| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Session | Specimen |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Challenging +1.2 This is a structured multi-part question on second-order differential equations with a real-world context. While it covers several techniques (forming DEs from verbal descriptions, finding particular integrals, applying initial conditions, and interpretation), each part is heavily scaffolded with guidance. Parts (i)-(iii) are standard Further Maths procedures, and parts (iv)-(vii) involve straightforward interpretation and adaptation rather than novel problem-solving. The question is moderately above average difficulty due to its length and Further Maths content, but the extensive scaffolding keeps it accessible. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (i) | d (cid:167) dm (cid:183) |
| Answer | Marks |
|---|---|
| dt2 dt | B1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 2.1 | Translate the given context into a correct |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (ii) | d2m dm |
| Answer | Marks |
|---|---|
| (cid:79)(cid:32)(cid:16)1 | M1 |
| Answer | Marks |
|---|---|
| [4] | i |
| Answer | Marks |
|---|---|
| 1.1 | m |
| Answer | Marks | Guidance |
|---|---|---|
| Follow through their auxiliary equation | Or complete the square | |
| 11 | (iii) | d2m dm |
| Answer | Marks |
|---|---|
| (cid:159)m(cid:32)e(cid:16)t(cid:11)40sin2t(cid:16)80cos2t(cid:12)(cid:14)80 | M1 |
| Answer | Marks |
|---|---|
| [7] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | Correct Particular Integral and |
| Answer | Marks |
|---|---|
| Final answer stated | Award first two marks if GS |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (iv) | S |
| As t(cid:111)(cid:102), m(cid:111)80 | B1 | |
| [1] | 3.4 | |
| 11 | (v) | E.g. There is a limiting process here. |
| Answer | Marks | Guidance |
|---|---|---|
| is only displayed after infinite time | B1 | |
| [1] | 3.5b | |
| 11 | (vi) | (a) |
| Answer | Marks |
|---|---|
| … when t(cid:32)1.107 (cid:159)m(cid:32)106 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1b |
| 3.4 | Differentiation, FT their expression |
| Answer | Marks |
|---|---|
| by 106 | OR By graphical calculator |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (vi) | (b) |
| [1] | 3.2a | n |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (vii) | e.g. |
| Answer | Marks | Guidance |
|---|---|---|
| dt2 dt | E1 | |
| [1] | 3.5c | e |
| Answer | Marks |
|---|---|
| dt2 dt a | Accept any sensible suggestion |
| Answer | Marks |
|---|---|
| 11(vi)(a) | n |
| Answer | Marks |
|---|---|
| 11(vi)(b) | e |
| Answer | Marks |
|---|---|
| 11(vii) | S |
Question 11:
11 | (i) | d (cid:167) dm (cid:183)
(cid:168)0.5 (cid:14)m(cid:184)(cid:32)k(cid:11)80(cid:16)m(cid:12)
dt(cid:169) dt (cid:185)
d2m dm
0.5 (cid:14) (cid:32)k(80(cid:16)m)
dt2 dt
d2m dm
(cid:159) (cid:14)2 (cid:32)2k(80(cid:16)m)
dt2 dt
d2m dm
(cid:159) (cid:14)2 (cid:14)2km(cid:32)160k
dt2 dt | B1
E1
[2] | 3.3
2.1 | Translate the given context into a correct
mathematical model
n
e
AG
11 | (ii) | d2m dm
(cid:14)2 (cid:14)2km(cid:32)160k
dt2 dt
(cid:159)A.E. n2(cid:14)2n(cid:14)2k (cid:32)0
(cid:16)2(cid:114) 4(cid:16)8k
(cid:159)n(cid:32) (cid:32)(cid:16)1(cid:114) 1(cid:16)2k
2
CF in the formAcos2t(cid:14)Bsin2t(cid:159) p
1(cid:16)2k(cid:31)0
so 2k(cid:16)1(cid:32)2(cid:159)k (cid:32) 5 S
2
(cid:79)(cid:32)(cid:16)1 | M1
A1
e
E1
A1ft
[4] | i
c
1.1
1.1
2.2a
1.1 | m
For auxiliary equation
Need not be simplified
AG
Follow through their auxiliary equation | Or complete the square
11 | (iii) | d2m dm
(cid:14)2 (cid:14)5m(cid:32)400
dt2 dt
dm d2m
P.I. Try m(cid:32)(cid:80)(cid:159) (cid:32) (cid:32)0
dt dt2
Substituting (cid:159)5(cid:80)(cid:32)400(cid:159)(cid:80)(cid:32)80
(cid:159)G.S. m(cid:32)e(cid:16)t(cid:11)Acos2t(cid:14)Bsin2t(cid:12)(cid:14)80
m(cid:32)0 when t(cid:32)0
(cid:159)A(cid:32)(cid:16)80
dm
(cid:32)160when t(cid:32)0
dt
dm
(cid:32)(cid:16)e(cid:16)t(cid:11)Acos2t(cid:14)Bsin2t(cid:12)(cid:14)
dt
2e(cid:16)t(cid:11)(cid:16)Asin2t(cid:14)Bcos2t(cid:12)
(cid:32)160 whent(cid:32)0
p
(cid:159)(cid:16)A(cid:14)2B(cid:32)160(cid:159)B(cid:32)40
(cid:159)m(cid:32)e(cid:16)t(cid:11)40sin2t(cid:16)80cos2t(cid:12)(cid:14)80 | M1
A1
M1
A1
M1
M1
e
A1
[7] | 3.1a
1.1
3.3
1.1
3.3
i
c1.1
1.1 | Correct Particular Integral and
substituting
n
General Solution including correct (cid:80)
e
Translate the given context into an
initial condition to find A
m
Translate the given context into an
initial condition to find B
Differentiation and substitution
Final answer stated | Award first two marks if GS
stated correctly
11 | (iv) | S
As t(cid:111)(cid:102), m(cid:111)80 | B1
[1] | 3.4
11 | (v) | E.g. There is a limiting process here.
According to the model the mass of the crate
is only displayed after infinite time | B1
[1] | 3.5b
11 | (vi) | (a) | m(cid:32)e(cid:16)t(cid:11)40sin2t(cid:16)80cos2t(cid:12)(cid:14)80
dm
(cid:159) (cid:32)e(cid:16)t(cid:11)160cos2t(cid:14)120sin2t(cid:12) (cid:32)0
dt
… when t(cid:32)1.107 (cid:159)m(cid:32)106 | M1
A1
[2] | 3.1b
3.4 | Differentiation, FT their expression
from part (iii)
Set (cid:32)0and solve BC, may be implied
by 106 | OR By graphical calculator
M1 draw m(cid:32)f(t)
A1 find value of m at the first
maximum
11 | (vi) | (b) | This isthe maximum mass displayed (to 3 sf) | E1
[1] | 3.2a | n
Must state that this is the maximum
value
11 | (vii) | e.g.
d2m dm
(cid:14)2 (cid:14)5m(cid:32)500
dt2 dt | E1
[1] | 3.5c | e
m
Any differential equation of the form
d2m dm
(cid:14)2 (cid:14)am(cid:32)bwithb (cid:32)100
dt2 dt a | Accept any sensible suggestion
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Y540 Mark Scheme June 20XX
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Y540 Mark Scheme June 20XX
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Y540 Mark Scheme June 20XX
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PPMMTT
A Level Further Mathematics A
Y540 Pure Core 1
Printed Answer Booklet
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Time allowed: 1 hour 30 minutes
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11(vii) | S11 A company is required to weigh any goods before exporting them overseas. When a crate is placed on a set of weighing scales, the mass displayed takes time to settle down to its final value.
The company wishes to model the mass, $m \mathrm {~kg}$, which is displayed $t$ seconds after a crate X is placed on the scales.\\
For the displayed mass it is assumed that the rate of change of the quantity $\left( 0.5 \frac { \mathrm {~d} m } { \mathrm {~d} t } + m \right)$ with respect to time is proportional to $( 80 - m )$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\frac { \mathrm { d } ^ { 2 } m } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} m } { \mathrm {~d} t } + 2 \mathrm {~km} = 160 \mathrm { k }$, where $k$ is a real constant.
It is given that the complementary function for the differential equation in part (i) is $\mathrm { e } ^ { \lambda t } ( A \cos 2 t + B \sin 2 t )$, where $A$ and $B$ are arbitrary constants.
\item Show that $k = \frac { 5 } { 2 }$ and state the value of the constant $\lambda$.
When X is initially placed on the scales the displayed mass is zero and the rate of increase of the displayed mass is $160 \mathrm {~kg} \mathrm {~s} ^ { - 1 }$.
\item Find $m$ in terms of $t$.
\item Describe the long term behaviour of $m$.
\item With reference to your answer to part (iv), comment on a limitation of the model.
\item (a) Find the value of $m$ that corresponds to the stationary point on the curve $m = \mathrm { f } ( t )$ with the smallest positive value of $t$.\\
(b) Interpret this value of $m$ in the context of the model.
\item Adapt the differential equation $\frac { \mathrm { d } ^ { 2 } m } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} m } { \mathrm {~d} t } + 5 m = 400$ to model the mass displayed $t$ seconds after a crate Y , of mass 100 kg , is placed on the scales.
\section*{END OF QUESTION PAPER}
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\hfill \mbox{\textit{OCR Further Pure Core 1 Q11 [19]}}