| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string with compression (spring) |
| Difficulty | Standard +0.3 This is a standard M2 elastic string/spring problem with equilibrium, energy conservation, and solving a quadratic. Part (a) is routine Hooke's law, part (b) is direct formula application, and part (c) requires energy conservation but the equation is given to show. The quadratic solving is straightforward. Slightly above average due to the multi-step energy work, but still a typical textbook M2 question. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(2g = \frac{49 \times x}{0.5}\) giving \(x = 0.2\) | M1, A1, A1 | |
| (b) \(EPE = \frac{49 \times (0.2)^2}{2 \times 0.5} = 1.96 \text{ (J)}\) | M1, A1 | |
| (c)(i) \(1.96 = \frac{49 \times x^2}{2 \times 0.5} + 0.8 \times 9.8 \times (0.2 + x)\) giving \(x^2 + 0.16x - 0.008 = 0\) | M1, A3, A1 | All terms attempted for M1; -1 EE from A3 |
| (ii) \(x = \frac{0.16 \pm \sqrt{0.16^2 + 4 \times 0.008}}{2}\) giving \(x = 0.04\) | M1, A1 | \(x = 0.04\) only identified |
**(a)** $2g = \frac{49 \times x}{0.5}$ giving $x = 0.2$ | M1, A1, A1 | | **Total: 3 marks**
**(b)** $EPE = \frac{49 \times (0.2)^2}{2 \times 0.5} = 1.96 \text{ (J)}$ | M1, A1 | | **Total: 2 marks**
**(c)(i)** $1.96 = \frac{49 \times x^2}{2 \times 0.5} + 0.8 \times 9.8 \times (0.2 + x)$ giving $x^2 + 0.16x - 0.008 = 0$ | M1, A3, A1 | All terms attempted for M1; -1 EE from A3 | **Total: 5 marks**
**(ii)** $x = \frac{0.16 \pm \sqrt{0.16^2 + 4 \times 0.008}}{2}$ giving $x = 0.04$ | M1, A1 | $x = 0.04$ only identified | **Total: 2 marks**
**Question 8 Total: 12 marks**
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**GRAND TOTAL: 75 marks**8 Two small blocks, $A$ and $B$, of masses 0.8 kg and 1.2 kg respectively, are stuck together. A spring has natural length 0.5 metres and modulus of elasticity 49 N . One end of the spring is attached to the top of the block $A$ and the other end of the spring is attached to a fixed point $O$.
\begin{enumerate}[label=(\alph*)]
\item The system hangs in equilibrium with the blocks stuck together, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{480a817d-074f-440d-829e-c8f8a9746151-8_385_239_669_881}
Find the extension of the spring.
\item Show that the elastic potential energy of the spring when the system is in equilibrium is 1.96 J .
\item The system is hanging in this equilibrium position when block $B$ falls off and block $A$ begins to move vertically upwards.
Block $A$ next comes to rest when the spring is compressed by $x$ metres.
\begin{enumerate}[label=(\roman*)]
\item Show that $x$ satisfies the equation
$$x ^ { 2 } + 0.16 x - 0.008 = 0$$
\item Find the value of $x$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M2 2007 Q8 [12]}}