Mark Scheme (Results) Summer 2013 GCE Mechanics M3 6679/01 Original Paper
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June 2013 6679 M3
Mark Scheme
Question Number Scheme Marks
1.
A
32
l
v
E P E gained 21 5 1 5
2 2 8mg mglll
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
M1A1
21 5 32 8 2
mgl mglmv + = M1A1ft
2 74glv =
1 72
v gl= oe A1
1 5
Question Number Scheme Marks
2
R α mg
( )R cosR mgα↑ = M1A1
( )2 215R sin ,
200mvR m
rα→ = = × M1,A1
Divide: 215tan
200gα = M1
6.548...α = 6.5, 6.55= (2 or 3 s.f. only) A1
1 6
Question Number Scheme Marks
3
(a) 2 2r r x′ = − B1
( )2 2
0d
rr x x xπ −∫ M1
2 2 4 4
02 4 4
rx r x rππ
⎡ ⎤= − =⎢ ⎥
⎣ ⎦ A1
4
32 34 3 8rx r rπ π= ÷ = * M1A1
(b) Mass 313
krπ ρ 32 33
rπ ρ× ( )31 63
r kπ ρ + B1
k 6 6k +
x 14
kr 38
− r x B1
( ) ( )21 3M 6 64 8
O k r r x k− × = + M1A1ft
( )( )
2 94 6k r
xk−
=+
Distance = ( )( )
2 9
4 6
k r
k
−
+ A1
(c) C of M must be at O 0x⇒ = M1
2 9 0k∴ − = ( )3 0k k= > A1
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Question Number Scheme Marks
4
A B
(a) ( )2 0.41 2.5
0.8A
xT x
+= = + * M1A1
(b) ( ) ( )2 0.41 2.5
0.8A
xT x
−= = − B1
0.2B AT T x− = M1
5 51 1 0.22 2
x x x⎛ ⎞− − + =⎜ ⎟⎝ ⎠
A1
5 0.2x x− = 25x x= − M1
∴ SHM period 25π
= * A1
(c) cos 0.3cos5x a t tω= =
0.3a = and 0.2x = − B1
2cos53
t = − M1
11 2cos5 3
t − ⎛ ⎞= −⎜ ⎟⎝ ⎠
Time to D: 0.460st = A1
(d) ( )2 2 2 2v a xω= − ( )( )22 2 25 0.3 0.2Dv = − ± M1
New oscillation: ( )2
2 2 25 0.22Dv a⎛ ⎞= −⎜ ⎟
⎝ ⎠ M1
( ) ( )2
2 2 2 2 2 25 0.2 5 0.3 0.22Dv a⎛ ⎞= − = −⎜ ⎟
⎝ ⎠ M1dep
0.374... 0.37a = = m A1
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Question Number Scheme Marks
5 (a)
Energy A to C: ( )2 21 1 1 cos2 2
mv mu mga θ− = − M1A1
NL2 along radius at C: ( )2
cos mvmg Ra
θ − = M1
2cosmga mvθ = A1
( )21 1cos 1 cos2 2
mg mu mgaθ θ− = − M1dep
2cos 2 2 cosag u ga gaθ θ− = −
( )21cos 23
u gaga
θ = + A1
(b) 4cos5
θ = 2 24 2or 5 5
v ga u ga⇒ = = B1
Vert. motion: Initial speed 6sin5 5
agv θ= = Energy Solution:
(can be worked from top)
2
2 6 925 5 5
ag aV g⎛ ⎞
= + ×⎜ ⎟⎜ ⎟⎝ ⎠
21 4 9 12 5 5 2
am ga mg mV⎛ ⎞ + × =⎜ ⎟⎝ ⎠
M1A1
2 36 18 486125 5 125
ag agV ag= + = 2 225
V ag= A1
Horiz. speed 4 8cos 25 5 5 5
ag agv θ= = × = M1A1
Dirn: 486 8 486tan125 645 5
φ⎛ ⎞
= ÷ =⎜ ⎟⎜ ⎟⎝ ⎠
8 1cos5 22
φ = × M1
70.05... 70φ = = ° 70φ = °
A1
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Question Number Scheme Marks
6
(a) 3
14
kxx
= − M1
3
1 d4 d
v kvx x= − M1dep
22
18 2
kv cx
= + M1depA1
11, 2 48 2
kx v c= = × = + M1A1
1 1 14,2 8 4 32
kx v c= = × = + A1
1 1 1 148 4 2 32
k⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
1k = M1A1
(b) 1 1 02 2
c c= + ⇒ =
22
4vx
= M1
2v (or v) is never zero ⇒ direction of motion never reversed A1
(c) 2
1 416 x
= M1
2 64 8x x= = A1
d 2dxvt x
= =
8
1 0d 2d
Tx x t=∫ ∫
[ ]82
01
22
Tx t⎡ ⎤
=⎢ ⎥⎣ ⎦
M1A1
63 22
T= 3154
T = s or 15.75 s A1
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Question Number Scheme Marks
7
60o
T
mg
(a) ( )R cos 60T mg↑ = 2T mg= B1
H.L. 62 mgxmgl
= M1
13
x l= A1
43
AB l= * A1
(b)
2sin 60T mrω= M1
4 sin 603
r l= B1
243
T mlω=
2423
mg mlω= M1
2 32gl
ω =
32gl
ω = A1
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