| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Horizontal road towing |
| Difficulty | Moderate -0.3 This is a standard A-level mechanics problem involving Newton's second law applied to connected particles. Part (a) requires applying F=ma to the system and trailer separately (routine calculation), part (b)(i) needs analyzing forces after the string breaks (standard technique), and part (b)(ii) uses basic kinematics. All steps are textbook applications with no novel insight required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors3.03o Advanced connected particles: and pulleys |
| 19 (b) (ii) Find \(h\) | Do not write outside the box |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolves the 2 N force to obtain either \(2\cos 40\) AWRT 1.53 or \(2\sin 40\) AWRT 1.29 | M1 | AO 1.1a; may be seen on diagram |
| Use \(F = ma\) for system: \(2\cos 40 - (0.8 + R) = 2.2(0.06)\) | M1 | AO 3.3; four-term equation for whole system; condone one incorrect sign |
| \(1.53 - 0.8 - R = 0.132\) | A1 | AO 3.3; fully correct equation for whole system, or two fully correct equations for engine and trailer e.g. \(2\cos 40 - T - 0.8 = 0.09\); \(T - R = 0.042\) |
| \(R \approx 0.6\) N | R1 | AO 2.1; completes reasoned argument AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Forms equation of motion without driving force for engine or combined system; PI by \(a = \dfrac{7}{11}\) or \(-\dfrac{7}{11}\) | M1 | AO 3.3 |
| Obtains one of: \(\pm(T - 0.6) = 0.7a\); \(\mp(0.8 + T) = 1.5a\); \(\pm(0.8 + 0.6) = 2.2a\); allow use of \(a = -\dfrac{7}{11}\) | M1 | AO 3.4 |
| \(-0.8 - T = 1.5a\); \(T - 0.6 = 0.7a\); correct pair of equations both moving same direction | A1 | AO 1.1b |
| \(T = \dfrac{17}{110}\) N; AWFW \([0.15, 0.16]\) | A1 | AO 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = -\dfrac{7}{11}\); AWRT \(-0.64\); condone missing/incorrect units | B1 | AO 3.1b |
| \(0 = 0.5^2 + 2ah\); selects appropriate equation of constant acceleration, substitutes \(u = 0.5\), \(v = 0\), and their \(a\); do not accept \(a = -g\) | M1 | AO 1.1a |
| \(h = \dfrac{11}{56} \approx 0.20\); AWRT 0.2; ISW | A1 | AO 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| States one appropriate modelling assumption about the rod e.g. "The rod is horizontal" | E1 | Accept "Rod is rigid" OE |
| Subtotal: 1 mark | Question 19 Total: 12 marks |
## Question 19(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolves the 2 N force to obtain either $2\cos 40$ AWRT 1.53 or $2\sin 40$ AWRT 1.29 | M1 | AO 1.1a; may be seen on diagram |
| Use $F = ma$ for system: $2\cos 40 - (0.8 + R) = 2.2(0.06)$ | M1 | AO 3.3; four-term equation for whole system; condone one incorrect sign |
| $1.53 - 0.8 - R = 0.132$ | A1 | AO 3.3; fully correct equation for whole system, or two fully correct equations for engine and trailer e.g. $2\cos 40 - T - 0.8 = 0.09$; $T - R = 0.042$ |
| $R \approx 0.6$ N | R1 | AO 2.1; completes reasoned argument AG |
**Subtotal: 4 marks**
---
## Question 19(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Forms equation of motion **without** driving force for engine or combined system; PI by $a = \dfrac{7}{11}$ or $-\dfrac{7}{11}$ | M1 | AO 3.3 |
| Obtains one of: $\pm(T - 0.6) = 0.7a$; $\mp(0.8 + T) = 1.5a$; $\pm(0.8 + 0.6) = 2.2a$; allow use of $a = -\dfrac{7}{11}$ | M1 | AO 3.4 |
| $-0.8 - T = 1.5a$; $T - 0.6 = 0.7a$; correct pair of equations both moving same direction | A1 | AO 1.1b |
| $T = \dfrac{17}{110}$ N; AWFW $[0.15, 0.16]$ | A1 | AO 1.1b |
**Subtotal: 4 marks**
---
## Question 19(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = -\dfrac{7}{11}$; AWRT $-0.64$; condone missing/incorrect units | B1 | AO 3.1b |
| $0 = 0.5^2 + 2ah$; selects appropriate equation of constant acceleration, substitutes $u = 0.5$, $v = 0$, and their $a$; do not accept $a = -g$ | M1 | AO 1.1a |
| $h = \dfrac{11}{56} \approx 0.20$; AWRT 0.2; ISW | A1 | AO 1.1b |
**Subtotal: 3 marks**
## Question 19(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| States one appropriate modelling assumption about the rod e.g. "The rod is horizontal" | E1 | Accept "Rod is rigid" OE |
**Subtotal: 1 mark | Question 19 Total: 12 marks**
---19 A wooden toy comprises a train engine and a trailer connected to each other by a light, inextensible rod.
The train engine has a mass of 1.5 kilograms.\\
The trailer has a mass 0.7 kilograms.\\
A string inclined at an angle of $40 ^ { \circ }$ above the horizontal is attached to the front of the train engine.
The tension in the string is 2 newtons.\\
As a result the toy moves forward, from rest, in a straight line along a horizontal surface with acceleration $0.06 \mathrm {~ms} ^ { - 2 }$ as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{de8a7d38-a665-4feb-854e-ac83f413d133-30_373_789_904_756}
As it moves the train engine experiences a total resistance force of 0.8 N\\
19
\begin{enumerate}[label=(\alph*)]
\item Show that the total resistance force experienced by the trailer is approximately 0.6 N\\
19
\item At the instant that the toy reaches a speed of $0.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ the string breaks.
As a result of this the train engine and trailer decelerate at a constant rate until they come to rest, having travelled a distance of $h$ metres.
It can be assumed that the resistance forces remain unchanged.\\
19 (b) (i) Find the tension in the rod after the string has broken.\\
\begin{center}
\begin{tabular}{|l|l|}
\hline
19 (b) (ii) Find $h$ & Do not write outside the box \\
\hline
\end{tabular}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{de8a7d38-a665-4feb-854e-ac83f413d133-33_2488_1716_219_153}
\end{center}
Nell and her pet dog Maia are visiting the beach.\\
The beach surface can be assumed to be level and horizontal.
Nell and Maia are initially standing next to each other.\\
Nell throws a ball forward, from a height of 1.8 metres above the surface of the beach, at an angle of $60 ^ { \circ }$ above the horizontal with a speed of $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
Exactly 0.2 seconds after the ball is thrown, Maia sets off from Nell and runs across the surface of the beach, in a straight line with a constant acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$
Maia catches the ball when it is 0.3 metres above ground level as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{de8a7d38-a665-4feb-854e-ac83f413d133-34_778_1287_1027_463}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2023 Q19 [12]}}