| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Connected particles with friction |
| Difficulty | Standard +0.3 This is a standard A-level mechanics problem involving connected particles on an inclined plane with friction. Part (a) requires setting up Newton's second law equations for both masses and using the given acceleration to find the coefficient of friction—straightforward but multi-step. Part (b) involves basic kinematics after the string breaks. The trigonometry is given, and all steps follow standard textbook methods with no novel insight required. Slightly easier than average due to the 'show that' format and clear structure. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks |
|---|---|
| 18(a) | Uses F = ma to form a four-term |
| Answer | Marks | Guidance |
|---|---|---|
| sine/cosine | 3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| of weight | 3.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| weight | 3.1b | B1 |
| Recalls F =µR | 1.2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains M1, B1, B1, B1, A1 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| substituted | 3.3 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| of motion with | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| exactly | 1.1b | A1 |
| 𝜇𝜇 = 0.17Subtotal | 8 |
| Answer | Marks |
|---|---|
| 18(b)(i) | Uses F = ma with T=0 to form a |
| Answer | Marks | Guidance |
|---|---|---|
| error condone one sign error | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AWRT -4.35 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| a ≠8.522928, 8.688 | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| CAO | 3.2a | A1 |
| Subtotal | 4 |
| Answer | Marks |
|---|---|
| 18(b)(ii) | Describes assumption in context |
| Answer | Marks | Guidance |
|---|---|---|
| String does not obstruct block | 3.5b | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Question Total | 13 | |
| Q | Marking Instructions | AO |
Question 18:
--- 18(a) ---
18(a) | Uses F = ma to form a four-term
equation with at least three
correct terms
condone consistent swapping of
sine/cosine | 3.3 | M1 | T −Wsinθ−Friction=ma
T −mgsinθ−µmgcosθ=ma
𝑇𝑇−0.2𝘨𝘨(0.28)−0.2𝜇𝜇𝘨𝘨(0.96)
543
= 0.2× 𝘨𝘨
625
543
2𝘨𝘨−𝑇𝑇 = 2× 𝘨𝘨
625
2𝘨𝘨−0.2𝘨𝘨(0.28)−0.2𝜇𝜇𝘨𝘨(0.96)
543
= 2.2× 𝘨𝘨
625
0.192𝜇𝜇 = 2−0.056− 1.911(36)
𝜇𝜇 = 0.17
Obtains perpendicular component
of weight | 3.1b | B1
Obtains parallel component of
weight | 3.1b | B1
Recalls F =µR | 1.2 | B1
Forms fully correct equation for
particle A
T −mgsinθ−µmgcosθ=ma
Obtains M1, B1, B1, B1, A1 | 1.1b | A1
Forms fully correct equation for
particle B
or with acceleration
substituted | 3.3 | B1
2E𝘨𝘨lim−in𝑇𝑇a=tes2 𝑎𝑎 from their equations
a≠0
of motion with | 3.4 | M1
𝑇𝑇
Shows at least one step leading to
given value AWRT 0.170
If exact values used throughout
with g=𝜇𝜇9.81 then
exactly | 1.1b | A1
𝜇𝜇 = 0.17Subtotal | 8
--- 18(b)(i) ---
18(b)(i) | Uses F = ma with T=0 to form a
three-term equation with at least
two correct terms
condone consistent sine/cosine
error condone one sign error | 3.4 | M1 | −0.2×9.81×(0.28)
− 0.2(0.17)×9.81
×(0.96)=0.2𝑎𝑎
Us𝑎𝑎e= −4.347792
2 2
𝑣𝑣 =𝑢𝑢 +2𝑎𝑎𝑠𝑠
0=0.25−8.696𝑠𝑠
m
𝑠𝑠 = 0.0288
Obtains a= -4.347792
AWRT -4.35 | 1.1b | A1
Uses with u=0.5 and
v=0 and2 the2ir value
𝑣𝑣 =𝑢𝑢 +2𝑎𝑎𝑠𝑠
Do not accept the following values
for a 𝑎𝑎
543
a≠0, g,8.51424
625
a ≠8.522928, 8.688 | 3.4 | M1
Obtains 0.0288 m
Must have units
CAO | 3.2a | A1
Subtotal | 4
--- 18(b)(ii) ---
18(b)(ii) | Describes assumption in context
of part (b)(i) eg No air resistance,
String does not obstruct block | 3.5b | E1 | does not reach the pulley before
coming to rest
𝐴𝐴
Subtotal | 1
Question Total | 13
Q | Marking Instructions | AO | Marks | Typical SolutionBlock $A$, of mass $0.2$ kg, lies at rest on a rough plane.
The plane is inclined at an angle $\theta$ to the horizontal, such that $\tan \theta = \frac{7}{24}$
A light inextensible string is attached to $A$ and runs parallel to the line of greatest slope until it passes over a smooth fixed pulley at the top of the slope.
The other end of this string is attached to particle $B$, of mass $2$ kg, which is held at rest so that the string is taut, as shown in the diagram below.
\includegraphics{figure_18}
\begin{enumerate}[label=(\alph*)]
\item $B$ is released from rest so that it begins to move vertically downwards with an acceleration of $\frac{543}{625}$ g ms$^{-2}$
Show that the coefficient of friction between $A$ and the surface of the inclined plane is $0.17$
[8 marks]
\item In this question use $g = 9.81\text{ ms}^{-2}$
When $A$ reaches a speed of $0.5\text{ ms}^{-1}$ the string breaks.
\begin{enumerate}[label=(\roman*)]
\item Find the distance travelled by $A$ after the string breaks until first coming to rest.
[4 marks]
\item State an assumption that could affect the validity of your answer to part (b)(i).
[1 mark]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2020 Q18 [13]}}