| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Two particles over pulley, vertical strings |
| Difficulty | Moderate -0.3 This is a standard connected particles problem requiring two equations of motion (Newton's second law for each mass) and solving simultaneously. Part (a) is routine mechanics bookwork, part (b) applies v=u+at with given acceleration, and part (c) asks for a standard assumption. While it requires careful setup and algebra, it follows a well-practiced template with no novel insight needed, making it slightly easier than average. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03f Weight: W=mg3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks |
|---|---|
| 19(a) | Uses F =ma to form a three- |
| Answer | Marks | Guidance |
|---|---|---|
| M or N with one side correct | 3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| modelling M or N | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| correct | 3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| term equations for M and N | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | 2.1 | R1 |
| Subtotal | 5 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 19(b) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| AWRT 0.89 | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| stated | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 19(c) | States reasonable assumption |
| Answer | Marks | Guidance |
|---|---|---|
| floor, the string breaks OE | 3.5b | E1 |
| Subtotal | 1 | |
| Question 19 Total | 8 | |
| Question Paper Total | 80 |
Question 19:
--- 19(a) ---
19(a) | Uses F =ma to form a three-
term equation modelling object
M or N with one side correct | 3.3 | M1 | Model for M
0.6g−T =0.6a
Model for N
T −0.5g =0.5a
So
0.1g =1.1a
1
a = g
11
Finds fully correct equation for
modelling M or N | 1.1b | A1
Uses F =ma to form a three-
term equation for modelling the
other object with one side
correct | 3.3 | M1
Eliminates T to find an equation
in terms of a using their three-
term equations for M and N | 1.1a | M1
Completes reasoned argument
1
to show a = g
11 | 2.1 | R1
Subtotal | 5
Q | Marking instructions | AO | Marks | Typical solution
--- 19(b) ---
19(b) | 1
Uses v=u+at with a= g or
11
AWRT 0.89 | 3.4 | M1 | Using v=u+at
1 g
v=0+ g(0.5)=
11 22
k = 22
Obtains k = 22
AWRT 22
g
Accept v= if no value of k
22
stated | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 19(c) ---
19(c) | States reasonable assumption
Accept M does not reach the
floor, the string breaks OE | 3.5b | E1 | N does not reach the peg.
Subtotal | 1
Question 19 Total | 8
Question Paper Total | 80Two objects, $M$ and $N$, are connected by a light inextensible string that passes over a smooth peg.
$M$ has a mass of 0.6 kilograms.
$N$ has a mass of 0.5 kilograms.
$M$ and $N$ are initially held at rest, with the string taut, as shown in the diagram below.
\includegraphics{figure_19}
$M$ and $N$ are released at the same instant and begin to move vertically.
You may assume that air resistance can be ignored.
\begin{enumerate}[label=(\alph*)]
\item It is given that $M$ and $N$ move with acceleration $a$ m s$^{-2}$
By forming two equations of motion show that
$$a = \frac{1}{11}g$$
[5 marks]
\item The speed of $N$, 0.5 seconds after its release, is $\frac{g}{k}$ m s$^{-1}$ where $k$ is a constant.
Find the value of $k$
[2 marks]
\item State one assumption that must be made for the answer in part (b) to be valid.
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2024 Q19 [8]}}