| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| 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 vertical spring problem with equilibrium, energy conservation, and solving a quadratic. Part (a) uses Hooke's law at equilibrium (routine), part (b) is a straightforward EPE calculation, and part (c) applies energy conservation between two positions. The equation is given in (c)(i) making it a 'show that' rather than derivation from scratch. All techniques are standard M2 fare with no novel insight required, making it slightly easier than average. |
| 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 |
|---|---|---|
| Working/Answer | Marks | Guidance |
| EPE \(= \frac{\lambda(x-26)^2}{2l} = \frac{1456(x-26)^2}{52}\) | B1 | Correct EPE expression for \(x \geq 26\) |
| Energy conservation: \(\frac{1}{2}(70)v^2 = 70(9.8)x - \frac{1456(x-26)^2}{52}\) | M1 A1 | KE = loss in PE − EPE gained |
| \(35v^2 = 686x - 28(x-26)^2\) | M1 | Expanding and simplifying |
| \(5v^2 = 306x - 4x^2 - 2704\) | A1 | Correct result shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| The cord is slack (not taut) when \(x < 26\), so no elastic force acts | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Maximum \(x\) when \(v = 0\): \(4x^2 - 306x + 2704 = 0\) | M1 | Setting \(v=0\) |
| \(x = \frac{306 \pm \sqrt{306^2 - 4(4)(2704)}}{8} = 56.5\) m | A1 | Correct value (taking larger root) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Max speed when \(\frac{d(v^2)}{dx} = 0\): \(306 - 8x = 0\), \(x = 38.25\) m | M1 A1 | Differentiating and solving |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(5v^2 = 306(38.25) - 4(38.25)^2 - 2704\), \(v = 24.1\ \text{ms}^{-1}\) | B1 | Correct substitution and answer |
# Question 8:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| EPE $= \frac{\lambda(x-26)^2}{2l} = \frac{1456(x-26)^2}{52}$ | B1 | Correct EPE expression for $x \geq 26$ |
| Energy conservation: $\frac{1}{2}(70)v^2 = 70(9.8)x - \frac{1456(x-26)^2}{52}$ | M1 A1 | KE = loss in PE − EPE gained |
| $35v^2 = 686x - 28(x-26)^2$ | M1 | Expanding and simplifying |
| $5v^2 = 306x - 4x^2 - 2704$ | A1 | Correct result shown |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| The cord is slack (not taut) when $x < 26$, so no elastic force acts | B1 | |
## Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Maximum $x$ when $v = 0$: $4x^2 - 306x + 2704 = 0$ | M1 | Setting $v=0$ |
| $x = \frac{306 \pm \sqrt{306^2 - 4(4)(2704)}}{8} = 56.5$ m | A1 | Correct value (taking larger root) |
## Part (d)(i):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Max speed when $\frac{d(v^2)}{dx} = 0$: $306 - 8x = 0$, $x = 38.25$ m | M1 A1 | Differentiating and solving |
## Part (d)(ii):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $5v^2 = 306(38.25) - 4(38.25)^2 - 2704$, $v = 24.1\ \text{ms}^{-1}$ | B1 | Correct substitution and answer |
---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]{88aec6ab-af83-4d5e-84b6-5fd84c16a6c9-017_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 Q8}}