| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work done against air resistance - vertical motion |
| Difficulty | Moderate -0.3 This is a straightforward multi-part energy question requiring standard formulas (KE = ½mv², PE = mgh) and the work-energy principle. Parts (a)-(c) involve direct substitution and simple algebra with no novel problem-solving. Part (d) requires basic conceptual understanding. Slightly easier than average due to its routine nature and clear structure. |
| Spec | 6.02c Work by variable force: using integration6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(KE = \frac{1}{2} \times 0.6 \times 14^2 = 58.8 \text{ J}\) | M1, A1 | 2 marks |
| (b) \(0.6 \times 9.8h = 58.8\) and \(h = \frac{58.8}{0.6 \times 9.8} = 10 \text{ m}\) | M1, A1, A1 | 3 marks |
| (c)(i) \(WD \text{ against resistance} = 58.8 - 0.6 \times 9.8 \times 8 = 11.76 - 11.8 \text{ J (to 3 sf)}\) | M1, A1, A1 | 3 marks |
| (ii) \(8F = 11.76\) and \(F = 1.47 \text{ N}\) | M1, A1 ft | 2 marks |
| (d) The magnitude of the force would vary with the speed of the ball. | B1 | 1 mark |
**(a)** $KE = \frac{1}{2} \times 0.6 \times 14^2 = 58.8 \text{ J}$ | M1, A1 | 2 marks | Use of KE formula; correct energy
**(b)** $0.6 \times 9.8h = 58.8$ and $h = \frac{58.8}{0.6 \times 9.8} = 10 \text{ m}$ | M1, A1, A1 | 3 marks | Two term energy equation involving PE and previous energy; correct equation; correct height. Note: Constant acceleration methods not accepted.
**(c)(i)** $WD \text{ against resistance} = 58.8 - 0.6 \times 9.8 \times 8 = 11.76 - 11.8 \text{ J (to 3 sf)}$ | M1, A1, A1 | 3 marks | Three term energy equation; correct equation; correct value
**(ii)** $8F = 11.76$ and $F = 1.47 \text{ N}$ | M1, A1 ft | 2 marks | Using work done = $Fd$ with $d = 8$; correct force (accept 1.48)
**(d)** The magnitude of the force would vary with the speed of the ball. | B1 | 1 mark | Appropriate explanation
**Total for Question 2: 11 marks**
---2 A ball of mass 0.6 kg is thrown vertically upwards from ground level with an initial speed of $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the initial kinetic energy of the ball.
\item Assuming that no resistance forces act on the ball, use an energy method to find the maximum height reached by the ball.
\item An experiment is conducted to confirm the maximum height for the ball calculated in part (b). In this experiment the ball rises to a height of only 8 metres.
\begin{enumerate}[label=(\roman*)]
\item Find the work done against the air resistance force that acts on the ball as it moves.
\item Assuming that the air resistance force is constant, find its magnitude.
\end{enumerate}\item Explain why it is not realistic to model the air resistance as a constant force.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2006 Q2 [11]}}