE
1 2 6mg ⎛ a ⎞ a mv + ⎜ ⎟ = mg sin 30° 2 2a ⎝ 12 ⎠ 12 Leading to
v=
√
6 ga ⎛ ga ⎞ ⎜ ⎟= 12 ⎝ 24 ⎠
M1 A1
(7) [12]
Alternative approaches to (b) are considered on the next page.
6679/01 GCE Mathematics January 2009
M1 A1=A1
4
Question Number 5
Scheme
Marks
Alternative approach to (b) using calculus with energy. Let distance moved by B be x
1 2 6mg 2 mv + x = mgx sin 30° 2 2a 6g 2 v 2 = gx − x a d 2 dv 12 g v ) = 2v = g − x=0 For maximum v ( a dx dx a x= 12 CE
M1 A1=A1
M1 A1
2
ga ⎛ a ⎞ 6g ⎛ a ⎞ v = g⎜ ⎟− ⎜ ⎟ = 24 ⎝ 12 ⎠ a ⎝ 12 ⎠ ⎛ ga ⎞ v= ⎜ ⎟ ⎝ 24 ⎠ 2
√
M1 A1
(7)
Alternative approach to (b) using calculus with Newton’s second law. As before, the centre of the oscillation is when extension is
mg sin 30° − T = mx&& ⎛a ⎞ 6mg ⎜ + x ⎟ 1 ⎝ 12 ⎠ = mx&& mg − 2 a 6g 6g && x=− x ⇒ ω2 = a a 6 g a ⎛ ⎞ ⎛ ga ⎞ vmax = ω a = ⎜ ⎟× = ⎜ ⎟ ⎝ a ⎠ 12 ⎝ 24 ⎠
a 12
M1 A1
N2L
√
6679/01 GCE Mathematics January 2009
√
5
M1 A1 A1 M1 A1
(7)
Question Number 6
(a)
Scheme
∫y
2
Marks
dx = ∫ ( 4 − x 2 ) dx = ∫ (16 − 8 x 2 + x 4 ) dx 2
= 16 x −
8 x3 x5 + 3 5
M1 A1
2
⎡ 8 x3 x5 ⎤ 256 16 x − + ⎥ = ⎢ 3 5 ⎦ 0 15 ⎣
∫ xy
2
M1 A1
dx = ∫ x ( 4 − x 2 ) dx = ∫ (16 x − 8 x 3 + x5 ) dx 2
= 8x2 − 2 x4 +
x6 6
M1 A1
2
⎡ 2 x6 ⎤ 32 4 8 x − 2 x + = ⎢ ⎥ 6 ⎦0 3 ⎣
M1A1
∫ xy dx = 32 × 15 = 5 x= ∫ y dx 3 216 8 2
2
(b)
¿
l 2 256 5 l π × = π ×16l × 15 8 2 A × x = ( π r 2l ) ×
Leading to l =
2√3 3
6679/01 GCE Mathematics January 2009
M1 A1 (10)
M1 A1 ft
accept exact equivalents or awrt 1.15
M1 A1 (4) [14]
6
Question Number 7
(a)
Scheme Let speed at C be u CE
(b)
Marks
1 1 ⎛ ag ⎞ mu 2 − m ⎜ ⎟ = mga (1 − cos θ ) 2 2 ⎝ 4 ⎠ 9 ga − 2 ga cos θ u2 = 4 mu 2 mg cos θ ( + R ) = a 9mg − 2mg cos θ eliminating u mg cos θ = 4 3 Leading to cos θ = ¿ 4
9 ga 3 3 − 2 ga × = ga 4 4 4 ⎛ 3ga ⎞ 3 ⎛ 27 ga ⎞ u x = u cos θ = ⎜ ⎟× = ⎜ ⎟ = 2.033 a ⎝ 4 ⎠ 4 ⎝ 64 ⎠ ⎛ 3 ga ⎞ √ 7 ⎛ 21ga ⎞ = u y = u sin θ = ⎜ ⎟× ⎜ ⎟ = 1.792 a ⎝ 4 ⎠ 4 ⎝ 64 ⎠ 21 7 245 v y2 = u y2 + 2 gh ⇒ v y2 = ga + 2 g × a = ga 64 4 64 v ⎛ 245 ⎞ tanψ = y = ⎜ ⎟ ≈ 3.012 … ux ⎝ 27 ⎠ ψ ≈ 72° awrt 72°
At C
( →)
(↓)
u2 =
√ √
√ √
√
Or
c
1.3
c
(1.2502 )
M1 A1
M1 A1 M1 M1 A1 (7)
B1 M1 A1ft M1 M1 A1 M1 A1
c
awrt 1.3
Alternative for the last five marks Let speed at P be v. CE
1 2 1 ⎛ ag ⎞ mv − m ⎜ ⎟ = mg × 2a or equivalent 2 2 ⎝ 4 ⎠ 17mga v2 = 4 ux 27 4⎞ ⎛ ⎛ 27 ⎞ cosψ = = ⎜ × ⎟= ⎜ ⎟ ≈ 0.315 v ⎝ 64 17 ⎠ ⎝ 272 ⎠ ψ ≈ 72° awrt 72°
√
Note: The time of flight from C to P is
6679/01 GCE Mathematics January 2009
√
235 − 21 ⎛ a ⎞ ⎜⎜ ⎟⎟ ≈ 1.38373 8 ⎝g⎠
7
⎛a⎞ ⎜⎜ ⎟⎟ ⎝g⎠
M1 M1 A1 M1 A1
(8) [15]