Mark Scheme (Results) January 2010
GCE
Mechanics M3 (6679)
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January 2010 Publications Code UA022968 All the material in this publication is copyright Š Edexcel Ltd 2010
January 2010 6679 Mechanics M3 Mark Scheme Question Number
Scheme
Marks
0.5a = 4 + cos (π t )
Q1.
Integrating
0.5v = 4t +
sin (π t )
π
B1
(+ C )
M1 A1
Using boundary values 3 = 4 + C ⇒ C = −1
M1 A1
When t = 1.5
0.5v = 6 −
1
π
−1
v ≈ 9.36 ( m s −1 )
M1
cao
A1
(7) [7]
GCE Mechanics M3 (6679) January 2010
Question Number Q2.
Scheme
2π
Marks
5π ( ≈ 2.62 ) 6 ω x = 0, t = 0 ⇒ x = a sin ωt
(a)
= 2.4 ⇒ ω =
M1 A1
⎛ √3 ⎞ a⎟ ⎜= ⎝ 2 ⎠ 25π 2 ⎛ 2 3a 2 ⎞ 48 2 2 2 2 v = ω ( a − x ) ⇒ 16 = ( ≈ 3.06 ) ⎜a − ⎟ ⇒ a= 36 ⎝ 4 ⎠ 5π vmax = aω = 8 (or awrt 8.0 if decimals used earlier) cao ⎛ 5π ⎞ when t = 0.4 , x = a sin ⎜ × 0.4 ⎟ ⎝ 6 ⎠
(b)
Alternative in (a) (a)
GCE Mechanics M3 (6679) January 2010
&& xmax = aω 2 =
2π
20π 3
5π 6 ω x = 0, t = 0 ⇒ x = a sin ωt x& = aω cos ωt ⎛ 5π ⎞ 4 = aω cos ⎜ × 0.4 ⎟ ⎝ 6 ⎠ 48 a= ( ≈ 3.06 ) or aω = 8 5π vmax = aω = 8 = 2.4 ⇒ ω =
awrt 21
M1 M1 A1 M1 A1
(7)
M1 A1
(2) [9]
M1 A1 M1 M1 A1 M1 A1
(7)
Question Number Q3.
Scheme
(a) Mass ratios x
s 8 3 2 × r 8 3
B 19 x
Marks
S 27 3 r 8
1 3 8 × r + 19 x = 27 × r 4 8 65 x= r 152
anything in correct ratio
B1 B1
M1 A1ft
*
A1
(5)
(b)
x
r
θ Mg
kMg
Mg × x sin θ = kMg × r cos θ 13 leading to k = 38
GCE Mechanics M3 (6679) January 2010
M1 A1=A1 M1 A1
(5) [10]
Question Number Q4.
Scheme
Marks
O
θ T 30 N P 40 N ↑ T cos θ = 40 → T sin θ = 30 leading to T = 50 E= HL
λ x2
= 10 2a λx T= = 50 a
leading to
x = 0.4
OP = 0.5 + 0.4 = 0.9 ( m )
M1 attempt at both equations M1 A1 A1 M1 A1 B1 M1 M1 A1 A1ft
(10) [10]
GCE Mechanics M3 (6679) January 2010
Question Number Q5.
Scheme
(a)
Marks
θ T 3a 2a mg
1 1 m × 2ag − mv 2 = mg ( 2a − 3a sin θ ) 2 2 leading to v 2 = 2 ga ( 3sin θ − 1) ¿ (b) minimum value of T is when v = 0 ⇒ sin θ = T = mg sin θ = maximum value of T is when θ =
↑
mg 3
π 2
mv 2 + mg 3a 7 mg = 3
T=
7 mg ⎞ ⎛ mg ≤T ≤ ⎜ ⎟ 3 ⎠ ⎝ 3
GCE Mechanics M3 (6679) January 2010
1 3
M1 A1=A1
cso
M1 A1
(5)
B1 M1 A1
(v
2
= 4ag ) M1 A1 A1
(6)
[11]
Question Number Q6.
Scheme
Marks
(a) R
µR mg
↑ R = mg Use of limiting friction, Fr = µ R
B1 B1
m282 120 2 28 2 µ= = 120 × 9.8 3
←
(b)
µR =
M1 A1
cao M1 A1
À
(6)
R α
µR α ↑ ←
mg
R cos α − µ R sin α = mg
µ R cos α + R sin α =
mv r
µ cos α + sin α v 2 = cos α − µ sin α rg 2 cos α + 3sin α 25 = 3cos α − 2sin α 24 leading to
GCE Mechanics M3 (6679) January 2010
tan α =
27 122
M1 A1
2
M1 A1
Eliminating R Substituting values awrt 0.22
M1 M1 M1 A1
(8) [14]
Question Number Q7.
Scheme
Marks
(a) 1 2 3mgx 2 mv + = mg ( a + x ) 2 4a 3gx 2 leading to v 2 = 2 g ( a + x ) − ¿ 2a
M1 A2 (1, 0)
cso
(b) Greatest speed is when the acceleration is zero λ x 3mgx 2a T= = = mg ⇒ x = a 2a 3
A1
(4)
M1 A1
2
2a ⎞ 3g ⎛ 2a ⎞ ⎛ 8ag ⎞ ⎛ v = 2g ⎜ a + ⎟ − × ⎜ ⎟ ⎜ = ⎟ 3 ⎠ 2a ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 2 v = √ ( 6ag ) accept exact equivalents 3 2
3gx 2 =0 (c) v = 0 ⇒ 2 g ( a + x ) − 2a 3 x 2 − 4ax − 4a 2 = ( x − 2a )( 3x + 2a ) = 0 x = 2a mx&& = mg −
At D,
λ × 2a
A1
(4)
M1
M1 A1
ft their 2a
a
M1
&x& = 2 g
M1 A1ft A1
(6) [14]
Alternative to (b) 3gx 2 v = 2g ( a + x) − 2a Differentiating with respect to x dv 3gx 2v = 2 g − dx a dv 2a =0 ⇒x= dx 3 2
M1 A1 2
2a ⎞ 3g ⎛ 2a ⎞ ⎛ 8ag ⎞ ⎛ v = 2g ⎜ a + ⎟ − × ⎜ ⎟ ⎜ = ⎟ 3 ⎠ 2a ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 2 v = √ ( 6ag ) accept exact equivalents 3 2
GCE Mechanics M3 (6679) January 2010
M1 A1
(4)
Question Number Q7.
Scheme
Marks
Alternative approach using SHM for (b) and (c) If SHM is used mark (b) and (c) together placing the marks in the gird as shown. Establishment of equilibrium position λ x 3mge 2a = = mg ⇒ e = T= a 2a 3 N2L , using y for displacement from equilibrium position 3 mg ( y + e ) 3g my&& = mg − 2 =− y 2a a 3g ω2 = 2a 2 Speed at end of free fall u = 2 ga
bM1 bA1
bM1 bA1
cM1
Using A for amplitude and v 2 = ω 2 ( a 2 − x 2 ) 2 3 g ⎛ 2 4a 2 ⎞ u 2 = 2 ga when y = − a ⇒ 2 ga = ⎜A − ⎟ 3 2a ⎝ 9 ⎠
4a 3 4a ⎛ 3g ⎞ 2 × ⎜ ⎟ = √ ( 6ag ) Maximum speed Aω = 3 ⎝ 2a ⎠ 3 4a 3 g Maximum acceleration Aω 2 = × = 2g 3 2a A=
√
GCE Mechanics M3 (6679) January 2010
cM1 cA1 cM1 cA1 cA1
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