Mark Scheme (Results) Summer 2010
GCE
GCE Mechanics M3 (6679/01)
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Summer 2010 Publications Code UA024475 All the material in this publication is copyright Š Edexcel Ltd 2010
Summer 2010 Mechanics M3 6679 Mark Scheme Question Number
Marks
Scheme
Q1
A
α
13l T B
5l
mg
12 13 T cos α = mg cosα =
(a)
R (↑)
T×
(b)
Eqn of motion
v=
5 2
gl 3
12 = mg 13 13 T = mg 12
B1 M1
oe
v2 5l 13mg 5 v2 × =m 12 13 5l 25 gl v2 = 12 T sin α = m
⎛ ⎞ gl 25 gl or or any other equiv) ⎟⎟ ⎜⎜ accept 5 12 12 ⎝ ⎠
A1
(3)
M1 A1 M1 dep
A1
(4) [7]
Question Number Q2
Marks
Scheme
k x2 k mg = ( − ) 2 R mgR 2 F= x2 F = (−)
(a)
M1 M1
*
mgR 2 x2 dv gR 2 v =− 2 x dx ⎛ gR 2 ⎞ 1 2 v = ∫ ⎜ − 2 ⎟ dx 2 ⎝ x ⎠ mx = −
(b)
x = R, v = 3U
x = 2R , v = U
1 2 gR 2 v = ( +c ) 2 x 9U 2 = gR + c 2 1 2 gR 2 9U 2 v = + − gR 2 x 2 1 2 gR 2 9U 2 U = + − gR 2 2R 2 gR U2 = 8 gR U= 8
A1
(3)
M1 M1 M1 dep on 1st M mark A1 M1 dep on 3rd M mark
M1 dep on 3rd M mark
A1
(7) [10]
GCE Mechanics M3 (6679) Summer 2010
Question Number
Marks
Scheme
Q3
R
F θ mg EPE lost =
R (↑)
λ × 0.62 2 × 0.9
−
λ × 0.12 ⎛
7 ⎞ ⎜= λ⎟ 2 × 0.9 ⎝ 36 ⎠
R = mg cos θ
M1 A1 M1
4 = 0.4 g 5 F = µ R = 0.15 × 0.4 g P.E. gained = E.P.E. lost − work done against friction = 0.5 g ×
0.5g × 0.7 sin θ =
λ × 0.62
λ × 0.12
− 0.15 × 0.4 g × 0.7 2 × 0.9 3 0.1944λ = 0.5 × 9.8 × 0.7 × + 0.15 × 0.4 × 9.8 × 0.7 5 λ = 12.70...... λ = 13 N or 12.7 2 × 0.9
−
M1 A1
M1 A1 A1
A1 [9]
GCE Mechanics M3 (6679) Summer 2010
Question Number Q4
(a)
Marks
Scheme
cone 4π l 3 mass ratio 3 4 dist from l O Moments:
container 68π l 3 3 68 x
cylinder 24π l 3 72 3l
4l + 68 x = 72 × 3l 212l 53 x= = l 68 17
M1 A1
B1 M1 A1ft
accept 3.12l
A1
(6)
(b)
G
X
θ GX = 6l − x
seen 2l 6l − x 2 × 17 = 49 θ = 34.75... = 34.8 or 35
tan θ =
GCE Mechanics M3 (6679) Summer 2010
M1 M1 A1
A1
(4) [10]
Question Number
Marks
Scheme
Q5
v C
mg (a)
(b)
Energy:
1 1 m × 5ag − mv 2 2 2 2 v = 5ag − 2ag sin θ
mga sin θ =
(3)
M1 A1 M1 A1
(4)
At C , θ = 90° T = mg ( 5 − 3) = 2mg T > 0 ∴ P reaches C
(d)
A1
Eqn of motion along radius: mv 2 T + mg sin θ = a m T = ( 5ag − 2ag sin θ ) − mg sin θ a T = mg ( 5 − 3sin θ )
(c)
M1 A1
Max speed at lowest point ( θ = 270° ; v 2 = 5ag − 2ag sin 270 ) v 2 = 5ag + 2ag v = √ ( 7 ag )
M1 A1 A1
(3)
M1 A1
(2) [12]
GCE Mechanics M3 (6679) Summer 2010
Question Number
Q6
Marks
Scheme
d2 x 3 =− 2 2 dt ( t + 1)
(a)
dx −2 = ∫ −3 ( t + 1) dt dt −1 = 3 ( t + 1) ( + c ) t = 0, v = 2 2 = 3 + c c = −1 dx 3 = −1 dt t + 1
t = 0, x = 0
*
⇒ c′ = 0 x = 3ln ( t + 1) − t v=0⇒ t=2
3 =1 t +1
x = 3ln 3 − 2 = 1.295... = 1.30 m (Allow 1.3)
GCE Mechanics M3 (6679) Summer 2010
M1 A1 M1
⎛ 3 ⎞ x = ∫⎜ − 1⎟ dt ⎝ t +1 ⎠ = 3ln ( t + 1) − t (+ c′)
(b)
M1
A1
(5)
M1 A1 B1
M1 A1 M1 A1
(7) [12]
Question Number
Marks
Scheme
Q7
A
3a
e
T O
R ( ↑ ) T = 2mg
(a)
2mg B1
6mge 3a 6mge 2mg = 3a e=a AO = 4a
Hooke's law: T =
M! A1
(3)
(b)
T
4a
x
x
H.L. Eqn. of motion
GCE Mechanics M3 (6679) Summer 2010
2mg 6mg ( a − x ) 2mg ( a − x ) T= = 3a a -2mg + T = 2mx 2mg ( a − x ) -2mg + = 2mx a 2mgx − = 2mx a g x=− x a a period 2π g
B1ft M1 M1
A1
*
A1
(5)
Question Number
v2 = ω 2 ( a2 − x2 )
(c)
vmax
2
vmax
(d)
Marks
Scheme
x=−
a 8
2 ⎞ g ⎛⎛ a ⎞ = ⎜⎜ ⎟ − 0⎟ ⎟ a ⎜⎝ ⎝ 4 ⎠ ⎠ 1 = √ ( ga ) 4
v2 =
g ⎛ a2 a2 ⎞ ⎜ − ⎟ a ⎝ 16 64 ⎠
3ag 64 2 2 v = u + 2as 3ag 0= − 2 gh 64 3a h= 128 a 3a 19a Total height above O = + = 8 128 128
M1 A1 A1
(3)
M1
=
M1 A1
A1
(4) [15]
GCE Mechanics M3 (6679) Summer 2010
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