| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| 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 question involving Newton's second law applied to connected particles. Part (a) requires setting up F=ma for the system, part (b) isolates the trailer to find tension, and part (c) uses constant acceleration equations. All steps are routine applications of core mechanics principles with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks |
|---|---|
| 15(a) | Models overall system as a single |
| Answer | Marks | Guidance |
|---|---|---|
| equation | 3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains fully correct equation | 1.1b | A1 |
| Obtains correct value for m | 1.1b | A1 |
| Answer | Marks |
|---|---|
| 15(b) | Models either tractor or trailer |
| Answer | Marks | Guidance |
|---|---|---|
| 11080β160βππ = 0.8Γππ | 3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone omission of units. | 1.1b | A1F |
| Answer | Marks |
|---|---|
| 15(c) | Models trailer using only resistance |
| Answer | Marks | Guidance |
|---|---|---|
| or finds s using energy. | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| (161.25) | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| consistent units | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| units CAO | 1.1b | A1 |
| Total | 9 |
Question 15:
--- 15(a) ---
15(a) | Models overall system as a single
particle using Newtonβs second law,
one side of equation correct.
If two separate equations used, must
eliminate T to obtain a single
equation | 3.3 | M1 | By ,
πΉπΉ = ππππ
11080β160β60 0 = 0.8(4ππ+ππ)
10320 = 4ππ
ππ = 2580
Obtains fully correct equation | 1.1b | A1
Obtains correct value for m | 1.1b | A1
--- 15(b) ---
15(b) | Models either tractor or trailer
separately using resistance force, T
and their m value.
Tractor:
11080β160βππ = 0.8Γππ | 3.3 | M1 | Trailer:
ππβ600 = ne0w.8t oΓns4 Γ2580
ππ = 8856
Obtains correct value for T using
their m value
Condone omission of units. | 1.1b | A1F
--- 15(c) ---
15(c) | Models trailer using only resistance
force = Β±600 and their 4m value to
find a from Newtonβs second law.
or finds s using energy. | 3.4 | M1 | β600= 10 3 2 0 ππ-2
m s
5
β ππ = β86
18 km h-1 = 5 m s-1
Using :
ππ = π’π’+πππ‘π‘
5
2.5 = 5+οΏ½β οΏ½π‘π‘
86
Time taken, seconds
π‘π‘ = 43
Finds the correct value of a or s
(161.25) | 1.1b | A1
Selects suitable suvat equation to
find required time, using their
calculated value for a or s and
consistent units | 3.4 | M1
Obtains correct value for t including
units CAO | 1.1b | A1
Total | 9A tractor and its driver have a combined mass of $m$ kilograms.
The tractor is towing a trailer of mass $4m$ kilograms in a straight line along a horizontal road.
The tractor and trailer are connected by a horizontal tow bar, modelled as a light rigid rod.
A driving force of $11080 \text{N}$ and a total resistance force of $160 \text{N}$ act on the tractor.
A total resistance force of $600 \text{N}$ acts on the trailer.
The tractor and the trailer have an acceleration of $0.8 \text{m s}^{-2}$
\begin{enumerate}[label=(\alph*)]
\item Find $m$.
[3 marks]
\item Find the tension in the tow bar.
[2 marks]
\item At the instant the speed of the tractor reaches $18 \text{km h}^{-1}$ the tow bar breaks.
The total resistance force acting on the trailer remains constant.
Starting from the instant the tow bar breaks, calculate the time taken until the speed of the trailer reduces to $9 \text{km h}^{-1}$
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2019 Q15 [9]}}