| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Beam on point of tilting |
| Difficulty | Moderate -0.3 This is a straightforward statics problem involving moments about a pivot. Part (a) requires a simple moment equation with one unknown. Part (b) involves finding the maximum number of objects before tipping, requiring careful consideration of the rod's center of mass position. The concepts are standard A-level mechanics with no novel insight required, though the multi-step nature and need to identify the correct pivot point elevates it slightly above pure recall. |
| Spec | 3.04b Equilibrium: zero resultant moment and force3.04c Use moments: beams, ladders, static problems |
| Answer | Marks |
|---|---|
| 14(a) | Finds a moment of a force about |
| Answer | Marks | Guidance |
|---|---|---|
| PI by fully correct equation | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0.06(0.28g + mg) | 3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | 1.1b | A1 |
| Answer | Marks |
|---|---|
| 14(b) | Forms a moments equatioππn for |
| Answer | Marks | Guidance |
|---|---|---|
| throughout part 14(b) | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0.06R = 0.00048ng + 0.1mg | 3.4 | A1F |
| Answer | Marks | Guidance |
|---|---|---|
| expression for R | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| CSO | 1.1b | A1 |
| Answer | Marks |
|---|---|
| 14(c) | States an assumption about the |
| Answer | Marks | Guidance |
|---|---|---|
| The rod is rigid | 3.5b | E1 |
| Total | 8 | |
| Q | Marking Instructions | AO |
Question 14:
--- 14(a) ---
14(a) | Finds a moment of a force about
any point. Must have the form
force x distance
Can be awarded for 6R
PI by fully correct equation | 1.1b | B1 | Take moments about A
πππ¨π¨Γ0.04= 0.28π¨π¨Γ0.03
= 0.21
ππ
Forms a fully correct moments
equation using the correct model
Must have included g on both sides
Moments about B gives (in metres)
0.28g(0.03) + 0.1mg =
0.06(0.28g + mg) | 3.3 | M1
Solves equation to show = 0.21
AG | 1.1b | A1
--- 14(b) ---
14(b) | Forms a moments equatioππn for
equilibrium of rod with correct
number of terms β can use m, 0.21
or their value for m from part 14(a)
Condone omission of g
throughout part 14(b) | 3.1b | M1 | Take moments about A
0.21π¨π¨Γ0.04= 0.048π¨π¨Γ0.05Γππ
ππ = 3.5
Maximum n
= 3
Forms a moments equation for
equilibrium of rod with term
involving n correct β can use m,
0.21 or their m value from part
14(a)
FT their incorrect m
Moments about B gives
0.06R = 0.00048ng + 0.1mg | 3.4 | A1F
Obtains a fully correct moments
equation with substituted
Moments abouππt B= g0i.v2e1s
0.06(0.21g + 0.048ng) =
0.00048ng + 0.1(0.21)g
Must have substituted correct
expression for R | 1.1b | A1
States n = 3
CSO | 1.1b | A1
--- 14(c) ---
14(c) | States an assumption about the
rod
Accept
The mass/weight of the rod acts in
the middle
The rod is in limiting equilibrium OE
The rod is rigid | 3.5b | E1 | The rod is uniform
Total | 8
Q | Marking Instructions | AO | Marks | Typical SolutionA metal rod, of mass $m$ kilograms and length 20 cm, lies at rest on a horizontal shelf.
The end of the rod, $B$, extends 6 cm beyond the edge of the shelf, $A$, as shown in the diagram below.
\includegraphics{figure_14}
\begin{enumerate}[label=(\alph*)]
\item The rod is in equilibrium when an object of mass 0.28 kilograms hangs from the midpoint of $AB$.
Show that $m = 0.21$ [3 marks]
\item The object of mass 0.28 kilograms is removed.
A number, $n$, of identical objects, each of mass 0.048 kg, are hung from the rod all at a distance of 1 cm from $B$.
Find the maximum value of $n$ such that the rod remains horizontal. [4 marks]
\item State one assumption you have made about the rod. [1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2019 Q14 [8]}}