1(a)(i M L T −2
)
Allow kg m s −2
B1
1
B1
M1
For M L−1
Powers of M: α + γ = 0
of L: α + β − γ = 0
of T: −2α = 1
M2
For three equations
Give M1 for one equation
α = − 12 , β = 1 , γ =
A2
Give A1 for one correct
(ii)
(T) = (M LT
−2 α
β
−1 γ
) (L) (ML )
1
2
6
(iii) kF1α l1β σ γ = kF2α l2 β σ γ
1
M1
1
F1− 2 l1 = F2 − 2 l2
A1
OR F α l β is constant
F is proportional to l 2
F2 = 90 ×
Equation relating F1 , F2 , l1 , l2
M1
A1
or equivalent
2.0 2
M1
1.2 2
= 250 ( N)
A1
4
(b)(i) 2π = 0.01
B1
ω
ω = 200π
Maximum speed is Aω = 0.018 × 200π
M1
A1
= 11.3 (m s −1 )
Accept 3.6π
3
(ii) Using v 2 = ω 2 ( A2 − x 2 )
M1
M1
A1
A1
82 = (200π ) 2 (0.018 2 − x 2 )
x = 0.0127 (m)
Substituting values
4
OR v = 3.6π cos(200π t ) = 8
when 200π t = 0.785
( t = 0.001249 )
x = 0.018 sin(200π t ) = 0.018 sin(0.785)
= 0.0127
M1
A1
M1
A1
Condone the use of degrees in
this part
2 (a)
ω=
2π
2.4 × 10
6
( = 2.618 × 10 − 6 )
Acceleration a = rω 2 ( or
v2
)
r
= 2.604 × 10 −3
Force is ma = 7.5 × 10 22 × 2.604 × 10 −3
= 1.95 × 10 20 ( N)
or v =
B1
2π × 3.8 × 108
2.4 × 10 6
( = 994.8 )
M1
M0 for F − mg = ma etc
Accept 1.9 × 10 20 or 2.0 × 10 20
M1
A1
4
Change in PE is mg (3.5 − 4 sin θ )
(b)(i) By conservation of energy
B1
mv 2 = mg (3.5 − 4 sin θ )
M1
v = 68.6 − 78.4 sin θ
A1
1
2
2
or as separate terms
Accept 7 g − 8 g sin θ
3
(ii)
0.2 × 9.8 sin θ − R = 0.2 ×
v2
4
Radial equation of motion (3
terms)
M1
1.96 sin θ − R = 0.05(68.6 − 78.4 sin θ )
R = 5.88 sin θ − 3.43
M1
A1
E1
Substituting from part (i)
4 Correctly obtained
(iii) When θ = 40°, v 2 = 18.21
v2
= 4.55 ( m s − 2 )
4
Tangential acceleration is 9.8 cos 40
Radial acceleration is
= 7.51 (m s −2 )
M1
or 0.2 g sin 40 − R = ma
A1
Accept 4.5 or 4.6
M1
A1
M0 for a = mg cos 40 etc
4
(iv) Leaves surface when R = 0
3.43
sin θ =
5.88
θ = 35.7°
M1
M1
A1 cao
Accept 36° , 0.62 rad
3
3 (i)
λ
15
× 0.8 = 12 × 9.8
M1
λ = 2205 ( N)
E1
2
(ii)
Equation of motion including
tension
M1
A1
2205
× 5 − 12 × 9.8 = 12a
15
a = 51.45 (m s − 2 )
A1
3 Accept 51 or 52
M1
A1
Calculating elastic energy
By conservation of energy
M1
F1
Equation involving EE and PE
OA = 20 − h = 4.375 (m)
A1
(iii) Loss of EE is
1
2
×
2205
× 52 ( = 1837.5 )
15
12 × 9.8 × h = 1837.5
h = 15.625
5
OR 12 × 9.8 × 5 + 12 × 12 × v 2 = 1837.5
Equation involving EE, PE and
KE
M1
v 2 = 208.25
(iv)
0 = 208.25 − 2 × 9.8 × H
H = 10.625
F1
OA = 15 − H = 4.375 (m)
A1
T=
12 × 9.8 −
λ
2205
(0.8 + x)
15
2205
d2 x
(0.8 + x) = 12 2
15
dt
2
d x
dt
2
or T = ( x0 + x)
B1
l
M1
Equation of motion with three
terms
A1
or mg − ( x0 + x) = m
λ
d2 x
l
dt 2
provided that mg =
= −12.25 x
E1
λ
l
x0 appears
somewhere
4 Correctly obtained
No marks for just writing
−
2205
d2x
x = 12 2 or just using
15
dt
the formula ω 2 =
λ
ml
If x is clearly measured
upwards, treat as a mis-read
(v)
x = 4.2 cos(3.5t )
Rope becomes slack when x = −0.8
4.2 cos(3.5t ) = −0.8
t = 0.504 (s)
For cos( 12.25 t ) or sin( 12.25 t )
M1
A1
M1
A1
Accept 0.50 or 0.51
4
4 (i)
[
2
⌠
2
1 3
∫ y dx = ⎮⌡0 (4 − x ) dx = 4 x − 3 x
]
2
(=
0
16
)
3
B1
2
⌠
2
∫ xy dx = ⎮⌡0 x(4 − x ) dx
[
= 2x −
x=
2
1
4
x
4
]
2
0
M1
(=4)
A1
4
16
3
= 0.75
M1
E1
∫
Correctly obtained
2
⌠
y 2 dx = ⎮ 12 (16 − 8 x 2 + x 4 ) dx
⌡0
1
2
[
= 8x −
4
3
x 3 + 101 x 5
]
2
0
(=
M1
128
)
15
A1
4
OR
⌠
∫ yx dy = ⎮⌡0 y 4 − y dy
M1
⎡
=⎢−
⎢⎣
y=
y(4 −
2
3
3
y) 2
− 154 ( 4 −
5
y) 2
128
15
16
3
Valid method of integration
4
⎤
⎥
⎥⎦ 0
4
3
5 ⎤
⎡
or ⎢ − 83 (4 − y ) 2 + 52 (4 − y ) 2 ⎥
⎥⎦ 0
⎣⎢
A1
M1
= 1.6
E1
Correctly obtained
9 SR If 1 is omitted, marks for y
2
are
(ii) x = 12 × 0 + 6.5 × 0.75 + 6.5 × 2.75
12 + 6.5 + 6.5
=
22.75
= 0.91
25
12 × 0 + 6.5 × 1.6 + 6.5 × 1.6
25
20.8
=
= 0.832
25
M1A0M0E0
For 6.5 × 0.75 + 6.5 × 2.75
Using ( ∑ m ) x = ∑ mx
M1
M1
A1
y=
Using
M1
( ∑ m ) y = ∑ my
A1
5
(iii)
tan θ =
2 − 0.91
1.09
(=
)
4 − 0.832
3.168
M1
M1
For CM vertically below A
For trig in a triangle containing
θ , or finding the gradient of AG
Correct expression for tan θ or
A1
tan(90 − θ )
θ = 19.0°
Accept 0.33 rad
A1
4
4763
Mark Scheme
June 2006
1(a)(i) [ Force ] = M L T −2
B1
[ Power ] = [ Force ] × [ Distance ] ÷ [Time ]
= [ Force ] × L T −1
M1
= M L2 T −3
or [ Energy ] = M L2 T −2
or [ Energy ] × T −1
A1
3
(ii)
[ RHS ] =
( L ) 3 ( L T −1 ) 2 ( M L−3 )
M L2 T − 3
=T
[ LHS ] = L so equation is not consistent
B1B1
For ( L T −1 ) 2 and ( M L−3 )
M1
A1
Simplifying dimensions of RHS
E1
With all working correct (cao)
5 SR ‘ ... L = 289 π T , so inconsistent ’
can earn B1B1M1A1E0
(iii) [ RHS ] needs to be multiplied by L T −1
which are the dimensions of u
28 π r 3 u 3 ρ
Correct formula is x =
M1
A1
A1 cao
9P
RHS must appear correctly
3
OR x = k r α u β ρ γ P δ
M1
A1
β =3
x=
28 π r 3 u 3 ρ
9P
(b)(i) Elastic energy is
1
2
Equating powers of one
dimension
A1
× 150 × 0.82
= 48 J
M1
A1
2
Treat use of modulus
λ = 150 N as MR
(ii) In extreme position,
length of string is 2 1.2 2 + 0.9 2 ( = 3 )
B1
for 1.2 2 + 0.9 2 or 1.5 or 3
× 150 × 1.4 2 ( = 147 )
M1
allow M1 for (2 ×) 12 × 150 × 0.7 2
M1
A1
Equation involving EE and KE
elastic energy is
1
2
By conservation of energy,
147 − 48 = 12 × m × 10 2
Mass is 1.98 kg
A1
5
4763
2
(a)(i)
Mark Scheme
Vertically, T cos 55° = 0.6 × 9.8
Tension is 10.25 N
June 2006
M1
A1
2
Radius of circle is r = 2.8 sin 55° ( = 2.294 )
B1
v2
2.8 sin 55°
M2
(ii)
Towards centre, T sin 55° = 0.6 ×
OR T sin 55° = 0.6 × (2.8 sin 55°) × ω 2
ω = 2.47
v = (2.8 sin 55°) ω
Speed is 5.67 m s −1
Give M1 for one error
M1
or T = 0.6 × 2.8 × ω 2
M1
Dependent on previous M1
A1
4
Tangential acceleration is r α = 1.4 × 1.12
(b)(i)
M1
F1 = 0.5 × 1.4 × 1.12
= 0.784 N
A1
2
Radial acceleration is r ω = 1.4 ω
F2 = 0.5 × 1.4 ω
2
M1
2
= 0.7 ω 2 N
A1
(ii) Friction F = F 2 + F 2
1
2
Normal reaction R = 0.5 × 9.8
About to slip when F = μ × 0.5 × 9.8
0.784 2 + 0.49ω 4 = 0.65 × 0.5 × 9.8
ω = 2.1
SR F1 = −0.784 , F2 = −0.7ω 2
4
penalise once only
M1
M1
A1 A1
For LHS and RHS
Both dependent on M1M1
A1 cao
5
(iii)
tan θ =
=
F1
F2
0.784
Allow M1 for tan θ =
M1
0.7 × 2.12
A1
Angle is 14.25°
A1
Accept 0.249 rad
3
F2
etc
F1
4763
3 (i)
Mark Scheme
1323
× 2 ( = 882 )
3
1323
=
× 2.5 ( = 735 )
4.5
− mg − TBP = 882 − 15 × 9.8 − 735 = 0
June 2006
TAP =
B1
TBP
B1
TAP
so P is in equilibrium
E1
3
OR
1323
1323
(AP − 3) =
(BP − 4.5) + 15 × 9.8 B2
3
4.5
AP + BP = 12 and solving, AP = 5
E1
Give B1 for one tension correct
(ii) Extension of AP is 5 − x − 3 = 2 − x
1323
(2 − x) = 441(2 − x)
3
Extension of BP is 7 + x − 4.5 = 2.5 + x
1323
TBP =
(2.5 + x) = 294(2.5 + x)
4.5
TAP =
E1
B1
B1
3
(iii)
441(2 − x) − 15 × 9.8 − 294(2.5 + x) = 15
d2x
dt 2
2π
dt
Equation of motion involving 3
forces
M1
A1
d2x
2
= −49 x
M1
2π
=
= 0.898 s
Motion is SHM with period
ω
7
(iv) Centre of motion is AP = 5
If minimum value of AP is 3.5, amplitude is
1.5
Maximum value of AP is 6.5 m
A1
Obtaining
4 Accept
d2x
dt 2
= −ω 2 x (+c)
2
π
7
B1
1
(v) When AP = 4.1, x = 0.9
Using v 2 = ω 2 ( A 2 − x 2 )
2
2
M1
2
v = 49(1.5 − 0.9 )
A1
Speed is 8.4 m s −1
A1
Accept ± 8.4 or − 8.4
3
OR x = 1.5 sin 7t
When x = 0.9 , 7t = 0.6435 (t = 0.0919)
v = 7 × 1.5 cos 7t
= 10.5 cos(0.6435)
= 8.4
M1
A1
A1
or x = 1.5 cos 7t
or 7t = 0.9273 (t = 0.1325)
or v = −7 × 1.5 sin 7t
= (−) 10.5 sin(0.9273)
4763
Mark Scheme
(vi)
June 2006
x = 1.5 cos 7t
M1
A1
For cos( 49 t ) or sin( 49 t )
or x = 1.5 sin 7t
M1A1 above can be awarded in
(v) if not earned in (vi)
When 1.5 cos 7t = 0.5
M1
or other fully correct method to
find the required time
e.g. 0.400 − 0.224 or
Time taken is 0.176 s
A1
0.224 − 0.049
4 Accept 0.17 or 0.18
4763
4 (i)
Mark Scheme
⌠
π may be omitted throughout
4
2
∫ π y dx = ⎮ π x dx
⌡1
=
∫π x y
2
[
June 2006
1
π
2
x2
]
4
1
M1
= 7.5π
A1
dx
4
⌠
= ⎮ π x 2 dx =
⌡1
[
1
π
3
x3
]
M1
4
1
( = 21π )
A1
21π
7.5π
= 2.8
x=
M1
A1
6
(ii)
Cylinder has mass 3π ρ
Cylinder has CM at x = 2.5
Or volume 3π
B1
B1
M1
A1
(4.5π ρ ) x + (3π ρ )(2.5) = (7.5π ρ )(2.8)
x =3
Relating three CMs
( ρ and / or π may be omitted)
or equivalent, e.g.
(7.5π ρ )(2.8) − (3π ρ )(2.5)
x=
7.5π ρ − 3π ρ
Correctly obtained
E1
5
(iii)(A) Moments about A, S × 3 − 96 × 2 = 0
S = 64 N
Vertically, R + S = 96
R = 32 N
M1
A1
Moments equation
M1
or another moments equation
Dependent on previous M1
A1
4
(B) Moments about A,
S × 3 − 96 × 2 − 6 × 1.5 = 0
Moments equation
M1
A1
Vertically, R + S = 96 + 6
R = 35 N , S = 67 N
A1
Both correct
3
OR Add 3 N to each of R and S
R = 35 N , S = 67 N
M1
A2
Provided R ≠ S
Both correct
4762
Mark Scheme
Q1
Jan 2007
mark
(i)
10 m s –1
sub
0
before
0.5 kg
29.5 kg
v2 m s –1
v1 m s –1
after
10 × 0.5 = 0.5v2 + 29.5v1
v1 − v2
= −0.8
0 − 10
v1 = 0.3 so V1 = 0.3
v2 = −7.7 so V2 = 7.7 m s –1
in opposite to original direction
M1
A1
PCLM and two terms on RHS
All correct. Any form.
M1
NEL
A1
A1
A1
F1
Any form
Speed. Accept ± .
Must be correct interpretation of clear working
7
(ii)
(A)
10 × 0.5 = 30V
so V =
1
6
M1
A1
PCLM and coalescence
All correct. Any form.
A1
Clearly shown. Accept decimal equivalence. Accept
no direction.
3
(B)
Same velocity
No force on sledge in direction of motion
E1
E1
Accept speed
2
(iii)
2ms
–1
2ms
–1
before
39.5 kg
after
V
2 × 40 = 0.5u + 39.5V
u – V = 10
Hence V = 1.875
0.5 kg
B1
u
M1
A1
B1
A1
PCLM, masses correct
Any form
May be seen on the diagram.
Accept no reference to direction.
5
17
52
4762
Mark Scheme
Q2
(i)
comment
mark
X = R cos 30
Y + R sin 30 = L
(1)
(2)
B1
M1
A1
Jan 2007
sub
Attempt at resolution
3
(ii)
ac moments about A
R – 2L = 0
Subst in (1) and (2)
3
so X = 3L
2
1
Y + 2 L × = L so Y + L = L and Y = 0
2
X = 2L
B1
M1
Subst their R = 2L into their (1) or (2)
E1
Clearly shown
E1
Clearly shown
4
(iii)
(Below all are taken as tensions e. g. TAB in
AB)
B1
B1
Attempt at all forces (allow one omitted)
Correct. Accept internal forces set as
tensions or thrusts or a mix
2
(iv)
↓ A
TAD cos 30 ( −Y ) = 0
so TAD = 0
M1
Vert equilibrium at A attempted. Y = 0
need not be explicit
E1
2
(v)
Consider the equilibrium at pin-joints
M1
At least one relevant equilib attempted
A → TAB − X = 0 so TAB = 3L (T)
B1
(T) not required
C ↓
L + TCE cos 30 = 0
B1
Or equiv from their diagram
so TCE
−2 L
2L ⎛ 2 L 3 ⎞
=
so
⎜=
⎟ (C)
3 ⎠
3
3⎝
B1
Accept any form following from their
C ← TBC + TCE cos 60 = 0
B1
⎛ 2 3L ⎞ 1
3L
(T)
so TBC = − ⎜ −
⎟× =
3 ⎠ 2
3
⎝
B1
equation. (C) not required.
Or equiv from their diagram
FT their TCE or equiv but do not condone
inconsistent signs even if right answer
obtained. (T) not required.
T and C consistent with their answers and
their diagram
F1
7
(vi)
↓ B
TBD cos 30 + TBE cos 30 = 0
M1
Resolve vert at B
so TBD = −TBE so mag equal and opp sense
E1
A statement required
2
20
53
4762
Mark Scheme
Q3
(i)
Jan 2007
mark
(10, 2, 2.5)
sub
B1
1
(ii)
By symmetry
x = 10,
y=2
(240 + 80) z = 80 × 0 + 240 × 2.5
B1
B1
B1
M1
A1
so z = 1.875
Total mass correct
Method for c.m.
Clearly shown
5
(iii)
x = 10 by symmetry
⎛x⎞
⎛ 10 ⎞
⎛ 10 ⎞
⎜ ⎟
⎜
⎟
⎜ ⎟
(320 + 80) ⎜ y ⎟ = 320 ⎜ 2 ⎟ + 80 ⎜ 4 ⎟
⎜z⎟
⎜1.875 ⎟
⎜3⎟
⎝ ⎠
⎝
⎠
⎝ ⎠
y = 2.4
z = 2.1
E1
Could be derived
M1
Method for c.m.
B1
y coord c.m. of lid
B1
E1
E1
z coord c.m. of lid
shown
shown
6
(iv)
P
5 cm
2.4 cm
X
2.1 cm
Award for correct use of dimensions 2.1 and 2.4 or
equivalent
B1
30°
40 N
c.w moments about X
40 × 0.024 cos 30 − 40 × 0.021sin 30
= 0.41138… so 0.411 N m (3 s. f.)
1st term o.e. (allow use of 2.4 and 2.1)
2nd term o.e. (allow use of 2.4 and 2.1)
Shown
[Perpendicular method: M1 Complete method:
A1 Correct lengths and angles
E1 Shown]
B1
B1
E1
4
(v)
0.41138… - 0.05P = 0
M1
Allow use of 5
P = 8.22768…… so 8.23 (3 s. f.)
A1
Allow if cm used consistently
2
18
54
4762
Q4
(i)
Mark Scheme
Jan 2007
mark
Fmax = μ R
R = 2g cos 30
so Fmax = 0.75 × 2 × 9.8 × cos 30 = 12.730...
so 12.7 N (3 s. f.)
M1
B1
sub
Must have attempt at R with mg resolved
A1
[Award 2/3 retrospectively for limiting friction seen
below]
either
Weight cpt down plane is 2gsin 30 = 9.8 N
so no as 9.8 < 12.7
or
Slides if μ < tan 30
But 0.75 > 0.577… so no
B1
E1
The inequality must be properly justified
B1
E1
The inequality must be properly justified
5
(ii)
(A)
Increase in GPE is
2 × 9.8 × (6 + 4 sin 30) = 156.8 J
M1
B1
A1
Use of mgh
6 + 4 sin 30
3
(B)
WD against friction is
4 × 0.75 × 2 × 9.8 × cos 30 = 50.9222... J
M1
A1
Use of WD = Fd
2
(C)
Power is 10 × (156.8 + 50.9222…)/60
M1
= 34.620… so 34.6 W (3 s. f.)
A1
Use P = WD/t
2
(iii)
0.5 × 2 × 92
M1
Equating KE to GPE and WD term. Allow sign errors
and one KE term omitted. Allow ‘old’ friction as
well.
B1
A1
A1
A1
Both KE terms. Allow wrong signs.
All correct but allow sign errors
All correct, including signs.
cao
= 2 × 9.8 × (6 + x sin 30)
+ 0.5 × 2 × 42
−90
so x = 3.8163…. so 3.82 (3 s. f.)
5
17
55
Report on the units taken in January 2007
4762 - Mechanics 2
General Comments
Many excellent scripts were seen in response to this paper with the majority of candidates able
to make some progress worthy of credit on every question. The majority of candidates seemed
to understand the principles required. However, diagrams in many cases were poor and not as
helpful to the candidate as they could have been and some candidates did not clearly identify
the principle or process being used. As has happened in previous sessions, those parts of the
questions that were least well done were those that required an explanation or interpretation of
results or that required the candidate to show a given answer. In the latter case some
candidates failed to include all of the relevant steps in the working.
Comments on Individual Questions
1
Many candidates gained significant credit on this question. Those that drew a
diagram were usually more successful than those who did not. Many candidates
did not indicate which direction was to be positive and this did lead to some
errors in signs or inconsistencies between equations.
(i)
The majority of candidates were able to gain some credit for this part of the
question. Sign errors occurred in a few cases in the use of Newton’s
experimental law and many candidates forgot to indicate the direction in which
the ball was travelling after the impact.
(ii)
(A) Almost all the candidates could gain full credit for this part of the question.
(B) Very few candidates could obtain any credit for this part. Many did not
appreciate the vector nature of the problem and merely stated (incorrectly)
that the sledge would speed up because the mass had decreased and
momentum had to be conserved. A small number of candidates appreciated
that there would be no change in the velocity of the sledge but could not
give a valid reason for this. Few mentioned that there was no force on the
sledge in the direction of motion.
(iii)
2
Many candidates were able to gain full credit for this part of the question. Of
those who did not, a significant minority drew an inadequately labelled diagram
or made errors with the masses. A small number of candidates did not
understand the significance of the velocity of the snowball being relative to the
sledge and assumed that the snowball had a speed of 10 m s-1.
Some excellent answers to this question were seen, with many candidates
gaining almost full credit. It was pleasing to see that there were fewer mistakes
made with inconsistent equations than has been the case in previous sessions.
(i)
This part was well done by almost all the candidates.
(ii)
Most candidates did well on this part of the question. Some arithmetic errors
were seen and, in a few cases, some ‘creative’ algebra to try to establish the
given results.
24
Report on the units taken in January 2007
(iii)
In most cases the standard of the diagrams was satisfactory and worth some
credit but a significant minority of candidates did not label the internal forces
and/or omitted one or more of the external forces. A few candidates obviously
changed the diagram in response to their answers as they worked through the
following parts of the question and this then led to mistakes.
(iv)
This part of the question caused few problems.
(v)
Many candidates scored well on this part of the question. Those who did not
were usually those who drew an inadequate diagram or who ignored their
diagram; these made mistakes with signs or produced equations that were
inconsistent with each other and with the diagram drawn in part (iii).
(vi)
This part of the question was not as well done as previous parts. Arguments
based purely on symmetry at B were common but few appreciated that the
vertical equilibrium had to be considered. Those candidates who attempted to
look at the vertical equilibrium often forgot to resolve the forces in BD and BE. A
common answer was to simply write down TBD = ± TBE without any supporting
argument or interpretation of the result.
3
Only the last two parts of this question caused any problems to the vast majority
of candidates. The principles behind the calculation of centres of mass
appeared to be well understood and candidates who adopted column vector
notation made fewer mistakes than those who calculated the co-ordinates
separately.
(i)
Most obtained full credit for this part.
(ii)
Few candidates had difficulty with this part.
(iii)
Most candidates were able to obtain full credit for this part. A minority of them
wrongly assigned a mass of 100 units to the lid and it was common to see the z
co-ordinate of the centre of mass of the lid assumed to be 2.5 cm. However,
many who made this error realised the mistake and clearly corrected it.
Unfortunately, there were some candidates who made both of the above errors,
completed the working and still stated 2.1 as the z component of the centre of
mass.
(iv)
Few candidates made much progress here. The most common mistake was to
ignore one of the components of the weight. Trigonometric errors were also
common.
(v)
Very few correct responses to this part were seen. Many of those candidates
who appreciated that the moment of P had to be equated to the clockwise
moment of the weight from the previous part had inconsistent units for the
distance of P from the pivot.
25
4763
Mark Scheme
June 2007
B1
1(a)(i) [ Velocity ] = L T −1
[ Acceleration ] = L T
−2
(Deduct 1 mark if answers given as
ms −1 , ms −2 , kg ms −2 etc)
B1
[ Force ] = M L T −2
B1
−3
[ Density ] = M L
B1
[ Pressure ] = M L−1 T −2
B1
5
(ii)
[ P ] = M L−1 T −2
[ 12 ρ v 2 ] = ( M L−3 )(L T −1 ) 2
M1
= M L−1 T − 2
−3
−2
−1
[ ρ g h ] = (M L )(L T )(L) = M L T
Finding dimensions of 2nd or 3rd
term
A1
A1
−2
All 3 terms have the same dimensions
E1
Allow e.g. ‘Equation is
4 dimensionally consistent’
following correct work
(b)(i)
(ii)
Period
2π
= 3.49
M1
For a ‘cos’ curve (starting at the
highest point)
A1
Approx correct values marked on
2 both axes
M1
ω
ω = 1 .8
2π
3.49
For h = c + a cos/ sin with either
c = 12 (1.6 + 2.2) or a = 12 ( 2.2 − 1.6)
Accept
A1
M1
h = 1.9 + 0.3 cos 1.8t
F1
4
(iii) When h = 1.7 , float is 0.2 m below centre
Acceleration is ω 2 x = 1.82 × 0.2
= 0.648 m s −2 upwards
M1A1
Award M1 if there is at most one
error
A1 cao
3
OR When h = 1.7 , cos 1.8t = − 23
( 1.8t = 2.30 , t = 1.28 )
Acceleration h = −0.3 × 1.82 cos 1.8t
= −0.3 × 1.8 2 × ( − 23 )
M1
A1
= 0.648 m s −2 upwards A1 cao
63
4763
2 (i)
Mark Scheme
R cos 60 = 0.4 × 9.8
M1
A1
Normal reaction is 7.84 N
(ii)
R sin 60 = 0.4 ×
2
v
2.7 sin 60
Speed is 6.3 m s −1
June 2007
Resolving vertically
(e.g. R sin 60 = mg is M1A0
2
R = mg cos 60 is M0 )
M1
Horizontal equation of motion
M1
A1
Acceleration
v2
v2
(M0 for
)
r
2.7
A1 cao
4
OR
R sin 60 = 0.4 × (2.7 sin 60)ω
ω = 2.694
v = (2.7 sin 60)ω
2
Speed is 6.3 m s −1
M1
A1
Horizontal equation of motion
or R = 0.4 × 2.7 × ω 2
M1
For v = rω (M0 for v = 2.7ω )
A1 cao
(iii) By conservation of energy,
1
2
2
M1
A1
Equation involving KE and PE
v 2 = 28.08 − 52.92 cos θ
A1
Any (reasonable) correct form
e.g. v 2 = 81 − 52.92(1 + cos θ )
3
v2
2.7
M1
A1
M1
A1
Radial equation with 3 terms
2
× 0.4 × (9 − v ) = 0.4 × 9.8 × ( 2.7 + 2.7 cos θ )
81 − v 2 = 52.92 + 52.92 cos θ
(iv)
R + 0.4 × 9.8 cos θ = 0.4 ×
0.4
(28.08 − 52.92 cos θ )
2.7
R + 3.92 cos θ = 4.16 − 7.84 cos θ
R = 4.16 − 11.76 cos θ
R + 3.92 cos θ =
(v)
E1
Leaves surface when R = 0
SR If θ is taken to the downward
vertical, maximum marks are:
5 M1A0A0 in (iii)
M1A1M1A1E0 in (iv)
M1
4.16
cos θ =
11.76
v 2 = 28.08 − 52.92 ×
Substituting expression for v 2
A1
4.16
( = 9.36)
11.76
M1
Dependent on previous M1
or using mg cos θ =
Speed is 3.06 m s −1
A1 cao
4
64
mv 2
r
4763
3 (i)
Mark Scheme
Tension is 637 × 0.1 = 63.7 N
Energy is
1
2
June 2007
B1
2
× 637 × 0.1
M1
= 3.185 J
A1
3
(ii) Let θ be angle between RA and vertical
cos θ = 135
B1
( θ = 67.4° )
T cos θ = mg
M1
5
63.7 × 13
= m × 9.8
Resolving vertically
A1
Mass of ring is 2.5 kg
E1
4
Loss of PE is 2.5 × 9.8 × (0.9 − 0.5)
EE at lowest point is
1
2
× 637 × 0.32 ( = 28.665)
By conservation of energy,
2. 5 × 9. 8 × 0. 4 +
1
2
× 2.5u 2 =
1
2
Considering PE
or PE at start and finish
Award M1 if not more than one
error
M1
A1
M1
(iii)
× 637 × 0.32 − 3.185
A1
Equation involving KE, PE and EE
M1
F1
9.8 + 1.25u 2 = 25.48
u 2 = 12.544
u = 3.54
A1 cao
7
(iv) From lowest point to level of A,
Loss of EE is 28.665
Gain in PE is 2.5 × 9.8 × 0.9 = 22.05
EE at ‘start’ and at level of A
PE at ‘start’ and at level of A
(For M2 it must be the same ‘start’)
Comparing EE and PE (or
equivalent,
e.g. 12 mu 2 + 3.185 = mg × 0.5 + 12 mv 2 )
Fully correct derivation
M1
M1
M1
Since 28.665 > 22.05 ,
Ring will rise above level of A
A1 cao
4
SR If 637 is used as modulus,
maximum marks are:
(i) B0M1A0
(ii) B1M1A1E0
(iii) M1A1M1A1M1F1A0
(iv) M1M1M1A0
65
4763
4 (a)
Mark Scheme
⌠
2
Area is ⎮ x 3 dx =
⌡0
[
x4
1
4
]
2
0
=4
June 2007
B1
2
⌠
⌠ 4
⎮ x y dx = ⎮ x dx
⌡
⌡0
[
=
x=
⌠
⎮
⌡
]
x5
2
0
= 6.4
A1
6.4
= 1.6
4
A1
2
1
2
⌠
y 2 dx = ⎮ 12 x 6 dx
⌡0
=
=
[
1
14
x7
]
2
0
=
M1
64
7
64
7
4
Condone omission of
1
2
A1
∫ 2 y dx
∫ y dx
2
1
y=
(b)(i)
1
5
M1
=
M1
16
7
Accept 2.3 from correct working
A1
8
π may be omitted throughout
2
⌠
⌠
⌡
⌡1
Volume is ⎮ π y 2 dx = ⎮ π (4 − x 2 ) dx
[
= π 4 x − 13 x 3
]
2
1
M1
= 53 π
A1
For
5
3
For
9
4
2
⌠
⌠
2
2
⎮ π x y dx = ⎮ π x(4 − x ) dx
⌡1
⌡
[
= π 2x2 −
1
4
x4
]
2
1
M1
A1
= 94 π
∫ π x y dx
x=
2
∫ π y dx
2
=
9
π
4
5
π
3
=
M1
27
= 1.35
20
6
B1
M1
F1
Height of solid is h = 2 3
(ii)
T h = mg × 0.35
F = T = 0.101mg , R = mg
Least coefficient of friction is
Must be fully correct
E1
F
= 0.101
R
Taking moments
A1
4
66
Must be fully correct
(e.g. A0 if m = 53 π is used)
4763
Mark Scheme
4763
January 2008
Mechanics 3
1(a)(i) [ Force ] = M L T −2
B1
[ Density ] = M L−3
B1
2
(ii)
[E]=
[ F ] [ l0 ]
(M L T −2 )(L)
=
[ A ] [ l − l0 ]
( L2 )(L)
= M L−1 T − 2
B1
for [ A ] = L2
M1
A1
Obtaining the dimensions of E
3
(iii)
T = Lα ( M L−3 ) β (M L−1 T −2 )γ
−2γ = 1 , β + γ = 0
γ = − 12
β=
B1 cao
1
2
F1
M1
A1
A1
α − 3β − γ = 0
α =1
Obtaining equation involving
α, β ,γ
5
(b)
AP = 1.7 m
F = T cos θ
R + T sin θ = 5 × 9.8
B1
M1
M1
T cos θ = 0.4(49 − T sin θ )
8
T
17
= 0.4(49 −
Resolving in any direction
Resolving in another direction
(M1 for resolving requires no
force omitted, with attempt to
resolve all appropriate forces)
Using F = 0.4 R to obtain an
equation involving just one force
(or k)
Correct equation Allow
T cos 61.9 etc
M1
15
T)
17
A1
T = 23.8
A1
T = k (1.7 − 1.5)
M1
Stiffness is 119 N m −1
or R = 28 or F = 11.2 May be
implied
λ
Allow M1 for T =
× 0.2
A1
1.5
8
41
If R = 49 is assumed, max
marks are
B1M1M0M0A0A0M1A0
4763
2(a)(i)
Mark Scheme
0.1 + 0.01 × 9.8 = 0.01 ×
January 2008
M1
A1
u2
0.55
Speed is 3.3 m s −1
Using acceleration u 2 / 0.55
A1
3
(ii)
1
2
m(v 2 − u 2 ) = mg (2 × 0.55 − 0.15)
1
2
(v 2 − 3.32 ) = 9.8 × 0.95
M1
A1
Using conservation of energy
v 2 = 29.51
( ft is v 2 = u 2 + 18.62 )
R − mg cos θ = m
R − 0.01 × 9.8 ×
v2
a
Forces and acceleration
towards centre
M1
0.4
29.51
= 0.01 ×
0.55
0.55
A1
Normal reaction is 0.608 N
A1
( ft is
5
(b)(i)
u 2 + 22.54
)
55
B1
T = 0.8 r ω 2
160
( r − 2)
2
80(r − 2) 100(r − 2)
ω2 =
=
0.8r
r
200
ω 2 = 100 −
< 100 , so ω < 10
r
T=
B1
E1
E1
4
(ii)
EE =
1 160
×
× (r − 2) 2 = 40(r − 2) 2
2
2
B1
KE = 12 m(rω ) 2
Use of
M1
= 12 × 0.8 × r 2 ×
1
2
mv 2 with v = rω
100(r − 2)
r
= 40r (r − 2)
A1
40r (r − 2) > 40(r − 2) 2
Since r > r − 2 ,
i.e. KE > EE
From fully correct working only
E1
4
(iii)
100(r − 2)
r
r = 3.125
When ω = 6 , 36 =
Obtaining r
M1
T = 80(r − 2) = 80(3.125 − 2)
M1
Tension is 90 N
A1 cao
3
42
4763
3 (i)
Mark Scheme
dx
= Aω cos ωt − Bω sin ωt
dt
2
d x
dt 2
= − Aω 2 sin ωt − Bω 2 cos ωt
B1
B1
Aω = −1.44
M1
A1 cao
Period is
2π
ω
=
Using
d2x
M1
A1 cao
or
dt 2
dx
= −1.44 when t = 0
dt
= −0.18 when t = 0 (or x = 2 )
A1 cao
ω = 0.3 , A = −4.8
(iii)
Fully correct completion
3 SR For x = − Aω cos ωt + Bω sin ωt
x = − Aω 2 sin ωt − Bω 2 cos ωt
award B0B1E0
B=2
− Bω 2 = −0.18
− 0.18 = −ω 2 (2)
Must follow from their x
B1 ft
= −ω 2 ( A sin ωt + B cos ωt ) = −ω 2 x E1
(ii)
January 2008
6
2π
= 20.94 = 20.9 s
0.3
E1
(3 sf)
Amplitude is
or 1.442 = 0.32 (a 2 − 22 )
A2 + B 2 = 4.82 + 2 2
= 5.2 m
M1
A1
3
(iv)
x = −4.8 sin 0.3t + 2 cos 0.3t
v = −1.44 cos 0.3t − 0.6 sin 0.3t
M1
Finding x when t = 12 and t = 24
A1
Both displacements correct
M1
M1
Considering change of direction
Correct method for distance
When
t = 12 , x = 0.3306
( v = 1.56 )
When
t = 24 , x = −2.5929
( v = −1.35 )
Distance travelled is
(5.2 − 0.3306) + 5.2 + 2.5929
= 12.7 m
A1
5 ft from their A , B , ω and amplitude:
Third M1 requires the method to be
comparable to the correct one
A1A1 both require
ω ≈ 0.3, A ≠ 0, B ≠ 0
Note ft from A = +4.8 is
x12 = −3.92 (v < 0) x24 = 5.03 (v > 0)
Distance is (5.2 − 3.92) + 5.2 + 5.03
= 11.5
43
4763
4 (i)
Mark Scheme
January 2008
π may be omitted throughout
8
−1
⌠
V = ⎮ π ( x 3 ) 2 dx
⌡1
M1
8
⎡ 1 ⎤
= π ⎢ 3 x 3 ⎥ = 3π
⎦1
⎣
A1
8
−1
⌠
V x = ⎮ π x ( x 3 ) 2 dx
⌡1
⎡
=π ⎢
⎣
x=
M1
8
3
4
4
45
⎤
x3 ⎥ =
π
4
⎦1
A1
45
π
4
3π
15
=
= 3.75
4
Dependent on previous M1M1
M1
A1
6
(ii)
8
⌠ −1
A = ⎮ x 3 dx
⌡1
⎡
=⎢
⎣
M1
8
3
2
2
9
⎤
x 3 ⎥ = = 4.5
⎦1 2
A1
8
−1
⌠
A x = ⎮ x ( x 3 ) dx
⌡1
⎡
=⎢
⎣
3
5
x=
x
5
3
93
⎤
⎥ = 5 = 18.6
⎦1
⎡
=⎢
⎣
3
2
y=
A1
18.6 62
=
( ≈ 4.13 )
4.5 15
8
⌠
Ay=⎮
⌡1
M1
8
1
2
x
) 2 dx
M1
3
⎤
⎥ = 2 = 1.5
⎦1
A1
(x
1
3
−1
A1
3
If
8
1.5 1
=
4.5 3
A1
8
44
1
2
omitted, award M1A0A0
4763
(iii)
Mark Scheme
M1
⎛ 62 ⎞ ⎛18.6 ⎞
⎛ 4.5 ⎞
⎛x⎞
⎟⎟
⎟⎟ = (4.5) ⎜⎜ 15 ⎟⎟ = ⎜⎜
(1) ⎜⎜ ⎟⎟ + (3.5) ⎜⎜
1
⎝ 0.25 ⎠
⎝ y⎠
⎝ 3 ⎠ ⎝ 1.5 ⎠
M1
A1
A1
x = 2.85
y = 0.625
45
January 2008
Attempt formula for CM of
composite body (one coordinate
sufficient)
Formulae for both coordinates;
signs must now be correct, but
areas (1 and 3.5) may be
wrong.
ft only if 1 < x < 8
4 ft only if 0.5 < y < 1
Other methods: M1A1 for x
M1A1 for y
(In each case, M1 requires a
complete and correct method
leading to a numerical value)
4763
Mark Scheme
June 2008
4763 Mechanics 3
B1
1(a)(i) [ Velocity ] = L T −1
[ Acceleration ] = L T
−2
(Deduct 1 mark if kg, m, s are
consistently used instead of M,
L, T)
B1
[ Force ] = M L T −2
B1
3
(ii)
[λ ]=
[ Force ]
2
[v ]
=
M L T −2
M1
(L T −1 ) 2
= M L−1
A1 cao
2
(iii)
(Condone constants left in)
⎡ U 2 ⎤ (L T −1 ) 2
=L
⎢
⎥=
L T −2
⎣⎢ 2 g ⎦⎥
⎡ λU 4
⎢
2
⎢⎣ 4mg
B1 cao
⎤ (M L−1 )(L T −1 ) 4
⎥=
M (L T − 2 ) 2
⎥⎦
3
=
ML T
−4
M L2 T − 4
M1
=L
A1 cao
[ H ] = L ; all 3 terms have the same
dimensions
E1
Dependent on B1M1A1
4
(iv) (M L−1 ) 2 (L T −1 ) α M β (L T −2 ) γ = L
β = −2
B1 cao
−2 + α + γ = 1
At least one equation in α , γ
One equation correct
M1
A1
− α − 2γ = 0
α =6
γ = −3
A1 cao
A1 cao
5
51
4763
(b)
Mark Scheme
EE is
1 2060 2
×
×6
2 24
( = 1545 )
June 2008
B1
(PE gained) = (EE lost) + (KE lost)
50 × 9.8 × h = 1545 + 12 × 50 × 12 2
M1
Equation involving PE, EE and
KE
Can be awarded from start to
point where string becomes
slack or any complete method
(e.g. SHM) for finding v 2 at
natural length
If B0, give A1 for v 2 = 88.2
correctly obtained
F1
or 0 = 88.2 − 2 × 9.8 × s
490h = 1545 + 3600
( s = 4.5)
Notes
h = 10.5
OA = 30 − h = 19.5 m
A1
1 2060
×
× 6 used as EE can
2 24
4 earn
B0M1F1A0
2060
× 6 used as EE gets B0M0
24
52
4763
Mark Scheme
June 2008
T cos α = mg
2 (i)
3.92 cos α = 0.3 × 9.8
cos α = 0.75
Angle is 41.4° (0.723 rad)
M1
Resolving vertically
A1
(Condone sin / cos mix for M
marks throughout this question)
2
(ii)
T sin α = m
M1
v2
r
3.92 sin α = 0.3 ×
B1
A1
v2
4.2 sin α
Speed is 4.9 m s −1
(iii)
T − mg cos θ = m
A1
v2
a
Force and acceleration towards
centre
(condone v 2 / 4.2 or 4.2ω 2 )
For radius is 4.2sin α (= 2.778)
Not awarded for equation in ω
unless v = (4.2 sin α )ω also
4 appears
M1
T − 0.3 × 9.8 × cos 60° = 0.3 ×
Forces and acceleration
towards O
2
8. 4
4. 2
A1
Tension is 6.51 N
A1
3
(iv)
1
2
mv 2 − mg × 4.2 cos θ = 12 m × 8.4 2 − mg × 4.2 cos 60°
v 2 − 82.32 cos θ = 70.56 − 41.16
M1
A1
Equation involving
1
2
mv 2 and PE
4
2
v
a
29.4 + 82.32 cos θ
( T ) − m × 9.8 cos θ = m ×
4.2
String becomes slack when T = 0
−9.8cos θ = 7 + 19.6 cos θ
( T ) − mg cos θ = m
For (−)mg × 4.2 cos θ in PE
E1
v 2 = 29.4 + 82.32 cos θ
(v)
M1
M1
M1
A1
Force and acceleration
towards O
Substituting for v 2
M1
7
cos θ = −
29.4
θ = 104° (1.81 rad)
Dependent on first M1
A1
5
53
No marks for v = 0 ⇒ θ = 111°
4763
3 (i)
Mark Scheme
TPB = 35( x − 3.2)
B1
M1
[ = 35 x − 112 ]
TBQ = 5(6.5 − x − 1.8)
= 5(4.7 − x)
June 2008
[ = 23.5 − 5 x ]
Finding extension of BQ
A1
3
(ii)
TBQ + mg − TPB = m
d2x
dt 2
5(4.7 − x) + 2.5 × 9.8 − 35( x − 3.2) = 2.5
160 − 40 x = 2.5
d2x
dt 2
M1
Equation of motion (condone
one missing force)
A2
Give A1 for three terms correct
d2x
dt 2
2
d x
dt 2
= 64 − 16 x
E1
4
(iii)
At the centre,
d2x
dt 2
=0
M1
A1
x=4
2
M1
(iv) ω 2 = 16
Period is
2π
16
= 12 π = 1.57 s
A1
Seen or implied (Allow M1 for
ω = 16 )
2 Accept
B1 ft
(v) Amplitude A = 4.4 − 4 = 0.4 m
Maximum speed is Aω
1
2
π
ft is 4.4 − (iii)
M1
A1 cao
= 0.4 × 4 = 1.6 m s −1
3
(vi) x = 4 + 0.4 cos 4t
v = (−) 1.6sin 4t
0.9
1.6
4t = π + 0.5974
M1
A1
For v = C sin ωt or C cos ωt
This M1A1 can be earned in (v)
M1
Fully correct method for finding
the required time
When v = 0.9 , sin 4t = −
Time is 0.935 s
e.g.
A1 cao
1
4
arcsin
0.9 1
+ period
1.6 2
4
2
2
2
OR 0.9 = 16(0.4 − y )
y = −0.3307
M1
y = 0.4 cos 4t
0.3307
0. 4
4t = π + 0.5974
A1
y = 0.4 cos 4t
cos 4t = −
Time is 0.935 s
Using v 2 = ω 2 ( A 2 − y 2 )
and y = A cos ωt or A sin ωt
For y = ( ± ) 0.331 and
M1
A1 cao
54
4763
4
(a)(i)
Mark Scheme
June 2008
π may be omitted throughout
Limits not required for M marks
throughout this question
8
⌠
⌠
V = ⎮ π x 2 dy = ⎮ π (4 − 12 y ) dy
⌡0
⌡
[
= π 4 y − 14 y 2
]
8
0
M1
= 16π
A1
⌠
V y = ⎮ π y x 2 dy
⌡
M1
8
⌠
= ⎮ π y (4 − 12 y ) dy
⌡0
[
= π 2 y 2 − 16 y 3
y=
]
8
0
A1
128
=
π
3
A1
128
π
3
16π
8
=
( ≈ 2.67 )
3
M1
Dependent on M1M1
A1
7
M1
M1
CM is vertically above lower corner
(ii)
2 2
3
tan θ = =
(= )
y 83
4
θ = 36.9°
A1
(= 0.6435 rad)
A1
4
Trig in a triangle including θ
Dependent on previous M1
Correct expression for tan θ or
tan(90 − θ )
Notes
tan θ =
2
implies M1M1A1
cand's y
cand's y
implies M1M1
2
1
without further
tan θ =
cand's y
tan θ =
evidence is M0M0
55
4763
Mark Scheme
June 2008
May use 0 ≤ x ≤ 2 throughout
(b)
A=
2
⌠
⎮ (8 − 2 x 2 ) dx
⎮
⌡ −2
2
64
= ⎡ 8 x − 23 x3 ⎤ =
⎣
⎦ −2
3
⌠
⌠
⌡0
⌡ −2
⌠
(8 − 2 x 2 ) 2 dx
1
2
4 − 12 y dy
A1
2
A y = ⎮⎮
8
or (2) ⎮⎮
M1
2
= ⎡ 32 x − 163 x3 + 52 x5 ⎤
⎣
⎦ −2
8
M1
or (2) ⎮⎮ y 4 − 12 y dy
M1
(M0 if ½ is omitted)
For 32 x − 163 x 3 + 52 x 5 Allow one
⌡0
error
3
5
32
or − 83 y (4 − 12 y ) 2 − 15
(4 − 12 y ) 2
=
3
1024
15
y=
1024
64
5
or − 643 (4 − 12 y ) 2 + 165 (4 − 12 y ) 2
A1
15
3
Dependent on first two M1’s
M1
16
=
= 3.2
5
A1
7
56
4763
Mark Scheme
January 2009
4763 Mechanics 3
1 (i) [ Force ] = M L T −2
B1
−3
B1
[ Density ] = M L
2
(ii)
[η ] =
[ F ][ d ]
(M L T −2 )(L)
=
[ A ][ v2 − v1 ] ( L2 )(L T −1 )
= M L−1 T −1
B1
for [ A ] = L2 and [ v ] = L T-1
M1
A1
Obtaining the dimensions of η
3
B1
⎤ L2 (M L−3 )(L T −2 )
= L T −1
⎥=
M L−1 T −1
⎥⎦
M1
For [ g ] = L T −2
Simplifying dimensions of RHS
which is same as the dimensions of v
E1
Correctly shown
(iii) ⎡ 2a 2 ρ g
⎢
⎢⎣ 9η
3
(iv) ( M L−3 ) Lα (L T −1 ) β (M L−1 T −1 )γ is dimensionless
γ = −1
−β − γ = 0
−3 + α + β − γ = 0
α = 1, β = 1
B1 cao
M1
M1A1
A1 cao
5
(v)
R=
ρ wv 0.4 × 25 × 150
=
( = 9.375 × 107 )
5
−
η
1.6 × 10
1.3 × 5v
=
1.8 × 10−5
Required velocity is 260 m s −1
M1
Evaluating R
A1
Equation for v
A1 cao
3
43
4763
2
(a)(i)
Mark Scheme
January 2009
M1
T cos α = T cos β + 0.27 × 9.8
A1
Resolving vertically (weight and at
least one resolved tension)
Allow T1 and T2
1.2 3
4
(α = 36.87°)
= , cos α =
2.0 5
5
1.2 12
5
sin β =
, cos β =
( β = 67.38°)
=
1.3 13
13
27
T = 2.646
65
B1
For cos α and cos β [ or α and β ]
M1
Obtaining numerical equation for T
e.g. T (cos 36.9 − cos 67.4) = 0.27 × 9.8
E1
(Condone 6.36 to 6.38)
sin α =
Tension is 6.37 N
5
(ii)
T sin α + T sin β = 0.27 ×
v2
1.2
3
12
v2
6.37 × + 6.37 × = 0.27 ×
5
13
1.2
2
v = 43.12
M1
Using v 2 /1.2
A1
Allow T1 and T2
Obtaining numerical equation for
M1
Speed is 6.57 m s −1
v2
A1
4
(b)(i)
M1
u2
1.25
2
u = 9.8 × 1.25 = 12.25
Speed is 3.5 m s −1
0.2 × 9.8 = 0.2 ×
Using acceleration u 2 /1.25
E1
2
(ii)
1
2
m(v 2 − 3.52 ) = mg (1.25 − 1.25cos 60)
M1
A1
Using conservation of energy
M1
With numerical value obtained by
using energy
(M0 if mass, or another term,
included)
v 2 = 24.5
Radial component is
24.5
1.25
= 19.6 m s
−2
A1
Tangential component is g sin 60
M1
= 8.49 m s −2
(iii)
A1
T + 0.2 × 9.8cos 60 = 0.2 × 19.6
M1
Tension is 2.94 N
A1 cao
For sight of (m) g sin 60° with no
6 other terms
Radial equation (3 terms)
This M1 can be awarded in (ii)
2
44
4763
3 (i)
Mark Scheme
980
y = 5 × 9.8
25
M1
Extension is 1.25 m
A1
January 2009
Using
λy
l0
(Allow M1 for
980 y = mg )
2
(ii)
T=
980
(1.25 + x)
25
5 × 9.8 − 39.2(1.25 + x) = 5
−39.2 x = 5
2
d x
dt 2
(ft) indicates ft from previous parts
as for A marks
Equation of motion with three
terms
B1 (ft)
M1
d2 x
F1
dt 2
d2 x
Must have x
In terms of x only
dt 2
= −7.84 x
E1
4
M2
(iii)
8.42 = 7.84( A2 − 1.252 )
Using v 2 = ω 2 ( A2 − x 2 )
A1
A1
Amplitude is 3.25 m
4
OR
M2
980 2
y = 5 × 9.8 y + 12 × 5 × 8.42
2 × 25
y = 4.5
Equation involving EE, PE and KE
A1
Amplitude is 4.5 − 1.25 = 3.25 m
A1
OR x = A sin 2.8t + B cos 2.8t
x = −1.25, v = 8.4 when t = 0
M2
A1
⇒ A = 3, B = −1.25
Amplitude is
A2 + B 2 = 3.25
A1
(iv) Maximum speed is A ω = 3.25 × 2.8
= 9.1 m s
(v)
Obtaining A and B
Both correct
M1
−1
A1
x = 3.25cos 2.8t
B1 (ft)
−1.25 = 3.25cos 2.8t
M1
or equation involving EE, PE and
KE
2 ft only if answer is greater than 8.4
or
or
or
or
x = 3.25sin 2.8t
v = 9.1cos 2.8t or v = 9.1sin 2.8t
x = 3.25sin(2.8t + ε ) etc
x = ±3sin 2.8t ± 1.25cos 2.8t
Obtaining equation for t or ε by
setting x = (± )1.25 or v = (± )8.4 or
solving
±3sin 2.8t ± 1.25cos 2.8t = 3.25
M1
A1 cao
Time is 0.702 s
Strategy for finding the required
time
1
1.25 1 2π
e.g.
sin −1
+ ×
2.8
3.25 4 2.8
2.8t − 0.3948 = 12 π or
4 2.8t − 1.966 = 0
45
4763
Mark Scheme
January 2009
B1B1B1 Three modelling assumptions
(vi) e.g. Rope is light
Rock is a particle
3
No air resistance / friction / external forces
Rope obeys Hooke’s law / Perfectly elastic /
Within elastic limit / No energy loss in rope
4 (a)
∫
a
1
2
⌠
y 2 dx = ⎮
⌡− a
=⎡
⎣
1
2
(a 2 − x 2 ) dx
1
2
a
(a 2 x − 13 x3 ) ⎤
⎦ −a
= 23 a 3
y=
=
2
3
a
For integral of (a 2 − x 2 )
M1
A1
3
1 π a2
2
Dependent on previous M1
M1
4a
3π
E1
4
π may be omitted throughout
(b)(i)
h
⌠
V = ∫ π y 2 dx = ⎮ π (mx) 2 dx
⌡0
For integral of x 2
or use of V = 13 π r 2 h and r = mh
M1
h
= ⎡ 13 π m 2 x3 ⎤ = 13 π m 2 h3
⎣
⎦0
⌠
A1
h
2
2
∫ π xy dx = ⎮ π x(mx) dx
For integral of x3
M1
⌡0
h
= ⎡ 14 π m 2 x 4 ⎤ = 14 π m 2 h 4
⎣
⎦0
x=
=
A1
1 π m2 h4
4
1 π m 2 h3
3
3
h
4
Dependent on M1 for integral of
M1
x3
E1
6
(ii)
m1 = 13 π × 0.7 2 × 2.4 ρ = 13 πρ × 1.176
VG1 = 1.8
m2 = 13 π × 0.42 × 1.1ρ = 13 πρ × 0.176
B1
VG 2 = 1.3 + 34 × 1.1 = 2.125
B1
For m1 and m2 (or volumes)
or 14 × 1.1 from base
M1
F1
Attempt formula for composite
body
(m1 − m2 )(VG) + m2 (VG 2 ) = m1 (VG1 )
(VG) + 0.176 × 2.125 = 1.176 × 1.8
A1
Distance (VG) is 1.74 m
5
(iii) VQG is a right-angle
VQ = VG cos θ where tan θ =
VQ = 1.7428 ×
M1
0.7
2.4
(θ = 16.26°)
M1
24
25
= 1.67 m
ft is VG × 0.96
A1
3
46
4763
Mark Scheme
June 2009
4763 Mechanics 3
M1
1 (i)
1
2
2
2
m(v − 1.4 ) = m × 9.8(2.6 − 2.6 cos θ )
Equation involving KE and PE
A1
2
v − 1.96 = 50.96 − 50.96 cos θ
v 2 = 52.92 − 50.96 cos θ
E1
3
(ii)
2
v
2.6
6.37 cos θ − R = 0.25(52.92 − 50.96 cos θ )
0.65 × 9.8cos θ − R = 0.65 ×
6.37 cos θ − R = 13.23 − 12.74 cos θ
R = 19.11cos θ − 13.23
Radial equation involving v 2 / r
M1
Substituting for v 2
Dependent on previous M1
A1
Special case:
4 R = 13.23 − 19.11cos θ earns
M1A0M1SC1
M1
(iii) Leaves surface when R = 0
13.23
9
cos θ =
(=
)
19.11
13
9
v 2 = 52.92 − 50.96 ×
13
M1
A1
( θ = 46.19° )
a
b
( ft if R = a + b cos θ and 0 < − < 1
A1
)
M1
Speed is 4.2 m s −1
Dependent on previous M1
A1
4
M1
A1
M1
A1
(iv)
T sin α + R cos α = 0.65 × 9.8
1.22
T cos α − R sin α = 0.65 ×
2.4
⎛ 1.22 ⎞
OR T − mg sin α = m ⎜⎜
⎟⎟ cos α
⎝ 2.4 ⎠
⎛ 1.22
mg cos α − R = m ⎜
⎜ 2.4
⎝
sin α =
⎞
⎟⎟ sin α
⎠
2.4 12
5
=
, cos α =
2.6 13
13
Resolving vertically (3 terms)
Horiz eqn involving v 2 / r or r ω 2
M1A1
M1A1
( α = 67.38° )
M1
M1
Tension is 6.03 N
Normal reaction is 2.09 N
A1
A1
60
Solving to obtain a value of T or
R
Dependent on necessary M1s
8 ( Accept 6 , 2.1 )
Treat ω = 1.2 as a misread,
leading to T = 6.744, R = 0.3764
for 7 / 8
4763
Mark Scheme
2 (i)
1
2
× 5000 x 2 = 12 × 400 × 32
Compression is 0.849 m
June 2009
M1
A1
Equation involving EE and KE
A1
Accept
3
(ii) Change in PE is 400 × 9.8 × (7.35 + 1.4) sin θ
M1
Or 400 × 9.8 × 1.4sin θ
A1
1
× 400 × 4.542
2
Or 784 + 4116
and
= 400 × 9.8 × 8.75 × 17
= 4900 J
Change in EE is
1
2
× 5000 × 1.4
= 4900 J
M1
(iii) WD against resistance is 7560(24 + x)
Change in KE is
1
2
1
2
M1M1A1 can also be given for a
correct equation in x
(compression):
2
Since Loss of PE = Gain of EE, car will be
at rest
Change in EE is
× 5000x 2
× 400 × 302
Change in PE is 400 × 9.8 × (24 + x) ×
1
7
E1
2500 x 2 − 560 x − 4116 = 0
4 Conclusion required, or solving
equation to obtain x = 1.4
B1
( = 181440 + 7560x )
B1
( = 2500x 2 )
B1
( = 180000 )
B1
( = 13440 + 560x )
OR Speed 7.75 m s −1 when it hits buffer, then
B1
WD against resistance is 7560x
B1
Change in EE is 12 × 5000x 2
Change in KE is
1
2
× 400 × 7.75
2
Change in PE is 400 × 9.8 × x ×
3 2
5
1
7
( = 2500x 2 )
B1
( = 12000 )
B1
( = 560x )
M1
Equation involving WD, EE, KE,
PE
−7560(24 + x) = 12 × 5000 x 2 − 12 × 400 × 302
−400 × 9.8 × (24 + x) × 17
F1
2
−7560(24 + x) = 2500 x − 180 000 − 560(24 + x)
−3.024(24 + x) = x 2 − 72 − 0.224(24 + x)
M1
A1
x 2 + 2.8 x − 4.8 = 0
2
−2.8 + 2.8 + 19.2
2
= 1.2
x=
Simplification to three term
quadratic
M1
A1
10
61
4763
Mark Scheme
3(a)(i) [ Velocity ] = L T −1
B1
−2
B1
−3
B1
[ Force ] = M L T
[ Density ] = M L
June 2009
Deduct 1 mark for m s −1 etc
3
−2
−3 α
−1 β
2 γ
(ii) M L T = (M L ) (L T ) (L )
α =1
β =2
−3α + β + 2γ = 1
γ =1
B1
B1
( ft if equation involves
α , β and γ )
M1A1
A1
5
2π
(b)(i)
ω
M1
= 4.3
2π
( = 1.4612 )
4.3
A1
θ 2 = 1.46122 (0.082 − 0.052 )
M1
F1
Angular speed is 0.0913 rad s −1
A1
ω=
Using ω 2 ( A2 − θ 2 )
For RHS
( b.o.d. for v = 0.0913 m s−1 )
5
OR θ = 0.08ω cos ωt
= 0.08 × 1.4612 cos 0.6751
= 0.0913
Or θ = (−) 0.08ω sin ωt
M1
F1
A1
(ii) θ = 0.08sin ωt
When θ = 0.05, 0.08sin ωt = 0.05
= (−) 0.08 × 1.4612sin 0.8957
B1
M1
ωt = 0.6751
t = 0.462
A1 cao
Time taken is 2 × 0.462
M1
or θ = 0.08cos ωt
Using θ = (±) 0.05 to obtain an
equation for t
B1M1 above can be earned in
(i)
or t = 0.613 from θ = 0.08cos ωt
or t = 1.537 from θ = 0.08cos ωt
Strategy for finding the required
time
( 2 × 0.462 or 12 × 4.3 − 2 × 0.613
A1 cao
5 or 1.537 − 0.613 ) Dep on first
M1
= 0.924 s
For θ = 0.05sin ωt , max
B0M1A0M0
( for
0.05 = 0.05sin ω t )
62
4763
4 (a)
Mark Scheme
June 2009
ln 3
ln 3
⌠
e x dx = ⎡ e x ⎤
⎣
⎦
0
⌡0
=2
Area is ⎮
M1
A1
ln 3
⌠
x
∫ x y dx = ⎮⌡0 x e dx
M1
ln 3
= ⎡ x ex − ex ⎤
⎣
⎦0
M1
A1
Integration by parts
For x e x − e x
A1
ww full marks (B4) Give B3 for
0.65
= 3ln 3 − 2
x=
3ln 3 − 2 3
= 2 ln 3 − 1
2
∫ 12 y
ln 3
2
⌠
dx = ⎮
⌡0
=⎡
⎣
1
4
1
2
(e x ) 2 dx
M1
ln 3
e2 x ⎤
⎦0
For integral of (e x )2
A1
=2
For
2
y = =1
2
If area wrong, SC1 for
A1
9
(b)(i)
1 2x
e
4
a
36
⌡2
x4
Volume is ∫ π y 2 dx = ⌠
⎮ π
dx
x=
3ln 3 − 2
2
and y =
area
area
π may be omitted throughout
M1
a
⎡ 12 ⎤
⎛ 3 12 ⎞
=π ⎢ − 3 ⎥ =π ⎜ − 3 ⎟
⎣ x ⎦2
⎝2 a ⎠
∫ π xy
2
A1
a
36
dx = ⌠
⎮ π 3 dx
⌡2 x
M1
a
⎡ 18 ⎤
⎛ 9 18 ⎞
=π ⎢ − 2 ⎥ =π ⎜ − 2 ⎟
⎣ x ⎦2
⎝2 a ⎠
x=
A1
2
∫ π xy dx
∫π y
2
dx
M1
⎛ 9 18 ⎞
π⎜ − 2⎟
2 a ⎠ 3(a 3 − 4a)
= ⎝
=
⎛ 3 12 ⎞
a3 − 8
π⎜ − 3⎟
⎝2 a ⎠
E1
6
M1
Since a > 2, 4a > 8
(ii)
so a 3 − 4a < a3 − 8
Hence x =
3(a3 − 4a )
a3 − 8
A1
<3
E1
i.e. CM is less than 3 units from O
OR As a → ∞ , x =
3(1 − 4a −2 )
1 − 8a
−3
→3
Since x increases as a increases,
x is less than 3
M1A1
E1
63
Condone ≥ instead of >
throughout
3 Fully acceptable explanation
Dependent on M1A1
Accept x ≈
3a3
→ 3 , etc
a3
( M1 for x → 3 stated, but A1
requires correct justification )
4763
Mark Scheme
January 2010
4763 Mechanics 3
1(a)
(i)
[ Density ] = M L−3
B1
[ Kinetic Energy ] = M L2 T −2
B1
2
[ Power ] = M L T
−3
( Deduct B1 for kg m −3 etc)
B1
3
(ii)
2
ML T
−3
−1 2
B1
For [ v ] = L T −1
Can be earned in (iii)
M1
Obtaining the dimensions of η
= [η ]L ( L T )
−1
[η ] = M L T
−1
A1
3
(iii)
M L2 T −3 = ( M L−3 )α Lβ ( L T −1 )γ
α =1
−3 = −γ
γ =3
2 = −3α + β + γ
β =2
B1 cao
M1
A1
M1
A1
A1
Considering powers of T
(No ft if γ = 0 )
Considering powers of L
Correct equation (ft requires 4 terms)
(No ft if β = 0 )
6
M1
(b)
EE at start is
EE at end is
1
2
1
2
1
2
k × 0.82
2
k × 0.3
k × 0.82 = 12 k × 0.32 + 5.5 × 9.8 × 3.5
Stiffness is 686 N m −1
A1
Calculating elastic energy
λ
λ
k may be
or
A1
M1
F1
Equation involving EE and PE
(must have three terms)
A1
( A0 for λ = 823.2 )
l
1.2
6
[18]
42
4763
2
(a)
Mark Scheme
January 2010
a


2
2
2
 π x y dx =  π x(a − x ) dx

0
=π 

1
2
a
a 2 x 2 − 14 x 4 
0
= 14 π a 4
x=
=
1π
4
2π
3
3
a
8
a
M1
Limits not required
A1
For
1
2
a 2 x 2 − 14 x 4
A1
4
M1
a3
E1
5
(b)
(i)
4

Area is  (2 − x ) dx
1
4
3 

=  2 x − 23 x 2 

1
(=
4
)
3
M1
Limits not required
A1
For 2x − 23 x 2
M1
Limits not required
A1
For x 2 − 52 x 2
3
4


 x y dx =  x(2 − x ) dx

1
4
5


=  x 2 − 52 x 2 

1
x=
13
4
5
=
3
(=
13
)
5
39
= 1.95
20
4
1 2

 2 y dx = 

1
1
2
A1
M1
(2 − x ) 2 dx
4
3


=  2 x − 43 x 2 + 14 x 2 

1
y=
5
12
4
3
=
5
5
= 0.3125
16
5 A2
(=
)
12

2
 (2 − x ) dx

3

or  (2 − y ) 2 − 1 y dy

(
For 2x − 43 x 2 + 14 x 2 or
)
3
2
y 2 − 43 y 3 + 14 y 4
Give A1 for two terms correct, or all
correct with ½ omitted
A1
9
(ii)
TC × 3 − W × 0.95 = 0
M1
A1
Moments equation (no force omitted)
Any correct moments equation
(May involve both TA and TC )
Accept Wg or W = 43 , 43 g here
TA + TC = W
M1
Resolving vertically (or a second
moments equation)
A1
Accept 0.68W , 0.32W
Taking moments about A
TA =
41
W,
60
TC =
19
W
60
4
[18]
43
4763
Mark Scheme
3 (i) By conservation of energy,
1
2
2
2
× 0.6 × 6 − × 0.6 v = 0.6 × 9.8(1.25 − 1.25cos θ )
1
2
M1
A1
January 2010
Equation involving KE and PE
36 − v 2 = 24.5 − 24.5cos θ
v 2 = 11.5 + 24.5cos θ
E1
3
(ii)
For acceleration
Substituting for v 2
T = 5.52 + 17.64 cos θ
M1
A1
String becomes slack when T = 0
M1
v2
1.25
T − 5.88cos θ = 0.48(11.5 + 24.5cos θ )
T − 0.6 × 9.8cos θ = 0.6 ×
(iii)
v2
r
M1
A1
cos θ = −
5.52
17.64
( θ = 108.2° or 1.889 rad )
5.52
v = 11.5 − 24.5 ×
17.64
2
4
A1
May be implied
M1
or 0.6 × 9.8 ×
5.52
v2
= 0.6 ×
17.64
1.25
or −0.6 × 9.8 ×
Speed is 1.96 m s −1 (3 sf)
v 2 − 11.5
v2
= 0.6 ×
24.5
1.25
A1 cao
4
(iv)
T1 cos θ = mg
T1 ×
1.2
= 0.6 × 9.8
1.25
(where θ is angle COP)
Tension in OP is 6.125 N
mv 2
0.35
0.35
0.6 × 1.42
6.125 ×
+ T2 =
1.25
0.35
T1 sin θ + T2 =
Tension in CP is 1.645 N
M1
A1
Resolving vertically
A1
M1
Horizontal equation (three terms)
F1B1
For LHS and RHS
A1
7
[18]
44
4763
Mark Scheme
4(i)
TAP
TBP
7.35
=
× 0.05 ( = 0.245 )
1.5
7.35
=
× 0.5 ( = 1.47 )
2.5
M1
Using Hooke’s law
A1
or
A1
Resultant force up the plane is
January 2010
7.35
(AP − 1.5)
1.5
7.35
(2.05 − AP)
or
2.5
M1
TBP − TAP − mg sin 30°
= 1.47 − 0.245 − 0.25 × 9.8sin 30°
= 1.47 − 0.245 − 1.225
=0
Hence there is no acceleration
E1
Correctly shown
5
(ii)
7.35
(0.05 + x) ( = 0.245 + 4.9 x )
1.5
7.35
=
(4.55 − 1.55 − x − 2.5)
2.5
= 2.94(0.5 − x)
= 1.47 − 2.94 x
TAP =
B1
TBP
M1
(iii)
TBP − TAP − mg sin 30° = m
d2 x
dt 2
(1.47 − 2.94 x) − (0.245 + 4.9 x ) − 1.225 = 0.25
d2 x
dt 2
d2 x
dt 2
2π
31.36
=
3
M1
Equation of motion parallel to plane
A2
Give A1 for an equation which is correct
apart from sign errors
E1
Must state conclusion. Working must be
fully correct (cao)
If a is used for accn down plane, then
a = 31.36 x can earn M1A2; but E1
requires comment about directions
5π
Accept
= −31.36 x
Hence the motion is simple harmonic
Period is
E1
2π
5.6
Period is 1.12 s (3 sf)
B1 cao
14
5
x = −0.05cos 5.6t
A1
v = 0.28sin 5.6t
−0.2 = 0.28sin 5.6t
OR 0.22 = 31.36(0.052 − x 2 )
For A sin ωt or A cos ωt
Allow ±0.05sin/ cos 5.6t
Implied by v = ±0.28sin/ cos 5.6t
M1
Using v = ±0.2 to obtain an equation for t
M1
Fully correct strategy for finding the
required time
M1
(iv)
x = (±) 0.035
0.035 = −0.05cos 5.6t
5.6t = π + 0.7956
Time is 0.703 s (3 sf)
M1
A1cao
5
[18]
45
GCE
Mathematics (MEI)
Advanced GCE 4763
Mechanics 3
Mark Scheme for June 2010
Oxford Cambridge and RSA Examinations
4763
1(a)(i)
Mark Scheme
June 2010
M1
AP  2.42  0.7 2  2.5
Tension T  70  0.35 (  24.5 )
A1
M1
Resultant vertical force on P is 2T cos   mg
2.4
 2  24.5 
 4.8  9.8
2.5
 47.04  47.04  0
Hence P is in equilibrium
Attempting to resolve vertically
2.4
For T 
( or T cos16.3 etc )
2.5
For 4.8  9.8
B1
B1
E1
Correctly shown
6
(ii) EE  1  70  0.352
2
M1
A1
Elastic energy is 4.2875 J
(iii) Initial KE  1  4.8  3.52
2
(M0 for
1
 70  0.35 )
2
2 Note If 70 is used as modulus instead
of stiffness: (i) M1A0M1B1B1E0
(ii) M1 A1 for 1.99
B1
By conservation of energy
4.8  9.8h  2  4.2875  12  4.8  3.52
M1
F1
Equation involving EE, KE and PE
A1
(A0 for 0.8)
47.04h  8.575  29.4
Height is 0.807 m (3 sf)
4
(b)(i) [ Force ]  M L T 2
B1
[ Stiffness ]  M T 2
ft is
2  (ii)  29.4
47.04
Deduct 1 mark if units are used
B1
2
(ii)
L T 1  M (M T 2 )  L
 1
B1

0  
1
2
 
B1
M1
1
2
A1
2
Considering powers of M
When [Stiffness] is wrong in (i), allow
4 all marks ft provided the work is
comparable and answers are non-zero
4763
2 (i)
Mark Scheme
R cos   mg [  is angle between OB and vertical ]
R  0.8  0.4  9.8
M1
Normal reaction is 4.9 N
A1
June 2010
Resolving vertically
A1
3
(ii)
R sin   m
v2
r
4.9  0.6  0.4 
v2
1.5
v2
or r  2
r
M1
For acceleration
A1
or 4.9  0.6  0.4  1.5  2
v 2  11.025
(iii)
Speed is 3.32 m s 1 (3 sf)
A1
By conservation of energy
1
mu 2  mg  2.5
2
M1
A1
Equation involving KE and PE
R  mg  m 
M1
Vertical equation of motion
(must have three terms)
R  3mg
E1
u 2  5g
(u 7)
u2
2.5
R  mg  2mg
(iv)
(v)
1
2
ft is 1.5 R
3
mv 2  mg  2.5cos 
Correctly shown
4 or R  11.76 and 3  0.4  9.8  11.76
Mark (iv) and (v) as one part
Equation involving KE, PE and an
angle (  is angle with vertical)
M1
A1
v 2  5 g cos 
[ 12 mv 2  mgh can earn M1A1, but
only if cos   h / 2.5 appears
somewhere ]
R  mg cos   m 
v2
2.5
M1
Equation of motion towards O
(must have three terms, and the weight
must be resolved)
M1
M1
Obtaining an equation for v
Obtaining an equation for 
These two marks are each dependent
on M1M1 above
When R  2mg (  7.84 ) ,
mv 2
2.5
mv 2

2.5
 0.16v 2
98

3
2mg  mg cos  
mv 2
5
7.84  0.08v 2
2mg 
v2
Speed is 5.72 m s 1 (3 sf)
cos  
v2 2

5g 3
A1
(   48.2 or 0.841 rad )
Tangential acceleration is g sin 
Tangential acceleration is 7.30 m s
[ g sin  in isolation only earns M1 if
the angle  is clearly indicated ]
M1
2
A1
(3 sf)
8
3
4763
3 (i)
Mark Scheme
5
 1
Volume is     dx
1  x 
June 2010
M1
 may be omitted throughout
Limits not required
A1
For 
M1
Limits not required
A1
For ln x
2
5
4
 1 
   
(  )
x
5

1
5


2
  x y dx   

1
1
x
2
1
x   dx
x
5


(   ln 5 )
   ln x 

1
x
(ii)
 ln 5 5ln 5

4
4
5
( 2.012 )
A1
5
1
Area is 
 dx
1 x
SR If exact answers are not seen,
deduct only the first A1 affected
5
M1
Limits not required
A1
For ln x
M1
Limits not required
5


  ln x 
(  ln 5 )

1
5


 x y dx  

1
x
4
ln 5
5
 
1
x   dx (   x   4 )
x
 
 1
A1
( 2.485 )
5
M1
 1 
For    dx
 x 
2
 1 

( )

x
2
5

1
A1
For 
2
5ln 5
A1
 11
1 2
 2 y dx   2   dx

1  x 
2
2
5
y
(iii)
2
5
ln 5

( 0.2485 )
1
2x
7
4 
 2
,
CM of R2 is 

 5ln 5 ln 5 
Do not penalise inexact answers in this
part
B1B1 ft
2
(iv)
 4 
 2 
(ln 5) 
 (ln 5) 
 (1) 12
ln 5 
5ln 5 


x
ln 5  ln 5  1
4.9
4.9 

,
CM is 
( 1.161 , 1.161 )

 2 ln 5  1 2 ln 5  1 
 
4
B1
For CM of R3 is (
M1
(one coordinate is sufficient)
Using  mx with three terms
M1
Using
 mx
m
in each sum
A1 cao
4
1
2
,
1
2
)
with at least two terms
4763
4 (i)
Mark Scheme
dx
 A cos t  B sin t
dt
d2 x
a  2   A 2 sin t  B 2 cos t
dt
  2 ( A sin  t  B cos  t )   2 x
v
June 2010
B1
Finding the second derivative
M1
Correctly shown
E1
3
(ii)
B  16
B1
  0.25
B1
B2
A  30
When A is wrong, give B1 for a correct
4 equation involving A [e.g. A  7.5 or
7.52   2 ( A2  B 2  162 ) ] or for
A  30
(iii)
Maximum displacement is ()
Or 7.52   2 (amp 2  162 )
Or finding t when v  0 and
substituting to find x
M1
A2  B 2
A1
Maximum displacement is 34 m
Maximum speed is () 34
Maximum acceleration is () 34 2
M1
For either (any valid method)
Maximum speed is 8.5 m s 1
F1
Only ft from   amp
F1
Only ft from  2  amp
Maximum acceleration is 2.125 m s
2
5
(iv)
v  7.5cos 0.25t  4sin 0.25t
When t  15 , v  7.5cos 3.75  4sin 3.75
 8.44
M1
Speed is 8.44 m s 1 (3 sf); downwards
A1
2
(v)
2
 25 s ,

so t  0 to t  15 is less than one period
Period
When t  15 , x  30sin 3.75  16 cos 3.75
 4.02
M1
Distance travelled is 16  34  34  4.02
M1
M1
Distance travelled is 88.0 m (3 sf)
A1 cao
Take account of change of direction
Fully correct strategy for distance
4
5
GCE
Mathematics (MEI)
Advanced GCE
Unit 4763: Mechanics 3
Mark Scheme for January 2011
Oxford Cambridge and RSA Examinations
4763
Mark Scheme
1(a)(i) [ Force ]  M L T 2
B1
[ Density ]  M L3
B1
[ Angular speed ]  T 1
B1
January 2011
Deduct one mark if given as kg m s 2
etc
3
(ii)
[ Breaking stress ] 
M L T 2
M1
L2
 M L1 T 2
For [ Force ]  L2
E1
2
(iii)
1
1.2 109 
 0.0254  0.0012
0.454
 67.1 lb in 1 ms 2
(iv)
M1
For 0.0012 or 
M1
For 
A1
(3 sf)
1
2
 
0.0254
0.454
All marks ft provided work is
comparable
B1
1
2
0.0012
8
3 5.6  10 implies M1M1A0
2.15  1016 implies M1M0A0
T 1  ( M L1 T 2 ) ( M L3 )  L

1
B1
0    3  
  1
Considering powers of L
M1
A1
4
(v)
1
3140  k (1.2  109 ) 2 (7800)
k  4.00
S
1
2
(0.5) 1
(3 sf)
2r2

81202  2700  0.22
42
k2
 4.44  108 Pa (3 sf)
M1
M1
Obtaining equation for k
Obtaining numerical value for k
M1
Obtaining equation for S
A1 cao
4
OR
2
 8120  2700  0.2 
S  1.2  109  

 

 3140  7800  0.5 
2
M1M1M1
 4.44  108
(vi)
1
A1
1
  4(630) 2 (70) 2 (15) 1
Obtaining equation for 
M1
A1 cao
 0.8
2
5
4763
2(a)(i)
Mark Scheme
V2
(  is angle APC )
r
8.4
3.52
T
 48 
30
8.4
T cos   m
January 2011
V2
r
M1
Equation of motion including
A1
Or T cos 73.7  ... or T sin16.3  ...
A1
Tension is 250 N
Resolving vertically (three terms)
M1
T sin   R  mg
250  0.96  R  48  9.8
A1
Normal reaction is 230.4 N
5
T sin   mg
(ii)
T  0.96  48  9.8
T  490
V2
r
V2
490  0.28  48 
8.4
V  4.9
M1
A1
Vertical equation with R  0
Or T sin 73.7  ... or T cos16.3  ...
M1
Obtaining equation for V
T cos   m
A1
M1
A1
(b)(i)
1
2
2
Allow T=490 obtained in (i) and used
4 correctly in (ii) for full marks
2
m(v  u )  m  9.8(2.5  2.5cos  )
Equation involving KE and PE
v 2  u 2  49(1  cos  )
v 2  u 2  49  49 cos 
E1
3
(ii)
v2
r
M1
A1
Radial equation (three terms)

48v 2
  R 
2.5

M1
Obtaining equation in R, u, v
mg cos   R  m
 u 2  49  v 2
48  9.8 

49

9.6u 2  470.4  9.6v 2  R  19.2v 2
R  470.4  9.6u 2  28.8v 2
A1
4
(iii)
u  1.63
Substituting R  0 and v  4.15
or other complete method leading to an
equation for u
M1
470.4  9.6u 2  28.8  4.152  0
(3 sf)
A1
2
6
(ft requires 0  u  4.15 )
4763
3 (i)
Mark Scheme
Tension is 180(10  7)
 540 N
January 2011
Using T  k  extension
M1
A1
2
(ii)
M1
A1
4  540  T  200  9.8
T  200
200
Extension is
(  2.5 )
80
Natural length is 5.5 m
Resolving vertically
M1
A1 cao
4
80(2.5  x)  200  9.8  4 180(3  x)  200
200  80 x  1960  2160  720 x  200
d2 x
dt 2
(iv)
For 180(3  x) or 80(2.5  x)
Equation of motion (condone one
missing force)
B1 ft
(iii)
M1
d2 x
A1
dt 2
d2 x
dt 2
 4 x
E1
4
M1
Maximum acceleration is  2 A
 4  2.2  8.8 m s 2
Condone 8.8
A1
2
(v)
When x  1.6 , v 2   2 ( A2  x 2 )
 4(2.22  1.62 )
(vi)
Speed is 3.02 m s 1 (3 sf)
A1
x  2.2 cos 2t
B1
When t  5, x  1.846
M1
Period is
2

  , 5 s is
5

Using v 2   2 ( A2  x 2 )
(or other complete method)
(Allow M1 if  2  2 or 16 used
but M0 if x  3.8 is used)
M1
2 Condone 3.02
Condone x  2.2 cos 2t
This B1 can be earned in (v)
Obtaining x when t  5
(from x  A cos t or x  A sin t )
 1.6 periods
M1
A1
Distance travelled is 6  2.2  (2.2  1.846)
 13.6 m (3 sf)
Correct strategy for finding distance
4
7
4763
4 (a)
Mark Scheme
January 2011

For  ( x 2  k 2 ) dx

4k

Volume is   y 2 dx    ( x 2  k 2 ) dx
k
 

1
3
4k
x3  k 2 x 
k
M1
(  18 k 3 ) A1
For
1
3
M1
For
 xy
A1A1
For
1
4
M1
Dependent on previous M1M1
4k

k
2
3
2
  xy dx    ( x  k x) dx
 

x
1
4
4k
x 4  12 k 2 x 2 
k
(
225 k 4
)
4
225
 k4
4
3
18 k
25k

 3.125k
8
(b)(i)
2a

Area is 
0
x3
a2
dx
x4
a2
2a

32
5
a3
4a

2
(
32a3
)
5
8a
 1.6a
5
2a
1
2

y dx  
0
2
x6
2a 4
2a
64
7
a3
4a
2

x 4 and  12 k 2 x 2
M1
 x3
For  2 dx
 a
A1
For
M1
For
A1
For
x4
4a 2
 xy dx
x5
5a 2
A1
dx
 x7 

4 
 14a  0
y
(  4a 2 )
dx
 x5 
 2 
 5a  0
x
dx
7
2a
2a
2
A1
 x4 

2 
 4a  0

 xy dx  0
x3  k 2 x
(
64a 3
)
7
16a
7
M1
For
A1
For
y
2
dx or
x7
14a
 (2a  x) y dy
2
4
or ay 2  73 a 3 y
A1
8
(ii) Centre of mass is vertically below A
2a
2a  x
5
 40
tan  
(  0.07 )
8a  y
a
7
Angle is 4.00 (3 sf)
M1
May be implied
M1
Condone reciprocal
A1
3
8
7
3
GCE
Mathematics (MEI)
Advanced GCE
Unit 4763: Mechanics 3
Mark Scheme for June 2011
Oxford Cambridge and RSA Examinations
4763
1 (i)
Mark Scheme
dx
 A cos t
dt
d2 x
  A 2 sin t
dt 2
  2 ( A sin t )   2 x
June 2011
B1
M1
Obtaining second derivative
E1
2
 dx 
2 2
2
2 2
2
 dt   A  cos t  A  (1  sin t )
 
Using cos 2 t  1  sin 2 t
M1
  2 ( A2  A2 sin 2 t )   2 ( A2  x 2 )
E1
5
(ii)
M1
1.22   2 ( A2  0.7 2 )
A1
0.752   2 ( A2  22 )
A2  0.49
2

Using v 2   2 ( A2  x 2 )
Two correct equations
(M0 if x  7.3 used, etc)
1.22
0.752
A 4
A2  0.49  2.56( A2  4)
M1
Eliminating 
or Eliminating A
or Substituting A  2.5 into both
equations
E1
Correctly shown
M1
Using
9.75  1.56 A2
A2  6.25
Amplitude is 2.5 m
1.44   2 (2.52  0.7 2 )
  0.5
Period is
2
2
 0.5
 4  12.6 s

2

A1
(3 sf)
6
(iii) Maximum speed is A  1.25 m s 1
B1
(iv) Magnitude is 0.52  1.6
 0.4 m s 2
Direction is upwards
M1
ft only if greater than 1.2
1
A1
B1
Accept 0.4
B0 for just ‘towards centre’
3
(v)
x  2.5sin(0.5t )
When x  2, t  1.855 (or 10.71)
When x  1, t  0.823 (or 13.39)
B1
Time taken is 0.823  (1.855)
M1
 2.68 s
or x  2.5cos(0.5t )
or t  ( ) 4.996
or t  ( ) 2.319
Correct strategy for finding time
(must use radians)
(ft is 1.3388 /  )
A1
(3 sf)
3
5
4763
Mark Scheme
June 2011
2(a)(i)
0.6  0.2  9.8  0.2 
For acceleration
M1
A1
2
u
3.2
Speed is 6.4 m s 1
u2
3.2
A1
3
(ii) (A)
1
m(v 2  u 2 )  m  9.8(3.2  3.2 cos 60)
2
M1
A1
Equation involving KE and PE
A1
( ft is 29.4 
M1
M1A0 for g cos 60
v 2  135.04
Radial component is
v2
 42.2 m s 2
3.2
Tangential component is g sin 60
 8.49 m s 2
(B)
A1
(3 sf)
T  mg cos 60  m a
T  0.2  9.8cos 60  0.2  42.2
Tension is 9.42 N
M0 for mg sin 60
5 If radial and tangential components
are interchanged, withhold first A1
M1
Radial equation (three terms)
(Allow M1 for T  mg  ma )
This M1 can be awarded in (A)
A1
A1 cao
ft dependent on M1 for energy in (A)
3
(b)(i)
T cos 36  0.75sin 36  0.2  9.8
M1
Tension is 1.88 N
A1
(3 sf)
u2
)
3.2
SC If 60 with upward vertical,
(A) M1A0A0 M1A1 (8.49)
(B) M1A1A1 (3.54)
Resolving vertically (three terms)
Allow sin/cos confusion, but both T and
R must be resolved
2
(ii)
Angular speed  
2
(  3.491 )
1.8
2
 2 
T sin 36  0.75cos 36  0.2 r 

 1.8 
r  0.204
r
Length of string is
sin 36
 0.347 m
(3 sf)
Or v 
M1
A1
Horiz eqn involving r  2 or v 2 / r
Equation for r (or l )
M1
Dependent on previous M1
A1 cao
5
6
2 r
1.8
B1
4763
3 (i)
Mark Scheme
Elastic energy is
573.3
 0.92
3.9
= 59.535 J
1
2

June 2011
Allow one error
M1
A1
(Allow 60
A0 for 59)
2
(ii)
Length of string at bottom is 2 1.82  2.42
1 573.3

 (2.12  0.92 )  m  9.8 1.8
2
3.9
324.135  59.535  17.64m
(6)
M1
Finding length of string (or half-string)
M1
B1B1
Equation involving EE and PE
For change in EE and change in PE
E1
Mass is 15 kg
5
(iii)
Length of string is 2 1.02  2.42  5.2
573.3
Tension T 
 1.3 (  191.1 )
3.9
1.0
 15  9.8
2T sin   mg  2  191.1
2.6
 147  147
 0, hence it is in equilibrium
M1
A1
Finding tension (via Hooke’s law)
M1
Finding vertical component of tension
Give A1 for T  191.1 obtained from
resolving vertically
E1
4
(iv)
[ 8 2 h3 ]  L3 , [ 8h3  ad 2 ]  L3
So
8 2 h3
8h3  ad 2
Condone ‘ L3 / L3  0 , dimensionless’
E1
is dimensionless
But E0 for
1
B1
(v)


T  M L (MLT
 
L3  L3

L3
0
)
B1
1
2
    0, so  
L3
For [  ]  M L T 2
2 
    0, so  
SC If 573.3 is used as stiffness:
(i) M1A0 (ii) M1M1B0B1E0
(iii) M1A1 (745.29) M1E0
1
2
1
2
If  is wrong but non-zero,
give B1 ft for     
B1
B1
4
(vi)
a  3.9,   573.3, d  4.8, h  2.6, m  15
Using formula with numerical  ,  , 
(must use the complete formula)
M1
2 3
Period is
8 h
3
8h  ad
2
m
1
2
a
1
2

 12
 1.67 s (3 sf)
A1 cao
2
7
4763
4 (i)
Mark Scheme
Area is
3
0
3
 xy dx  0


x
3
x3  5 x 
0
(  24 )
( x 3  5 x ) dx
1
4
2
3
x 4  52 x 2 
0
3
1
0 2
dx  


(ii)
1
3
(
171
)
4
42.75 57

 1.78125
24
32
 12 y
y
M1
( x 2  5) dx


1
10
June 2011

A1
For
1
3
M1
For
 xy dx
A1
For
1
4
M1
For
y
A2
For
1
10
25
2
3
x
0
(  106.8 )
106.8 89

 4.45
24
20
x 4  52 x 2
2
dx
x5  53 x3  25
x
2
9
M1

For  ( y  5) dy

A1
For 

M1
For
x
A1
For
1
3
M1
Dependent on previous M1M1
14
 

1
2
14
y2  5 y 
5
(  40.5 )
14

5
2
2
  x y dy    ( y  5 y) dy
1
3
x3  5 x
Give A1 for two terms correct
A1

Volume is   x 2 dy    ( y  5) dy
5
 

( x 2  5) dx
A1
( x 4  10 x 2  25) dx
x5  53 x3 
For
14
y 3  52 y 2 
5
(  445.5 )
445.5
40.5
 11
y
1
2
2
14
y2  5 y 
5
y dx
y 3  52 y 2
A1
6
(iii) Volume of whole cylinder is   32  14  126
Using formula for composite body
M1
A1
126  7  40.5  11  (126  40.5 )  y A
126  7  40.5  11
126  40.5
97

 5.105 (4 sf)
19
yA 
A1 cao
3
8
4763
Mark Scheme
Question
1 (iv)
Answer


(v)
Guidance
2 
T  M L (M L T )
   12
    0,     0
  12 ,   12
1
Marks
June 2012
1
1
B1
CAO
M1
Considering powers of M or L
FT    ,    (provided non-zero)
A1A1
[4]
0.718  k (8) 2 (0.4) 2 (125)
k  4.4875
1
Obtaining equation for k
Or using ratio and powers
M1
Obtaining expression for new time
 75  2  3  2  20 
Or     
 

 8   0.4   125 
A1
CAO
No penalty for using b  1.2 and b  9
M1
2
1
t
1
(4.4875)(75) 2
1
1
(3) 2 (20) 2
New time is 15.1 s (3 sf)
2
(a)
(i)
R cos18  800  9.8
R sin18  800 
( R  8243 )
v2
45
v2
45  9.8
Speed is 12.0 ms 1
[3]
M1
M1
Resolving vertically
Horizontal equation of motion
1
1
Might also include F
Might also include F
A1
tan18 
2
(a)
(ii)
(3 sf)
R cos18  F sin18  800  9.8
A1
[4]
M1
A1
M1
152
R sin18  F cos18  800 
45
A1
Frictional force is 1380 N (3 sf)
Normal reaction is 8690 N (3 sf)
M1
A1
A1
[7]
Resolving vertically (three terms)
Horizontal equation (three terms)
Obtaining a value for F or R
6
Dependent on previous M1M1
2
4763
Mark Scheme
Question
2 (b)
Answer
1 m(7 2
2
 2.82 )  mg (a  a cos  )
June 2012
Marks
M1
Equation involving KE and PE
A1
Correct equation involving a and θ
M1
Radial equation of motion
A1
Correct equation involving a and θ
Eliminating  or a
x   A sin(t   )
M1
A1
A1
[7]
B1

x   A 2 cos(t   )
M1
Obtaining second derivative

x   ( x  c)
E1
[3]
B1
B1
Correctly shown
M1
Accept A  6
2
Using

A1
Accept   
Guidance
a(1  cos )  2.1
mg cos  m 
a cos  0.8
2.8
a
2
Length of string is 1.3 m
Angle with upward vertical is 52.0 (3 sf)
3
(i)
2
3
(ii)
c  10
A6
2
 10



5
x  16 when t  3  3    0

3
5
M1
h  2.1 implies M1
a is length of the string
(Might use angle with downward
vertical or horizontal)
Might also involve T
Dependent on previous M1M1
A0 for 128° or 38°

5
Obtaining simple relationship between 
and ω .
NB   3 is M0
A1
NB other values possible
If exact values not seen, give A0A1 for both
  0.63 and   1.9
[6]
Max 5/6 if values are not consistent
7
Allow one error
Or other complete method for
finding ω
2
Allow
etc
10
Or x  10  6cos{

5
(t  3) }
7
13
, 
,
5
5

8
x  10  6cos( t 
) etc
5
5
e.g.   
4763
Question
3 (iii)
3
(iv)
Mark Scheme
Answer
Maximum speed is A
6
or 3.77 ms 1 (3 sf)
Maximum speed is
5
When t  0 , height is 8.15 m (3 sf)
v
6
 t 3
sin(
)

5
5
5
When t  0 , velocity is 3.59 ms 1
3
(v)
When t  0, x  8.146
June 2012
Marks
Guidance
Or e.g. evaluating x when t  5.5
M1
A1
FT is A
[2]
B1
FT is c  A cos 
(must be positive)
(provided 4  x  16 )
Must use radians
2
M1
(3 sf)
A1
[3]
 
Or v 2    (62  1.8542 )
5
FT is A sin  (must be positive)
When t  14, x  14.854
M1
Finding x when t  14
(16  8.146)  12  12  (16  14.854)
Distance is 33 m
M1
M1
A1
[4]
(16  14.854) used
Fully correct strategy
CAO
8
Allow one error in differentiation
(   3 gives x  4.06, v  0.532 )
Correct (FT) value, or evidence of
substitution, required
(   3 gives x  15.3 )
Requires 4  x(14)  16
Also requires 4  x (0)  16
4763
Question
4 (a)
Mark Scheme
Answer
June 2012
Marks
Guidance
9

A   (3  x ) dx
0
M1
9
3


  3x  23 x 2 

0
(9)
3
A1
For 3x  23 x 2
M1
For
 xy dx
A1
For
3
2
9

A x   xy dx   x(3  x ) dx
0



x
9
3
2
5

x 2  52 x 2 
0
(  24.3 )
24.3
 2.7
9
A1
9

A y   12 y 2 dx   12 (3  x )2 dx
0



y
9
2
x
3
2x 2
6.75
 0.75
9
5
x 2  52 x 2
M1
For  ... y 2 dx
M1
Expanding (three terms) and integrating
(allow one error)
A1
For
9

 x 
0
1
4
2
(  6.75 )
A1
[9]
9
9
2
3
x  2 x 2  14 x 2
(3)

Or  (3  y ) 2 y dy
(0)
Or
9
2
y 2  2 y 3  14 y 4
4763
Mark Scheme
Question
4 (b) (i)
Answer
Marks
5

V    (25  x 2 ) dx
2
Guidance
M1
For  ... (25  x 2 ) dx
A1
For 25x  13 x3
M1
For
 xy
A1
For
25
2
A1
Accept 3.1 from correct working
5
   25 x  13 x3  (  36 )

2
5

V x    xy 2 dx    x(25  x 2 ) dx
2
5
441
   25
x 2  14 x 4  ( 
)
2

2
4
x
441 
4
36

49
16
(  3.0625 )
June 2012
2
dx
x 2  14 x 4
[5]
4
(b)
(ii)
sin  sin 25

5
x
  43.6
M1
CG is vertical (may be implied)
M1
Using triangle OGC or equivalent
Lenient, if CG drawn.
Needs to be quite accurate if CG
not drawn
M1
A1
Accept art 43° or 44° from correct work
 2.113 
FT is sin 1 

 x 
[4]
10
Provided 2.113  x  5
4763
Mark Scheme
Question
1 (a)
Answer
A  5.1
4.52   2 ( A2  62 )
A1
4.52  5.12  36 2
  0.4
2
Period (
) is 5  15.7 s (3 sf)
M1
Amplitude (A) is 12.75 m
A1
[6]
[ F ]  M L T 2
B1

1 (b)
(i)
 Fd 2
[G ] 
 m1m2
 M L T 2 L2

M2

 M 1 L3 T 2
1 (b)
(ii)
Marks
B1
M1
Using v 2   2 ( A2  x 2 )
Guidance
Eliminating A or ω
Allow 5π
A1
M1
January 2013
Obtaining dimensions of G
A1
[3]
T 1  (M 1 L3 T 2 ) M  L
  12
    0
  12
3    0
M1

A1
 32
B1
Considering powers of M
A1
M1
Considering powers of L
All marks FT from wrong [G] if
comparable. No FT within part (ii).
[5]
5
4763
Mark Scheme
Question
1 (b) (iii)
Answer
Marks
M1M1
 4.86  1014
  2.0  106  
 2500
OR
1
1
 2  30000  2
  

  50 
1
2.0  106  k  G 2  2500 2  50
A1
 4.86  10
For 
 2500 


 30000 
and 

 50 
3
2
Requires   0,   0
2
1
  1.414  105  (4.86  1014 ) 2  30000
Correct equation for ω
FT if comparable
M1 Requires   0 or   0
3
1
3
2
Angular speed is 6.0  105 rad s 1
(i)
A1
[4]
M1
1 mv 2
2
2
Guidance
1
14   2
3
kG 2  1.414  105
2 (a)
January 2013
 12 m(1.2)2  mg (0.8  0.8cos 16  )
M1 Requires   0 and   0
Condone the use of any value for G
(including G  1 )
A1 Correct equation for ω
FT if comparable
CAO
Equation involving initial KE, final KE
and attempt at PE
A1
v  3.5407
 v2 
Radial component 
is 4.43 ms 2 (3 sf)
 0.8 


A1
() mg sin 16   maT
M1
Tangential component is 4.9 ms 2
A1
[5]
Allow M1 for cos 16  used instead of
sin 16  ; but M0 for aT  mg sin 16 
Allow
6
1
2
g
4763
Question
2 (a) (ii)
Mark Scheme
Answer
1 mv 2
2
 12 m(1.2)2  mg (0.8  0.8cos  )
Marks
M1
M1
mg cos  R 
mv 2
0.8
Leaves surface when R  0
v2
v  1.44  2  9.8  0.8( 1 
)
7.84
2
2 (b)
January 2013
Guidance
θ between OP and upward vertical
Equation involving initial KE, final KE
Allow mgh for PE if h is linked to θ
and attempt at PE in general position
in later work
Equation involving resolved component
of weight and v 2 / r
A1
R may be omitted
M1
May be implied
e.g. Implied by mg cos  
M1
Obtaining equation in v or θ
Dependent on previous M1M1M1
107
 0.728
147
  0.756 rad or 43.3
ˆ  53.1,   SQC
ˆ  16.3
  RQC
mv 2
0.8
cos 
Speed is 2.39 ms 1
A1
[6]
TR sin   TS sin   mg
M1
Resolving vertically (three terms)
A1
Allow sin 53.1 , etc
M1
Horizontal equation of motion
Three terms, and v 2 / r
Obtaining TR or TS
Dependent on previous M1M1
0.8TR  0.28TS  0.9  9.8 (  8.82 )
2
v
r
52
0.6TR  0.96TS  0.9 
(  9.375 )
2.4
TR cos   TS cos   m
Tension in string RQ is 9.737 N
Tension in string SQ is 3.68 N
A1
M1
A1
A1
[7]
7
4763
Question
3 (i)
3 (ii)
Mark Scheme
Answer
Length of each string is 7.8 m
728
(7.8  6.4)
T
6.4
Tension is 159.25 N
2T cos   mg
5
2  159.25   m  9.8
13
122.5
m
 12.5 kg
9.8
Marks
B1
M1
January 2013
Guidance
Using Hooke’s law
Must use extension
M1
Resolving vertically
ˆ  67.4
  XPM
A1
FT
E1
Working must lead to 12.5 to 3 sf
A1
[3]
[3]
3 (iii)
New length of each string is 7.5 m
M1
728
(7.5  6.4) (  125.125 )
T
6.4
mg  2T cos   ma
12.5  9.8  2  125.125  0.28  12.5a
3 (iv)
Hooke’s law with new extension
A1
Acceleration is 4.19 ms 2 downwards (3 sf)
M1
A1
A1
[5]
Vertical equation of motion (3 terms)
FT for incorrect T
Some indication of downwards required
At maximum speed, acceleration is zero
Acceleration is zero in equilibrium position
B1
Mention of zero acceleration
[1]
8
Reference to v 2   2 ( A2  x 2 ) ,
SHM, etc, will usually be B0
4763
Question
3 (v)
Mark Scheme
Answer
Change of PE is 12.5  9.8  3 (  367.5 )
728  0.82
(  72.8 )
2  6.4
728  1.42
(  222.95 )
Final EE is 2 
2  6.4
Initial EE is 2 
1 (12.5)v 2
2
 367.5  222.95  72.8
Maximum speed is 5.90 ms 1 (3 sf)
4 (a)
January 2013
Marks
B1
Guidance
B1
Allow one string (36.4)
B1
Allow one string (111.475)
M1
Equation involving KE, PE and EE
FT from any B0 above
All signs must be correct
CAO
A1
A1
[6]
h
1

V    ( y 4 ) 2 dy
0
1
M1
For  ... ( y 4 ) 2 dy
A1
For
2 32
y
3
M1
For
x
A1
For
2 52
y
5
h
2 3
 2 3 
   y2  (   h2 )
3
 3
0
h
1

V y    x 2 y dy    y 2 y dy
0
h
2 5
 2 5 
   y2  (   h2 )
5
 5
0
5
y
2  h2
5
2
3
3
h2
3
 h
5
A1
[5]
9
2
y dy
All terms must be non-zero
4763
Question
4 (b) (i)
Mark Scheme
Answer
Marks
4

A   ( x  x )dx
0
4
2 3 
1
  x2  x 2 
3
 2
0
(
40
)
3
Guidance
M1
For  ( x  x ) dx
A1
For
1 2 2 32
x  x
2
3
M1
For
 xy dx
A1
For
1 3 2 52
x  x
3
5
4

A x   xy dx   x( x  x ) dx
0
4
2 5 
1
  x3  x 2 
5
3
0
x
512
15
40
3

(
512
)
15
64
 2.56
25
E1
4

A y   12 y 2 dx   12 ( x  x ) 2 dx
0
4
2 5 1
1

  x3  x 2  x 2 
5
4
6
0
y
412
15
40
3

103
 2.06
50
January 2013
(
M1
412
)
15
A1A1
For  ... y 2 dx
For
A1
[9]
10
1 3 2 52 1 2
x  x  x
6
5
4
Give A1 for two correct terms
4763
Mark Scheme
Question
4 (b) (ii)
Answer
40 32
Area of B is 24 

3
3
x
2.56

 2
32   40 
24






 
3 y
3 2.06
 


 3
Marks
 x   1.3 
 

 y   4.175 
OR

 14


1
 32
or 

or

1 4 y 1

2
y dy or

 x(6  x 
 12 (6  x 
Guidance
M1
M1
CM of composite body
Correct strategy
A1
A1
CAO
FT requires 0  y  6
x )dx
4
1  4 y  1 dy
M1 For any one of these
x )(6  x  x ) dx
M1 For one successful integration
A1A1
x  1.3, y  4.175
January 2013
[4]
11
(One coordinate sufficient)
FT is 6.75  1.25 y A
No FT from wrong area
4763
Mark Scheme
Question
1 (a)
(i)
1
(a)
(ii)
Answer
(b)
(i)
Marks
T sin   mr 2
r  3.2sin 
M1
T sin   (1.5)(3.2sin  )(2.5) 2
A1
Tension is 30 N
A1
[4]
M1
Equation involving r 2 or l 2
B1
T cos   mg
30cos   1.5  9.8
Angle is 60.7° (3 sf)
1
June 2013
[ k ]  (M L T 2 ) L1  M T 2
T  (1.5)(3.2)(2.5) 2 with no wrong
working earns M1B1A1
Resolving vertically
A1
[2]
or 1.06 rad
E1
Can use u 
4kd 2

or k 
3m
l
[1]
1
(b)
(ii)



1
2 2
4kd 2   M T 2 L
1

  L T
3m  
M


 u   L T 1 , so eqn is dimensionally consistent
M1
Obtaining dimensions of RHS
E1
Condone circular argument
[2]
1
(b)
(iii)
2 

T  (M T ) L M
   12
 0
   0
  12

B1
B1
M1
Considering powers of M
A1
FT from wrong non-zero 
[4]
1
Guidance
All marks in (a) can be earned anywhere
in (i) or (ii)
4763
Mark Scheme
Question
1 (b)
(iv)
Answer
u
4  60  0.7 2
 2.8 ms 1
3 5
Elastic energy is
1 (5)(2.8) 2
2
1
2
 60  0.7 2 (  14.7 )
 12 (5)v 2  14.7
Speed is 1.4 ms 1
June 2013
Marks
Guidance
B1
M1A1
M1A0 if one error
Equation involving initial KE, final KE and
EE
M1
No FT in any part of Q1 except (iii)
A1
[5]
2
(i)
Equation involving initial KE, final KE and
PE
M1
1 m(8.4) 2
2
2
 12 mv 2  mg (a  a cos  )
v  70.56  19.6a (1  cos )
( m  0.25 )
A1
A1
M1
T  mg cos   m
v2
a
T  2.45cos   0.25(
T  2.45cos  
T
2
(ii)
Using acceleration
v2
a
A1
70.56
 19.6  19.6cos  )
a
M1
Equation relating T, a, θ
Dependent on previous M1M1
In terms of θ or when   
17.64
 4.9  4.9cos 
a
17.64
 7.35cos   4.9
a
E1
If a  0.9, T  14.7  7.35cos 
T  0 for all θ, so P moves in a complete circle
[7]
M1
E1
M1
Expression for T when a  0.9
Any correct explanation
Using   0 or   
Maximum tension is 14.7  7.35  22.05 N
Minimum tension is 14.7  7.35  7.35 N
A1
Both correct
[4]
2
4763
Question
2 (iii)
2
(iv)
Mark Scheme
Answer
If P just completes the circle,
T  0 when   
17.64
 7.35  4.9  0
a
a  1.44
June 2013
Marks
Guidance
M1
A1
Condone a  1.44 etc
For 1.44
If a 1.6, T  6.125  7.35cos 
String becomes slack when T  0
A1
[3]
M1
M1
6.125
5
  [   2.56 rad or 146 ]
7.35
6
2
v  70.56  19.6  1.6(1  56 )
M1
Obtaining an equation for v
Using expression for T when a  1.6
cos  
Or  mg ( 56 )  m
Speed is 3.61 ms 1 (3 sf)
Dependent on previous M1M1
v2
1.6
No FT in any part of Q2
A1
[4]
3
(i)
686(2.2  l )
 18  9.8
l
Natural length is 1.75 m
3
(ii)
686
(0.45  x)
1.75
 176.4  392x
145
x (  58 x )
Thrust in BP is
2.5
Tension in AP is
M1
Using Hooke’s law
A1
[2]
M1
E1
Allow 58x
B1
[3]
3
Condone thrust / tension confusion
4763
Mark Scheme
Question
3 (iii)
Answer
Marks
M1
18  9.8  (176.4  392 x)  58 x  18
176.4  176.4  450 x  18
d2 x
dt 2
3
d2 x
dt 2
Guidance
2 forces from (ii), mg and ma
Equation of motion
A1
Correct LHS equated to 18a
E1
Fully correct derivation
No FT
d2 x
dt 2
 25 x
[3]
2
 1.26 s (3 sf)
5
A  3.4
3.4
) is 0.68 m
Amplitude ( A 
5
(iv)
June 2013
B1
Period is
Allow
2
5
M1
A1
[3]
3
v  3.4cos5t
When t  2.4, v  2.87
(v)
M1
Magnitude of velocity is 2.87 m s 1 (3 sf)
Since v  0 the direction is downwards
OR
When t  2.4, x  0.3649
2
A1
[3]
M1 Using v   ( A  x )
x  0.68sin 5t
When t  2.4, x  0.3649
2.4
 1.91 periods (between 1¾ and 2)
2.4 s is
1.26
Distance is 8  0.68  0.3649
Distance is 5.08 m (3 sf)
‘Downwards’ is sufficient
2
(3 sf)
Magnitude of velocity is 2.87 m s
Between 1¾ and 2 periods; hence downwards
(vi)
Dependent on M1A1
Earns B1M1 from (vi)
2
1
3
cos 52  t is M0
A1
v  25(0.68  0.3649 )
2
Using cos t or sin t
B1
M1
2
2
2
A1
No FT
A1 Dependent on M1A1
FT (from wrong amplitude)
Must be justified
B1M1 can be earned in (v)
8 A  xt 2.4 with xt 2.4  0
M1
A1
[4]
FT is 7.463A
4
Strictly, only for this
4763
Mark Scheme
Question
4 (a)
(i)
Answer
Marks
Guidance
4

V    x 2 (4  x ) dx
0
4
1
 4

   x3  x 4 
4
 3
0
(
64
)
3
M1
For
 x
A1
For
4 3 1 4
x  x
3
4
M1
For
 xy
A1
For x 4 
M1
Dependent on previous M1M1
A1
[6]
M1
Taking moments
A1
FT Correct equation for required angle
A1
FT is tan 1
4

Vx    xy 2 dx    x3 (4  x ) dx
0
4
1 

   x 4  x5 
5 0

x
51.2
64 
3
 2.4
4
(a)
(  51.2 )
(ii)
W (2.4sin  )  W (4cos  )
tan  
June 2013
2
4 x

2
dx
π may be omitted throughout
dx
1 5
x
5
W (2.4cos  )  W (4sin  ) is A0 unless
  90   also appears
4
5

2.4 3
  59.0 (3 sf)
[3]
5
4
x
FT requires x  4
4763
Mark Scheme
Question
4 (b)
Answer
x  2
June 2013
Marks
1
y3
Guidance
B1
8
8
1
3 4 


A   (2  y 3 ) dy   2 y  y 3  (  28 )
4
0

0
FT
M1
For
B2
FT for 2 y 
8
1
2
 

Ax   12 x 2 dy   12  4  4 y 3  y 3  dy
0 

Or 32   14 ( x  2)4 
B1

x
2
49.6 62

 1.77 (3 sf)
28
35
x
Or
1 ( x  2)5
5
Give B1 for one minor slip in integration,
or if ½ omitted
Or
1
4
Or
1 x5
5
y
832
7
28

208
 4.24 (3 sf)
49
 12 ( x  2)4
1 ( x  2)5
x( x  2) 4  20
 32 x 4  4 x3  4 x 2
Must be x
A1
CAO
M1
For
A1
3 7
FT for y  y 3
7
A1
CAO
Must be y
B2B2
For integrals, as above
8
3 7 
832

  y2  y 3  ( 
)
7
7

0
2
3 43 3 53
y  y
2
10
8
4
 

Ay   xy dy    2 y  y 3  dy

0 
4
Or 32  2   xy dx
dy
8
3 4 3 5 

  2 y  y 3  y 3  (  49.6 )
2
10

0
 xy dy
2
4
Or 32  4   ( 12 ) y 2 dx
2
Or B2 for
1 ( x  2) 7
14
Give B1 for one minor slip in
integration, or if ½ omitted
[9]
OR
Region under curve has CM (3.6,
16
7
)
B1 (for 28)
M1
A1
28 x  4  3.6  32  2
x  1.77
28 y  4  16
 32  4
7
M1
y  4.24
A1
6
4
2