| Answer | Marks | Guidance |
|---|---|---|
| 21(a) | States that the resultant force is | |
| 80g – T and states F = ma | 3.4 | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 21(b) | States 50g cos60 = 25g OE | 3.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 21(c) | Resolves parallel to the slope to |
| Answer | Marks | Guidance |
|---|---|---|
| OE | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| for friction | 3.3 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| errors | 3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| where µ=1 is stated | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone use of T = 80a + 80g | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | 2.1 | R1 |
| Subtotal | 6 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 21(d) | States any one of the following |
| Answer | Marks | Guidance |
|---|---|---|
| • No air resistance | 3.5b | E1 |
| Subtotal | 1 | |
| Question 21 Total | 9 | |
| Question Paper Total | 100 |
Question 21:
--- 21(a) ---
21(a) | States that the resultant force is
80g – T and states F = ma | 3.4 | E1 | 80g−T is the resultant force and as
F =ma, 80g – T = 80a
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 21(b) ---
21(b) | States 50g cos60 = 25g OE | 3.1b | B1 | Resolve perpendicular to slope
R=50gcos60=25g
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 21(c) ---
21(c) | Resolves parallel to the slope to
obtain 50g sin 60 or better PI
OE | 1.1b | B1 | parallel to slope
T – 50g sin60 – F = 50a
F =25gµ
T – 50g sin60 – 25gµ= 50a
Using given equation for N:
80g – T = 80a
80g – 50g sin60 – 25gµ = 130a
80g−50gsin60−25gµ
a =
130
( )
16−5 3−5µ g
a =
26
Acceleration will be at its minimum
when µ=1
Therefore
(11−5 3)g
a≥
26
Obtains µ×25gfor friction OE
Or
States µ=1 and obtains 25g
for friction | 3.3 | B1
Forms a three or four-term
equation of motion for M using
F = ma parallel to slope
Condone cos60 OE and sign
errors | 3.3 | M1
Obtains single correct equation
µ
or inequality with
Or
T – 50g sin60 – 25g = 50a
where µ=1 is stated
Or
T – 50g sin60 – 25g ≤50a
where µ=1 is stated | 1.1b | A1
Eliminates T using the equation
in 21 (a) and their equation or
µ
inequality with
Condone use of T = 80a + 80g | 1.1a | M1
Completes reasoned argument
(11−5 3)g
to show a≥
26
Must include clear reason for
using µ=1 or uses 0≤µ≤1or
uses µ≤1
AG | 2.1 | R1
Subtotal | 6
Q | Marking instructions | AO | Marks | Typical solution
--- 21(d) ---
21(d) | States any one of the following
valid assumptions
• Rope is light
• rope inextensible
• pulley is fixed
• No air resistance | 3.5b | E1 | Rope has no mass
Subtotal | 1
Question 21 Total | 9
Question Paper Total | 100Two heavy boxes, $M$ and $N$, are connected securely by a length of rope.
The mass of $M$ is 50 kilograms.
The mass of $N$ is 80 kilograms.
$M$ is placed near the bottom of a rough slope.
The slope is inclined at 60° above the horizontal.
The rope is passed over a smooth pulley at the top end of the slope so that $N$ hangs with the rope vertical.
The boxes are initially held in this position, with the rope taut and running parallel to the line of greatest slope, as shown in the diagram below.
\includegraphics{figure_21}
When the boxes are released, $M$ moves up the slope as $N$ descends, with acceleration $a$ m s$^{-2}$
The tension in the rope is $T$ newtons.
\begin{enumerate}[label=(\alph*)]
\item Explain why the equation of motion for $N$ is
$$80g - T = 80a$$
[1 mark]
\item Show that the normal reaction force between $M$ and the slope is $25g$ newtons.
[1 mark]
\item The coefficient of friction, $\mu$, between the slope and $M$ is such that $0 \leq \mu \leq 1$
Show that
$$a \geq \frac{(11 - 5\sqrt{3})g}{26}$$
[6 marks]
\item State one modelling assumption you have made throughout this question.
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2024 Q21 [9]}}