| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: projected from equilibrium or other point |
| Difficulty | Standard +0.3 This is a standard M2 elastic string problem requiring Hooke's law for equilibrium, EPE calculation, and energy conservation. All steps follow routine procedures with clear signposting. The algebra is straightforward and the question structure is typical of textbook exercises, making it slightly easier than average A-level standard. |
| 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 |
| Newton's second law: \(72g - 240v = 72\frac{dv}{dt}\) | M1 | Applying N2L with weight and resistance, allow sign errors |
| \(720 - 2400v = 720\frac{dv}{dt}\), divide by \(-240\): \(-\frac{3}{10}\frac{dv}{dt} = v - 2.94\) | A1 | Correct equation shown, using \(g = 9.8\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Separate variables: \(\int \frac{dv}{v - 2.94} = -\frac{10}{3}\int dt\) | M1 | Separating variables correctly |
| \(\ln | v - 2.94 | = -\frac{10}{3}t + c\) |
| At \(t=0\), \(v=30\): \(c = \ln(27.06)\) | M1 | Using initial condition |
| \(v - 2.94 = 27.06\,e^{-\frac{10}{3}t}\) | A1 | Correct equation |
| \(v = 2.94 + 27.06\,e^{-\frac{10t}{3}}\) | A1 | Final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Graph starts at \(v=30\) when \(t=0\) | B1 | Correct starting point marked |
| Curve decreasing asymptotically toward \(v = 2.94\) | B1 | Correct shape with asymptote indicated |
# Question 7:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Newton's second law: $72g - 240v = 72\frac{dv}{dt}$ | M1 | Applying N2L with weight and resistance, allow sign errors |
| $720 - 2400v = 720\frac{dv}{dt}$, divide by $-240$: $-\frac{3}{10}\frac{dv}{dt} = v - 2.94$ | A1 | Correct equation shown, using $g = 9.8$ |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Separate variables: $\int \frac{dv}{v - 2.94} = -\frac{10}{3}\int dt$ | M1 | Separating variables correctly |
| $\ln|v - 2.94| = -\frac{10}{3}t + c$ | A1 | Correct integration both sides |
| At $t=0$, $v=30$: $c = \ln(27.06)$ | M1 | Using initial condition |
| $v - 2.94 = 27.06\,e^{-\frac{10}{3}t}$ | A1 | Correct equation |
| $v = 2.94 + 27.06\,e^{-\frac{10t}{3}}$ | A1 | Final answer |
## Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Graph starts at $v=30$ when $t=0$ | B1 | Correct starting point marked |
| Curve decreasing asymptotically toward $v = 2.94$ | B1 | Correct shape with asymptote indicated |
---7 A particle, of mass 10 kg , is attached to one end of a light elastic string of natural length 0.4 metres and modulus of elasticity 100 N . The other end of the string is fixed to the point $O$.
\begin{enumerate}[label=(\alph*)]
\item Find the length of the elastic string when the particle hangs in equilibrium directly below $O$.
\item The particle is pulled down and held at a point $P$, which is 1 metre vertically below $O$.
Show that the elastic potential energy of the string when the particle is in this position is 45 J .
\item The particle is released from rest at the point $P$. In the subsequent motion, the particle has speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it is $x$ metres below $\boldsymbol { O }$.
\begin{enumerate}[label=(\roman*)]
\item Show that, while the string is taut,
$$v ^ { 2 } = 39.6 x - 25 x ^ { 2 } - 14.6$$
\item Find the value of $x$ when the particle comes to rest for the first time after being released, given that the string is still taut.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M2 Q7}}