4764
1(i)
Mark Scheme
June 2006
m = 43 π r 3 ρ
M1
Expression for m
dm
dr
= 4π r 2 ρ
dt
dt
M1
Relate
dr
dt
λ ⋅ 4π r 2 = 4π r 2 ρ
M1
dr λ
= =k
dt ρ
r = r0 + kt
4
πρ (r0
3
m=
+ kt )
3
dm
dr
to
dt
dt
dm
Use of
proportional to surface area
dt
E1
Accept alternative symbol for constant if used
correctly (here and subsequently)
M1
Integrate and use condition
A1
6
(ii)
d
( mv ) = mg
dt
M1
N2L
mv = ∫ mg dt = ∫ 43 πρ (r0 + kt )3 g dt
M1
Express mv as an integral
= 43 πρ g ⎡⎣ 41k (r0 + kt ) 4 + c ⎤⎦
M1
Integrate
M1
Use condition
M1
Substitute for m
t = 0, v = 0 ⇒ c = − 41k r0 4
4
πρ (r0
3
v=
g
4k
3
+ kt ) v =
4
πρ g ⋅ 41k
3
⎡ (r0 + kt ) − r0
⎣
⎦
4
4⎤
⎡
r0 4 ⎤
⎢ r0 + kt −
⎥
(r0 + kt )3 ⎦⎥
⎣⎢
A1
6
2(i)
AP = 2a cos θ
M1
PB = a − 2a cos θ
E1
V = − mg ⋅ PB − mg ⋅ PA cos θ
M1
5
2
= −mg
(
5
2
)
Attempt AP in terms of θ
Attempt V in terms of θ
a − 2a cos θ − mg ( 2a cos θ ) cos θ
(
= − mga 2 cos 2 θ − 2 cos θ + 52
)
E1
4
(ii)
dV
= mga sin θ ( 4 cos θ − 2 )
dθ
dV
= 0 ⇒ sin θ = 0 or cos θ =
dθ
⇒ θ = 0 or ± 13 π
d 2V
dθ 2
1
2
= mga sin θ ( −4sin θ ) + mga cos θ ( 4 cos θ − 2 )
θ =0⇒
d 2V
dθ
θ = ± 13 π ⇒
2
= 2mga > 0 ⇒ stable
2
dV
dθ 2
M1
Differentiate
M1
Solve
A1
For 0 and either of
M1
A1
Differentiate again
M1
Consider sign of V ′′ in one case
Correct deduction for one value of θ
Correct deduction for another value of θ
N.B. Each F mark is dependent on both M
marks.
To get both F marks, the two values of θ must
be physically possible (i.e. in the first or fourth
quadrant) and not be equivalent or symmetrical
positions.
F1
F1
= −3mga < 0 ⇒ unstable
1
π
3
or − 13 π
8
72
4764
3(i)
Mark Scheme
P = Fv = mv
dv
v
dx
June 2006
M1
Use of P = Fv
A1
Or equivalent
M1
Separate variables
M1
Integrate
v 2 = A e0.0008 x − 10 000
x = 0, v = 0 ⇒ A = 10 000
M1
Rearrange
M1
Use condition
0.0008 x
A1
(
)
dv
= 0.0004 10 000v + v3
dx
v
∫ 10 000 + v 2 dv = ∫ 0.0004 dx
v2
1
ln
2
10 000 + v 2 = 0.0004 x + c
v = 100 e
−1
x = 900 ⇒ v = 102.7 > 80 so successful
or v = 80 ⇒ x = 618.37 < 900 so successful
E1
Show that their v implies successful take off
8
(ii)
(
)
dv
v = 0.0004 10 000v + v3
dt
1
∫ 10 000 + v 2 dv = ∫ 0.0004 dt
1
100
tan −1
( 1001 v ) = 0.0004t + k
t = 0, v = 0 ⇒ k = 0
⇒ v = 100 tan(0.04t )
v → ∞ at finite time suggests model invalid
(iii)
F1
Follow previous DE
M1
Separate variables
M1
A1
M1
E1
B1
Integrate
B1
M1
M1
M1
At least 3sf
Attempt to calculate maximum P
Use solution in (i) to calculate x
Set up DE for t . 11 .
Constant acceleration formulae ⇒ M0.
Separate variables and integrate
Follow their maximum P (condone no constant)
Use condition on x, v (not v = 0 , not x = 0
unless clearly compensated for when making
conclusion).
Constant acceleration formulae ⇒ M0.
Relevant calculation. Must follow solving a DE.
All correct (accept 2sf or more)
7
t = 11 ⇒ v = 47.0781
Hence maximum P = 230.049m
v = 47.0781 ⇒ x = 250.237
dv
v2
= 230.049
dx
v = 47.0781, x = 250.237 ⇒ B = −22786.3
M1
F1
M1
v = 80 ⇒ x = 840.922 or x = 900 ⇒ v = 82.0696
so successful
M1
A1
1 3
v
3
Use condition
Clearly shown
= 230.049 x + B
9
73
4764
4(i)
Mark Scheme
Considering elements of length δ x ⇒ I = ∫
2a
ρ x 2 dx
0
=
M
2
3
∫0 ( 5ax − x ) dx
2a
2
8a
2a
M
= 2 ⎡⎣ 53 ax3 − 14 x 4 ⎤⎦
0
8a
2
7
= 6 Ma
2a
0
M
ρ x dx
2
∫0 ( 5ax − x ) dx
2a
2
M1
Set up integral
M1
Substitute for ρ in predominantly correct
integral
M1
Integrate
E1
Considering elements of length δ x ⇒ Mx = ∫
=
June 2006
8a
2a
M
= 2 ⎡⎣ 52 ax 2 − 13 x3 ⎤⎦
0
8a
11
x = 12 a
M1
Set up integral
M1
Substitute for ρ in predominantly correct
integral
M1
Integrate
E1
8
(ii)
1
2
M1
B1
M1
11
Iθ& 2 = Mg ⋅ 12
a (1 − cos θ )
11g
(1 − cos θ )
7a
θ& =
KE term in terms of angular velocity
11
± Mg ⋅ 12
a cos θ seen
energy equation
A1
4
(iii)
J1
B
A
J2
θ = 12 π ⇒ θ& =
11g
7a
⎛
11g ⎞
2a ⋅ ( − J1 ) = I ⎜⎜ 0 −
⎟
7 a ⎟⎠
⎝
F1
Their θ& at θ = 12 π
M1
Use of angular momentum
A1
Correct equation (their θ& )
1
J1 = 12
M 77ag
E1
J2 =
B1
1
12
M 77 ag
Correct answer or follow their J1
5
(iv)
J3
J4 = J2
M1
Consider horizontal impulses
F1
Follow their J 2
M1
M1
Vertical impulse-momentum equation
Use of rθ&
A1
cao
⎛ 1 M 77 ag ⎞
⎛J ⎞
⎟
angle = tan −1 ⎜ 3 ⎟ = tan −1 ⎜ 21
⎜ 1 M 77ag ⎟
⎝ J4 ⎠
⎝ 12
⎠
M1
Must substitute
= tan −1
A1
cao (any correct form)
J4
=
11
J 3 + J1 = M ⋅ 12
a
J3 =
1
M
21
1
12
M 77 ag
11g
7a
77ag
( 74 ) ≈ 0.519 rad ≈ 29.7°
7
74
4764
1(i)
Mark Scheme
x = PB
M1
2
x= a +y
2
May be implied
A1
V = 12 kx 2 − mgy
(
June 2007
)
= 12 k a 2 + y 2 − mgy
M1
M1
EPE term
GPE term
A1
cao
5
(ii)
dV
= ky − mg
dy
equilibrium ⇒
⇒ y=
dV
=0
dy
mg
k
d 2V
=k >0
dy 2
⇒ stable
M1
Differentiate their V
B1
Seen or implied
A1
cao
M1
Consider sign of V ′′ (or V ′ either side)
E1
Complete argument
5
(iii)
ˆ = k ⋅ PB ⋅ a
R = T sin PBA
PB
M1
= ka
A1
Use Hooke’s law and resolve
2
2(i)
d
( mv ) = 0 ⇒ mv constant
dt
hence mv = m0 u
M1
Or no external forces ⇒ momentum
conserved, or attempt using δ terms.
A1
dm
=k
dt
⇒ m = m0 + kt
B1
B1
dm
= k seen
dt
m0 + kt stated or clearly used as mass
Complete argument (dependent on all
previous marks and m0 + kt derived, not
just stated)
m0 u
m0 u
=
m
m0 + kt
E1
m0 u
dt
m0 + kt
M1
Integrate v
A1
cao
M1
Use condition
m0 u ⎛ m0 + kt ⎞
ln ⎜
⎟
k
⎝ m0 ⎠
A1
cao
v = 12 u ⇒ m0 + kt = 2m0
M1
Attempt to calculate value of m or t
v=
x=∫
m0 u
ln ( m0 + kt ) + A
k
mu
x = 0, t = 0 ⇒ A = − 0 ln m0
k
=
x=
9
(ii)
⇒x=
m0 u ⎛ 2m0 ⎞
ln ⎜
⎟
k
⎝ m0 ⎠
M1
Substitute their m or t into x
⇒x=
m0 u
ln 2
k
F1
t=
m0
or m = 2m0 in their x
k
3
68
4764
3(i)
Mark Scheme
June 2007
M1
A1
Set up integral
Or equivalent
M1
Use mass per unit length in integral or I
M1
Integrate
= 16 ma 2 − − 16 ma 2
M1
Use limits
1
ma 2
3
E1
Complete argument
a
I = ∫ ρ x 2 dx
−a
m
2a
m ⎡ 1 3 ⎤a
I=
x
2a ⎣ 3 ⎦ − a
ρ=
6
(ii)
I rod =
1
× 1.2 × 0.42
3
+ 1.2 × 0.4
2
I sphere = 52 × 2 × 0.12 + 2 × 0.92
I = I rod + Isphere = 1.884
1
ma 2
3
or
4
3
2
M1
Use
A1
M1
M1
A1
M1
A1
Rod term(s) all correct
Use formula for sphere
Use parallel axis theorem
Sphere terms all correct
Add moment of inertia for rod and sphere
cao
M1
M1
M1
A1
M1
F1
Use energy
KE term
Reasonable attempt at GPE terms
All terms correct (but ignore signs)
Rearrange
Only follow an incorrect I
ma
7
(iii)
1
2
Iθ 2 − 1.2 g × 0.4 cos θ − 2 g × 0.9 cos θ
= −1.2 g × 0.4 cos α − 2 g × 0.9 cos α
θ 2 =
4.56 g
( cos θ − cos α )
1.884
6
(iv)
4.56 g
− sin θ θ
2θ θ =
1.884
or Iθ = −1.2 g × 0.4sin θ − 2 g × 0.9sin θ
(
)
M1
F1
sin θ ≈ θ ⇒ θ = −11.86θ
M1
i.e. SHM
T≈
2π
11.86
E1
≈ 1.82
F1
Differentiate, or use moment = Iθ
Equation for θ (only follow their I or
θ2 )
Use small angle approximation (in terms
of θ )
All correct (for their I) and make
conclusion
2π
their ω
5
69
4764
4(i)
Mark Scheme
dv
= 2 − 8v 2
dx
v
∫ 1 − 4v 2 dv = ∫ dx
M1
A1
N2L
M1
Separate
− 18 ln 1 − 4v 2 = x + c1
A1
LHS
x = 0, v = 0 ⇒ c1 = 0
M1
M1
Use condition
Rearrange
E1
Complete argument
M1
2
Substitute given v into F
A1
cao
M1
Set up integral of F
A1
cao
M1
Integrate
A1
Accept
2v
2
1 − 4v = e
v2 =
(ii)
June 2007
1
4
−8 x
(1 − e )
−8 x
(
2
F = 2 − 8v = 2 − 2 1 − e
−8 x
7
)
= 2 e −8 x
Work done =
2
∫0 F dx
2
= ∫ 2 e−8 x dx
0
= ⎡⎣ − 14 e −8 x ⎤⎦
=
1
4
2
0
(1 − e )
−16
1
4
or 0.25 from correct working
6
(iii)
dv
2 = 2 − 8v 2
dt
1
1
dv = ∫ d t
4∫ 1
− v2
4
1
2
1
ln
4
1
2
+v
= t + c2
−v
t = 0, v = 0 ⇒ c2 = 0
1
2
1
2
+v
−v
= e 4t
M1
N2L
M1
Separate
A1
LHS
M1
Use condition
M1
Rearrange (remove log)
M1
Rearrange (v in terms of t)
E1
Complete argument
1 + 2v = e 4t (1 − 2v )
(
)
2v 1 + e 4t = e 4t − 1
⎛ e 4t − 1 ⎞
v = 12 ⎜⎜ 4t ⎟⎟ =
⎝ e +1⎠
(v)
−4t
⎛
1 1− e
⎜
2⎜
−4 t
⎝ 1+ e
⎞
⎟⎟
⎠
7
t = 1 ⇒ v = 0.4820
t = 2 ⇒ v = 0.4997
Impulse = mv2 − mv1
= 0.0353
B1
B1
M1
A1
Use impulse-momentum equation
Accept anything in interval [0.035, 0.036]
4
70
4764
Mark Scheme
June 2008
4764 Mechanics 4
1(i)
If δ m is change in mass over time δ t
PCLM mv = (m + δm)(v + δv) + δm (v − u )
δm < 0 ]
(m + δm)
[N.B.
δv
δm
dv
dm
+u
=0⇒m
= −u
δt
δt
dt
dt
dm
= −k ⇒ m = m0 − kt
dt
dv
⇒ (m0 − kt )
= uk
dt
M1
Change in momentum over time δt
M1
A1
Rearrange to produce DE
Accept sign error
M1
Find m in terms of t
E1
Convincingly shown
5
(ii)
uk
v=∫
dt
m0 − kt
M1
Separate and integrate
= −u ln(m0 − kt ) + c
t = 0, v = 0 ⇒ c = u ln m0
A1
M1
cao (allow no constant)
Use initial condition
⎛ m0 ⎞
v = u ln ⎜
⎟
⎝ m0 − kt ⎠
A1
All correct
M1
A1
A1
Find expression for mass or time
Or t = 2m 0 / 3k
4
(iii)
m = 13 m0 ⇒ m0 − kt = 13 m0
⇒ v = u ln 3
3
2(i)
P = Fv
dv
= mv v
dx
dv
⇒ mv 2
= m k 2 − v2
dx
(
M1
M1
)
v2
dv
=1
k − v dx
⎛ k2
⎞ dv
⇒ ⎜ 2 2 − 1⎟ = 1
⎜ k −v
⎟ dx
⎝
⎠
⎛ k2
⎞
∫ ⎜⎜ k 2 − v 2 − 1⎟⎟ dv = ∫ dx
⎝
⎠
k
+
v
⎛
⎞
1
k ln ⎜
⎟−v = x+c
2
⎝ k −v ⎠
x = 0, v = 0 ⇒ c = 0
⎛k +v⎞
x = 12 k ln ⎜
⎟−v
⎝ k −v ⎠
⇒
2
2
Used, not just quoted
Use N2L and expression for
acceleration
A1
Correct DE
M1
Rearrange
E1
Convincingly shown
M1
Separate and integrate
A1
LHS
M1
Use condition
A1
cao
9
(ii)
Terminal velocity when acceleration zero
⇒v=k
⎛ 1.9 ⎞
1
v = 0.9k ⇒ x = 12 k ln ⎜
⎟ − 0.9k = ( 2 ln19 − 0.9 ) k ≈
⎝ 0.1 ⎠
0.572k
M1
A1
F1
Follow their solution to (i)
3
57
4764
3(i)
Mark Scheme
M1
M1
M1
A1
a
M = ∫ k ( a + r ) 2π r dr
0
= 2kπ ⎡⎣ 12 ar 2 + 13 r 3 ⎤⎦
a
0
= 53 kπ a3
June 2008
Use circular elements (for M or I)
Integral for mass
Integrate (for M or I)
For […]
E1
a
I = ∫ k ( a + r ) 2π r ⋅ r dr
2
M1
Integral for I
A1
For […]
= 109 kπ a 5
A1
cao
2
E1
Complete argument (including mass)
B1
M1
M1
A1
Seen or used (here or later)
Use angular momentum
Use moment of impulse
cao
0
= 2kπ ⎡⎣ 14 ar 4 + 15 r 5 ⎤⎦
=
27
50
Ma
a
0
9
(ii)
I = 13.5
0.625 × 50 = I ω
⇒ ω ≈ 2.31
(iii)
4
30 − 2.31
θ=
≈ 1.38
20
M1
Find angular acceleration
Couple = Iθ
≈ 18.7
M1
F1
Use equation of motion
Follow their ω and I
Iθ = −3θ
B1
Allow sign error and follow their I
(but not M)
3
(iv)
dθ
= −3θ
dt
dθ
3
∫ θ = ∫ − I dt
t
ln θ = −
+c
4.5
M1
Set up DE for θ (first order)
M1
Separate and integrate
B1
ln multiple of θ seen
θ = A e−t / 4.5
M1
I
(v)
t = 0, θ = 30 ⇒ A = 30
M1
θ = 30 e−t / 4.5
A1
Model predicts θ never zero in finite time.
B1
(
)
Rearrange, dealing properly with
constant
Use condition on θ
7
1
58
4764
4(i)
Mark Scheme
2
⎛ mg ⎞
V = 12 ⎜
⎟ ( aθ ) + mga cos θ (relative to centre
⎝ 10a ⎠
of pulley)
dV 1 ⎛ mg ⎞
2
= ⎜
⎟ ⋅ 2a θ − mga sin θ
dθ 2 ⎝ 10a ⎠
dV
= mga 101 θ − sin θ
dθ
(
(ii)
)
(
M1
EPE term
B1
M1
A1
Extension = aθ
GPE relative to any zero level
(± constant)
M1
Differentiate
E1
6
dV
1
θ =0⇒
= mga 10
( 0 ) − sin 0 = 0
dθ
hence equilibrium
d 2V
= mga 101 − cos θ
dθ 2
V ′′(0) = −0.9mga < 0
hence unstable
(
June 2008
)
M1
E1
M1
A1
)
M1
E1
dV
Consider value of
dθ
Differentiate again
Consider sign of V ′′
V ′′ must be correct
6
(iii)
(iv)
If the pulley is smooth, then the tension in the
string is constant.
Hence the EPE term is valid.
B1
B1
2
Equilibrium positions at θ = 2.8 ,
θ = 7.1
and θ = 8.4
B1
B1
From graph, V ′′(2.8) = mgaf ′ ( 2.8 ) > 0
hence stable at θ = 2.8
V ′′(7.1) = mgaf ′ ( 7.1) < 0 ⇒ unstable at θ = 7.1
M1
A1
A1
V ′′(8.4) = mgaf ′ ( 8.4 ) > 0 ⇒ stable at θ = 8.4
A1
One correct
All three correct, no extras
Accept answers in [2.7,3.0), [7,7.2],
[8.3,8.5]
Consider sign of V ′′ or f ′
Accept no reference to V ′′ for one
conclusion but other two must relate
to sign of V ′′ , not just f ′ .
6
(v)
P
(vi)
B
If θ < 0 then expression for EPE not valid
hence not necessarily an equilibrium position.
B1
P in approximately correct place
B1
B in approximately correct place
2
M1
A1
2
59
4764
Mark Scheme
June 2009
4764 MEI Mechanics 4
B1
1(i)
Seen or implied
M1 Expand
M1
Use
M1 Separate variables (oe)
E1
5
M1 Integrate
(ii)
A1
LHS
M1 Use condition
A1
B1
5
(iii) As
So
gets large,
M1
gets large
A1
(i.e. constant)
Complete argument
2
B1
2(i)
GPE
M1 Reasonable attempt at EPE
A1 EPE correct
M1 Differentiate
E1
Complete argument
5
M1 Set derivative to zero
(ii)
M1 Solve
A1
Both
M1
Second derivative (or
alternative method)
M1 Consider sign
unstable
A1
One correct conclusion validly
shown
A1
Complete argument
stable
7
64
4764
3(i)
Mark Scheme
Mass of ‘ring’
June 2009
B1
May be implied
M1
Set up integral
A1
All correct
M1
Integrate
M1
Use relationship between
and
E1
Complete argument
6
(ii)
M1
Use parallel axis theorem
E1
Convincingly shown
B1
LHS
B1
M1
RHS
2
(iii)
M1
small
E1
, i.e. SHM
F1
Period
Expression for
Use small angle
approximation
Complete argument and
conclude SHM
Follow their SHM equation
6
(iv)
e.g.
C
A
B1
Show PAC in straight line (in
any direction)
M1
Moments or
(oe)
A1
Method may be implied
P
M1
E1
Convincingly shown
5
(v)
KE
M1
Attempt to find KE
A1
M1
A1
Work-energy equation
Correct equation
A1
5
65
4764
Mark Scheme
4(i) At terminal velocity,
June 2009
M1
Equilibrium of forces
E1
Convincingly shown
2
M1
(ii)
N2L
A1
M1
Separate and integrate
A1
LHS
M1
Rearrange, dealing properly
with constant
Use condition
E1
Complete argument
M1
7
(iii)
B1
WD against
M1
Integrate
A1
B1
Loss in energy
M1
GPE
M1
KE
Convincingly shown (including
signs)
E1
WD against
7
B1
(iv)
1
M1
(v)
N2L
A1
M1
Separate and integrate
A1
M1
M1
A1
Use condition (or limits)
Calculate
7
66
4764
1(i) Mark Scheme
(ii) A1 Accept sign errors in M1 Form DE E1 Complete argument (including signs) 4
M1 Integrate 1
A1 M1 A1 Fuel burnt when Impulse = change in momentum M1 M1 A1 A1 So M1 June 2010
Use condition 8
4764
Mark Scheme
2(i) M1 A1 M1 A1 (ii) Rearrange M1 Integrate M1 Use condition (iii) The speed decreases, tending to zero B1 The displacement tends to 2
Separate and integrate M1 Use condition A1 N2L M1 E1 June 2010
B1 Cv (10/k) 7
3
2
4764
Mark Scheme
3(i) M1 and unstable 3
Differentiate 6 Solve M1 Second derivative (or other valid method) A1 Any correct form M1 Substitute A1 Deduce unstable M1 Substitute other value A1 A1 EPE term E1 (A) M1 A1 (ii) GPE term M1 M1 A1 June 2010
4764
Mark Scheme
stable (B) stable (C) gives only (from both factors) M1 Consider second derivative A1 Complete argument M1 Consider solutions M1 Consider second derivative A1 Complete argument M1 Consider solutions 10 4 M1 Valid method Hence stable A1 Complete argument A1 A1 only June 2010
4
4 4764
Mark Scheme
4(i) Mass of slice So A1 A1 M1 E1 Mass of small cone ft 8
M1 Moment of inertia of small cone 5
E1 M1 B1 Mass of frustum A1 (ii) M1 So M1 M1 June 2010
4
4764
Mark Scheme
(iii) A1 M1 M1 A1 June 2010
(iv) Centre of mass: OG i.e. G is face m from the small circular Radius at G is A1 Any distance which locates G M1 Moment of impulse = ang. momentum Moment of impulse = ang. momentum B1 M1 A1 A1 (v) M1 A1 6
4
3
5
4764
Mark Scheme
1(i) (ii) AG (iii) 2(i) M1 A1 2 B1 N2L M1 Expand derivative M1 Substitute A1 M1 Separate and integrate A1√ M1 Use condition E1 Convincingly shown Accept substituted into their expression in part (i) 8 A1 Any correct form 2 M1 for E = 12 kx A1 A1 M1 E1 Convincingly shown 5 M1 A1 M1 E1 Convincingly shown 4 M1 E1 Convincingly shown B1 3 M1 June 2011
2
AG (ii) (iii) Hence (as continuous) between So equilibrium. Stable as . and . 1
4764
Mark Scheme
M1 N2L with A1 M1 Separate M1 Partial fractions A1 M1 Use condition A1 AEF, condone m E1 M1 Rearrange A1 Cao without m 10 M1 M1 A1 M1 A1 E1 Separate and integrate Use condition Any correct form 6 M1 N2L A1 M1 A1 M1 Separate and integrate Use condition M1 Rearrange A1 CAO Accept t = 45 (time for this part of motion) 3(i) (ii) AG (iii) June 2011
B1 2
8 4764
Mark Scheme
4(i) AG (ii) Mass of slice June 2011
B1 M1 Use perpendicular axes theorem B1 M1 Use parallel axes theorem M1 Use E1 Complete argument 6 M1 B1 Deal correctly with mass/density M1 A1 M1 Substitute for M1 A1 AG E1 Complete argument 8 M1 Energy equation B1 A1 F1 Position of centre of mass KE term GPE terms ft their CoM only E1 Complete argument 5 M1 Differentiate or use (iii) (iv) torque A1 for small M1 Use E1 B1 5 Hence SHM Period 3
4764
Mark Scheme
Question
(i)
1
Answer
mv  (m   m)(v   v)  ( m)(v  u )
Marks
M1
A1
(note  m  0 )
mv  mv  v m  m v   m v  v m  u m
v
m
v
m
u
m
0
t
t
t
dv
dm
m  u
dt
dt
dm
 k
dt
m  m0  kt
dv
( m0  kt )  uk
dt
M1
June 2012
Guidance
Attempt at momentum equation
Condone wrong sign of  m
Simplify and divide by  t
B1
B1
SOI
E1
All correct including sign of  m
[6]
1
(ii)
uk
 dv   m0  kt dt
v  u ln( m0  kt )  c
t  0, v  v0  v0  u ln m0  c
c  v0  u ln m0
 m0 
v  v0  u ln 

 m0  kt 
M1
Separate and integrate
A1
M1
A1
Use condition
A1
aef
[5]
2
(i)
Let equilibrium extension be e
dv
mv  mg  k (e  x)
dx
At equilibrium, mg  ke
dv
So mv   kx
dx
M1
N2L
A1
B1
All terms correct
oe
E1
With evidence of working
[4]
5
4764
Mark Scheme
Question
2 (ii)
Answer
Marks
 mv dv   kx dx
1
2
M1
mv   kx  c
2
1
2
2
1
2
mv  12 k  a  x
2
Guidance
Solutions from SHM acceptable
A1
x  a, v  0  0   ka  c
2
June 2012
1
2
2
2
M1

k 2
a  x2 

m
( v  0 when moving up)
v
A1
oe
E1
AG Complete argument including justification for square root.
[5]
2
(iii)

1
dx   
k
dt
m
M1*
a2  x2
k
 x
arcsin    
t  c2
m
a
x  a, t  0  12   c2
3
(i)
A1
DM1

 k 
k 
x  a sin  12  
t   a cos 
t 
m 

 m 
A1
l  2a sin 
[4]
M1
V 

2a
Solutions from SHM NOT acceptable
(2a sin   a) 2
A1
M1
Differentiate
M1
Use trigonometric identities as necessary
M1
dV

 2mga sin 2   2a sin   a   2a cos 
d
a
 4mga sin  cos  2 a cos  2sin   1
 2a cos θ (2 λ sin θ  2mg sin θ  λ)
E1
[7]
6
Either form
OE eg a 2  2cos 2
λ
(a 2  2cos 2θ  a) 2
EPE OE eg
2a
Both terms
GPE OE eg mga sin( 12   2 )
A1
...  mga cos 2
Accept c2 = arcsin 1
4764
Question
3 (ii)
Mark Scheme
Answer
Marks
dV
 0  ()  0
d
hence equilibrium
  12  
E1
d 2V
 2a  2  2mg   
d 2
d 2V
So   2mg  2  0  stable
d
If cos   0
dV
 0  2 sin   2mg sin     0
d
  12  
(iii)
M1
A1
Here or in (iii) or use sign method
M1
Use V '' or equivalent method
E1

M1
Consider other solutions
Attempt at showing not valid
Must consider both ends
If   2mg ,  12  as before
V   0 so unstable
M1
E1
[9]
B1
B1
or sin  
E1
2  2mg
But   2mg  2  2mg  
 sin   1 or sin   0
So no valid solutions
3
Guidance
M1
d 2V
 2a sin   2 sin   2mg sin    
d 2
2a cos   2 cos  2mg cos 
 sin  
June 2012
and
1
2


2  2mg

2   2 mg
 1 so gives valid solution
E1
  sin 1  2  2 mg  or   sin 1  2  2 mg 
B1
and V   0  2a cos 2   2  2mg 
M1
   ve   ve  so stable (in both cases)
A1
[7]
7
For both
4764
Question
(i)
4
Mark Scheme
Answer
 I  2 r r  r

June 2012
Marks
2
Guidance
B1
m
 a2
B1
a
m 3
r dr
2a 2
0
I disc  

M1
for k r 3 dr
M1
k 14 r 4
A1
k
4
a
m 1 
 2  r4 
2a  4  0
1
 ma 2
2
4
(ii)
+
1
2
ma 2  2
 a2
2 a 2  2 ah
2 ah
m1  M
2 a 2  2 ah
  a 2  2 ah 
So I  Ma 2 

2
 2 a  2 ah 
I
a
0
with limits
a4
E1
I  m1a 2
mM
 
[6]
M1
Curved surface
M1
Combine
m  πρa 2
B1
m1  2πahρ
Substitute
E1
[6]
4
(iii)
1
 0.5  0.6 
I   8  0.52 
  1.375
2
 0.5  0.3 
I (  0)  Ja
1.375  55  0.5
  20 rad s1
B1
M1
A1
[3]
8
+
B1
M1
1
 a  2h 
Ma 2 

2
 ah 
2πhρa 3
Impulse/momentum
I M
1
2
ρπa 4  2
πρa 4  2πρa 3
2πρa 2  2πρah
4764
Mark Scheme
Question
4 (iv)
Answer
4
(v)
Marks
d
I
 2 2
dt
2
1.375 d   2dt

June 2012
Guidance
B1
1.375
 2t  c

M1
Separate
M1
Integrate
A1
M1
Use condition
t  0,  20  c  0.06875
1.375
t 5 
 10  0.06875

   0.137 (3sf)
M1
 0.137 
I
  0.03
 t 
M1
Complete method
t  6.26 s
A1
A1
[3]
with correct acceleration (or both sides +ve)
awfw [6.25, 6.3] CAO
A1
[7]
9
4764
Question
1
(i)
Mark Scheme
Answer
Marks
dm
 k  m  kt  c
dt
conditions  m  m0  kt
d

  mv   0  
 dt

v
B1
mv  m0 v0
M1 A1
m0 v0
mv
 0 0
m
m0  kt
m0 v0
k
Momentum equation
Or derive a differential
equation in only two
variables
A1
M1
ln  m0  kt 
conditions  0 
Guidance
B1
 mv
x   0 0 dt
 m0  kt

June 2013
(+ c2)
Integrate their expression for v
A1
m0 v0
ln m0  c2
k
M1
m0 v0
 ln  m0  kt   ln m0 
k
 m  kt  m0 v0 
mv
kt 
ln 1 
 0 0 ln  0


k
k
 m0 
 m0 
Use initial conditions
so x 
E1
[9]
1
(ii)
kt  2m0  t 
x
2
(i)
2m0
 v  13 v0
k
m0 v0
ln 3
k
BC  2  0.5sin
B1
Follow through their v = f(t)
B1
Ft
[2]
E1
[1]
1

2
5
SC If kt = m0 Award B1
either correct on follow
through
4764
Question
2
(ii)
Mark Scheme
Answer
Marks
V  0.5 g  0.25sin  
1
2
June 2013
M1
M1
 2(BC  0.5) 2  12  2(BD  0.5) 2
BD 2  12  0.52  2  1 0.5cos   1.25  cos 
V  1.225sin   (sin 12   0.5) 2 
Guidance
GPE
At least one EPE term
B1
A1
oe
M1
M1
A1
Differentiate
Use of chain rule
one EPE term correct
E1
Complete argument
  1.2 and 4.1
[8]
B1
Both
Stable and unstable respectively
B1
( 1.25  cos   0.5) 2
dV
 1.225cos   2(sin 12   0.5)( 12 cos 12  ) 
d
sin 


2 1.25  cos   0.5 

 2 1.25  cos  


 1.225cos   sin 12  cos 12   0.5cos 12  
sin  
0.5sin 
1.25  cos 
 1.5sin   1.225cos  
2
(iii)
at   1.2 ,
0.5sin 
1.25  cos 
 0.5cos 12 
dV
increasing because the graph shows that
d
f ( ) is positive
so V minimum hence stable
at   4.1
M1
Consider gradient, relating f to
A1
Clear evidence from the graph
dV
decreasing because the graph shows that
d
f ( ) is negative, so max. so unstable
[4]
6
Allow B1M1A1 from 1.1
and/or 4.05
dV
d
4764
Question
3
(i)
Mark Scheme
Answer
Marks
dv 2v  4v

 6v
dt
v
dv
 v 2  3v  2  (1  v)(2  v)
dt
1

dv   dt

 (1  v)(2  v)
3
M1
A1
2
Guidance
N2L
E1
1 
 1


 dv   dt
 1 v 2  v 
 ln 1  v  ln 2  v  t  c
t  0, v  0  ln 2  c
t  ln(2  v)  ln(1  v)  ln 2
 ln
June 2013
2v
2(1  v )
M1
Separate
M1
Partial fractions
A1
A1
M1
M1
LHS
RHS
Use condition
Rearrange
E1
[10]
3
(ii)

v
3
(iii)
2v
 e t  2  v  2 et  2 e t v
2(1  v )
M1
2(et  1)
A1
2 et  1
[2]
E1
v  0.8  P  2  0.83  4  0.8  4.224
2  0.8
t  ln
 ln 3  1.10
2(1  0.8)
B1
[2]
7
Rearrange
4764
Question
3
(iv)
Mark Scheme
Answer
Marks
dv 4.224

 6v
v
dt
2

 4.224  6v dv   dt
 v
2
v

dv   dt

 2.112  3v 2
1
 ln 2.112  3v 2  t  c2
6
2.112  3v  A e
2
Separate
A1
LHS
A1
RHS
Rearrange to make v the subject
A1
Correct
A1
[10]
Ft their expression for v
B1
Or let ρ = 1 without lose of generality
a 2  3
element with radius x and ‘width’  x :
 m   x 3  x   I   x 3  x  x 2
2m
M1
M1
A1
M1
m
1
2
N2L
Alternate
6ln 3
v  0.704  46.656 e 6t
t    v  0.704  0.839

B1
6t
t  ln 3, v  0.8  2.112  3  0.8  A e
A  139.968
(i)
Guidance
M1
2
4
June 2013
Use condition to find constant
M1
x3 x
A1
Ft their ρ
a
2m 3
I 
 2 x dx
0 a
M1
Integrate

a2
a
2m  x 4 
 2  
a  4 0
A1
2m  a 4

a 2  4
E1

 1 2
  ma
 2
[6]
8
t    2.112  3v 2  0
4764
Question
4
(ii)
Mark Scheme
Answer
Marks
2a
(iii)
4
(iv)
   cos  mg    cos
1
2
I 2  mg
1
2
ma 2 2  2mg
 2 
Guidance
B1

4
June 2013
2a
2a
2

3
[1]
M1
A1
A1
OE
 2a   cos  12 
4g
 2 cos   1
a
RHS: or mg  2a  cos 13 
E1
[4]
M1
Max  when cos   1
12 g
  2 
A1
M1
a
Speed max. furthest from axis,
so max speed  a
Energy
Two correct terms
All correct
12 g
a

12 ag
A1

oe
[4]
4
4
(v)
(vi)
4g
2 
2sin  
a
4g
   sin 
a


M1
A1
[2]
M1
J  x  I    I  
J  34 a 
1
2
2 1
1
ma

2
2
   12 ma 2   
A1

B1
  () 13    2 
J m

4g
8g
2  12   1   
a
a
8ag
A1

[4]
9
Differentiate with respect to time
Or use C = I 
May be seen in (iv)
May be seen in (iv)
4764
Question
4
(vii)
Mark Scheme
1
2
I

1
2
 
8g
a
  mg 
2
cos 1 14
Answer
2a

 1.32
 cos 13   mg  2a  cos
June 2013
Marks
Guidance
M1
A1
A1
[3]
10
CAO