| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Projectile energy - basic KE/PE calculation |
| Difficulty | Moderate -0.8 This is a straightforward application of energy conservation with standard formulas (KE = ½mv², PE = mgh). All parts are direct substitutions requiring minimal problem-solving: students simply apply memorized formulas in sequence with clearly signposted steps. The multi-part structure guides students through the solution methodically, making it easier than a typical A-level mechanics question. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(KE = \frac{1}{2}mv^2 = \frac{1}{2}(3)(4)^2\) | M1 | Correct formula used |
| \(= 24\) J | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(PE = mgh = 3 \times 9.8 \times 51\) | M1 | Correct formula used |
| \(= 1499.4\) J | A1 | Accept \(1500\) J or use of \(g = 9.81\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(KE_{final} = KE_{initial} + PE_{lost}\) | M1 | Using energy conservation |
| \(KE_{final} = 24 + 1499.4 = 1523.4\) J | A1 | |
| \(\frac{1}{2}(3)v^2 = 1523.4\) | M1 | Using \(KE = \frac{1}{2}mv^2\) |
| \(v = \sqrt{\frac{2 \times 1523.4}{3}} \approx 31.9\) ms\(^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| No air resistance (or rock treated as particle) | B1 | Accept equivalent statements |
## Question 2:
**(a)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $KE = \frac{1}{2}mv^2 = \frac{1}{2}(3)(4)^2$ | M1 | Correct formula used |
| $= 24$ J | A1 | |
**(b)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $PE = mgh = 3 \times 9.8 \times 51$ | M1 | Correct formula used |
| $= 1499.4$ J | A1 | Accept $1500$ J or use of $g = 9.81$ |
**(c)(i) and (ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $KE_{final} = KE_{initial} + PE_{lost}$ | M1 | Using energy conservation |
| $KE_{final} = 24 + 1499.4 = 1523.4$ J | A1 | |
| $\frac{1}{2}(3)v^2 = 1523.4$ | M1 | Using $KE = \frac{1}{2}mv^2$ |
| $v = \sqrt{\frac{2 \times 1523.4}{3}} \approx 31.9$ ms$^{-1}$ | A1 | |
**(d)**
| Answer | Mark | Guidance |
|--------|------|----------|
| No air resistance (or rock treated as particle) | B1 | Accept equivalent statements |
---2 John is at the top of a cliff, looking out over the sea. He throws a rock, of mass 3 kg , horizontally with a velocity of $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
The rock falls a vertical distance of 51 metres to reach the surface of the sea.
\begin{enumerate}[label=(\alph*)]
\item Calculate the kinetic energy of the rock when it is thrown.
\item Calculate the potential energy lost by the rock when it reaches the surface of the sea.
\item \begin{enumerate}[label=(\roman*)]
\item Find the kinetic energy of the rock when it reaches the surface of the sea.
\item Hence find the speed of the rock when it reaches the surface of the sea.
\end{enumerate}\item State one modelling assumption which has been made.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{3ffa0a2b-aa7d-46eb-b92b-3e3ee59f235c-05_2484_1709_223_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2010 Q2 [9]}}