| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Year | 2022 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Work done by non-constant force integration |
| Difficulty | Standard +0.3 This is a straightforward application of work-energy principles with integration. Part (a) requires integrating F with respect to x from 0 to 0.2, which is routine calculus. Part (b) uses conservation of energy (initial KE = work done) to solve for A. The setup is clear, the mathematics is standard A-level integration, and no novel insight is required—slightly easier than average for Further Maths mechanics. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)6.02c Work by variable force: using integration6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| \(WD = \int_0^{0.2}(Ax + 9000x^2)\,dx = \left[\frac{Ax^2}{2} + \frac{9000x^3}{3}\right]_0^{0.2}\) | M1 | Uses integration to find work done; condone missing or incorrect limits |
| \(= \frac{A}{50} + 24\) | A1 | Obtains correct simplified expression for work done in terms of \(A\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2} \times 10000 \times 0.3^2 = \frac{A}{50} + 24 \Rightarrow 450 = \frac{A}{50} + 24\) | M1 | Uses their expression for work done from (a) and the correct KE to form an equation in terms of \(A\) |
| \(A = 21300\) | A1F | Obtains correct value for \(A\) from their expression in part (a), provided their expression is only in terms of \(A\) |
## Question 5(a):
$WD = \int_0^{0.2}(Ax + 9000x^2)\,dx = \left[\frac{Ax^2}{2} + \frac{9000x^3}{3}\right]_0^{0.2}$ | M1 | Uses integration to find work done; condone missing or incorrect limits
$= \frac{A}{50} + 24$ | A1 | Obtains correct simplified expression for work done in terms of $A$
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## Question 5(b):
$\frac{1}{2} \times 10000 \times 0.3^2 = \frac{A}{50} + 24 \Rightarrow 450 = \frac{A}{50} + 24$ | M1 | Uses their expression for work done from (a) and the correct KE to form an equation in terms of $A$
$A = 21300$ | A1F | Obtains correct value for $A$ from their expression in part (a), provided their expression is only in terms of $A$
---5 A train of mass 10000 kg is travelling at $0.3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it collides with a buffer. The buffer brings the train to rest.
As the buffer brings the train to rest it compresses by 0.2 metres.\\
When the buffer is compressed by a distance of $x$ metres it exerts a force of magnitude $F$ newtons, where
$$F = A x + 9000 x ^ { 2 }$$
where $A$ is a constant.
5
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $A$, the work done in compressing the buffer by 0.2 metres.\\
5
\item Find the value of $A$
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2022 Q5 [4]}}