| Exam Board | AQA |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Horizontal elastic string on rough surface |
| Difficulty | Challenging +1.8 This is a substantial Further Maths mechanics problem requiring elastic string tensions, friction analysis, and differential equations. Part (a) is routine verification. Part (b) requires setting up forces and deriving a second-order DE with friction. Part (c) involves solving a damped harmonic motion problem with particular integral—challenging but follows standard Further Maths techniques without requiring exceptional insight. |
| Spec | 4.10g Damped oscillations: model and interpret6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| 15(a) | Uses a rigorous argument to | |
| obtain the required result | 2.1 | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| Total | 1 | |
| Q | Marking Instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 15(b) | Obtains one correct tension | 1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| terms | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| equation | 2.1 | R1 |
| Total | 3 | |
| Q | Marking Instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 15(c) | Obtains correct 2nd order DE | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| three term Auxiliary Equation | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Complementary Function | 1.1b | A1F |
| Answer | Marks | Guidance |
|---|---|---|
| Particular Integral ACF | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| unknowns) | 2.2a | A1F |
| Answer | Marks | Guidance |
|---|---|---|
| obtain correctA ACF | 3.3 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| t =0 | 1.1a | M1 |
| Obtains correct B ACF | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| ACF | 2.1 | R1 |
| Total | 9 | |
| Question total | 13 | |
| Paper total | 100 |
Question 15:
--- 15(a) ---
15(a) | Uses a rigorous argument to
obtain the required result | 2.1 | R1 | Tension in AP = 24m(0.5)=12m
Tension in BP = 10m(1.2)=12m
So tensions are equal
Total | 1
Q | Marking Instructions | AO | Marks | Typical solution
--- 15(b) ---
15(b) | Obtains one correct tension | 1.1a | B1 | 2
d 𝑥𝑥
𝑚𝑚 2 = 10𝑚𝑚(1.2−𝑥𝑥)−24𝑚𝑚(0.5+𝑥𝑥)
d𝑡𝑡
+6.664𝑚𝑚
2
d 𝑥𝑥
2
d𝑡𝑡 +34𝑥𝑥 = 6.664
Uses Newton’s second law to
form a four term differential
equation with at least two terms
correct (allow equivalent
notation for derivatives)
Condone sign errors on the
terms | 3.1b | M1
Completes a rigorous argument
to give the required differential
equation | 2.1 | R1
Total | 3
Q | Marking Instructions | AO | Marks | Typical solution
--- 15(c) ---
15(c) | Obtains correct 2nd order DE | 2.2a | B1 | But
So 𝑚𝑚𝑥𝑥̈ = 10𝑚𝑚𝑣𝑣+6.664𝑚𝑚−34𝑚𝑚𝑥𝑥
𝑣𝑣 = −𝑥𝑥̇
λ2 + 10λ+34=0
𝑥𝑥̈ +10𝑥𝑥̇ + 34𝑥𝑥 = 6.664
λ=−5±3i
CF:
x = Ae−5tcos3t+Be−5tsin3t
PI:
Gen𝑥𝑥e=ral0 S.1o9l6ution:
x = Ae−5tcos3t+Be−5tsin3t+ .
0 196
𝑡𝑡−5=tc 0 , 𝑥𝑥 = 0 . 4 ⟹−5ts 𝐴𝐴 = 0.204
−5Ae o s 3t − 3 A e in 3t
𝑥𝑥 ̇ =
−5Be−5tsin3t+3Be−5tcos3t
0 = −5𝐴𝐴+3𝐵𝐵
𝐵𝐵 = 0.34
−5𝑡𝑡 −5𝑡𝑡
𝑥𝑥 = 0.204e cos3𝑡𝑡+0.34e sin3𝑡𝑡
+0.196
Obtains correct solution to their
three term Auxiliary Equation | 1.1a | M1
Obtains their correct
Complementary Function | 1.1b | A1F
Obtains correct
Particular Integral ACF | 1.1b | B1
Obtains correct general solution
(ft their CF, but must have two
unknowns) | 2.2a | A1F
Uses x =0.4 when t =0 to
obtain correctA ACF | 3.3 | B1
Sets their correct x =0 when
t =0 | 1.1a | M1
Obtains correct B ACF | 1.1b | A1
Obtains correct final equation
ACF | 2.1 | R1
Total | 9
Question total | 13
Paper total | 100In this question use $g = 9.8$ m s$^{-2}$
A particle $P$ of mass $m$ is attached to two light elastic strings, $AP$ and $BP$.
The other ends of the strings, $A$ and $B$, are attached to fixed points which are 4 metres apart on a rough horizontal surface at the bottom of a container.
The coefficient of friction between $P$ and the surface is 0.68
• When the extension of string $AP$ is $e_A$ metres, the tension in $AP$ is $24me_A$
• When the extension of string $BP$ is $e_B$ metres, the tension in $BP$ is $10me_B$
• The natural length of string $AP$ is 1 metre
• The natural length of string $BP$ is 1.3 metres
\includegraphics{figure_15}
\begin{enumerate}[label=(\alph*)]
\item Show that when $AP = 1.5$ metres, the tension in $AP$ is equal to the tension in $BP$.
[1 mark]
\item $P$ is held at the point between $A$ and $B$ where $AP = 1.9$ metres, and then released from rest.
At time $t$ seconds after $P$ is released, $AP = (1.5 + x)$ metres.
\includegraphics{figure_15b}
Show that when $P$ is moving towards $A$,
$$\frac{\mathrm{d}^2 x}{\mathrm{d}t^2} + 34x = 6.664$$
[3 marks]
\item The container is then filled with oil, and $P$ is again released from rest at the point between $A$ and $B$ where $AP = 1.9$ metres.
At time $t$ seconds after $P$ is released, the oil causes a resistive force of magnitude $10mv$ newtons to act on the particle, where $v$ m s$^{-1}$ is the speed of the particle.
Find $x$ in terms of $t$ when $P$ is moving towards $A$.
[9 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 1 2021 Q15 [13]}}