| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Car towing trailer, horizontal |
| Difficulty | Standard +0.3 This is a standard AS-level mechanics problem requiring Newton's second law applied to a two-body system. Students must write F=ma equations for both vehicles and solve simultaneously—a routine technique covered in all AS mechanics courses. The algebraic manipulation is straightforward, and part (b) asks for a standard modelling assumption. Slightly above average difficulty only due to the multi-step nature and need to handle two connected bodies, but no novel insight required. |
| Spec | 3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks |
|---|---|
| 18(a) | Uses F = m a to form at least |
| Answer | Marks | Guidance |
|---|---|---|
| least three terms. | 3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| by kD – 18 at any point. | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| equation. | 3.3 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| equation with D and T | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| OE | 1.1b | A1 |
| Subtotal | 5 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 18(b) | Describes any valid assumption. |
| Answer | Marks | Guidance |
|---|---|---|
| constant, tow bar breaks. | 3.5b | B1 |
| Subtotal | 1 | |
| Question 18 Total | 6 | |
| Question Paper Total | 80 |
Question 18:
--- 18(a) ---
18(a) | Uses F = m a to form at least
one equation modelling the van,
car or both combined with at
least three terms. | 3.3 | M1 | D – R – T = 2780 × 0.6
T – 0.6R = 1620 × 0.6
Eliminating R
0.6D – 1.6T = 28.8
0.6
T = (D – 48)
1.6
3
k =
8
Obtains a fully correct equation.
Other examples:
D – R – T = 1668
T – 0.6R = 972
D – 1.6R = 2640
NB T may have been replaced
by kD – 18 at any point. | 1.1b | A1
Forms a second fully correct
equation. | 3.3 | B1
Eliminates R to form an
equation with D and T | 3.4 | M1
3
Obtains k =
8
OE | 1.1b | A1
Subtotal | 5
Q | Marking instructions | AO | Marks | Typical solution
--- 18(b) ---
18(b) | Describes any valid assumption.
For example:
• Tow bar has negligible
mass.
• The car is directly behind
the van.
• The masses include
drivers.
• Tow bar is rigid.
• Tow bar is inextensible.
•
Do not accept any reference to
resistances, tension being
constant, tow bar breaks. | 3.5b | B1 | Tow bar is horizontal
Subtotal | 1
Question 18 Total | 6
Question Paper Total | 80A rescue van is towing a broken-down car by using a tow bar.
The van and the car are moving with a constant acceleration of $0.6 \text{ m s}^{-2}$ along a straight horizontal road as shown in the diagram below.
\includegraphics{figure_18}
The van has a total mass of 2780 kg
The car has a total mass of 1620 kg
The van experiences a driving force of $D$ newtons.
The van experiences a total resistance force of $R$ newtons.
The car experiences a total resistance force of $0.6R$ newtons.
\begin{enumerate}[label=(\alph*)]
\item The tension in the tow bar, $T$ newtons, may be modelled by
$$T = kD - 18$$
where $k$ is a constant.
Find $k$
[5 marks]
\item State one assumption that must be made in answering part (a).
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2023 Q18 [6]}}