| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | January |
| 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). Part (a) is direct substitution, part (b)(i) adds gravitational PE to initial KE, and part (b)(ii) reverses the KE formula. The 'show that' removes problem-solving difficulty. Stating assumptions is routine. Easier than average due to minimal steps and no conceptual challenges. |
| Spec | 3.02h Motion under gravity: vector form6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| \(KE = \frac{1}{2}mv^2 = \frac{1}{2}(6)(12)^2\) | M1 | Correct formula used |
| \(= 432 \text{ J}\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Using energy conservation: \(KE = 432 + mgh = 432 + 6(9.8)(4)\) | M1 | Use of energy conservation with GPE |
| \(= 432 + 235.2 = 667.2 \approx 667 \text{ J}\) | A1 | Correct answer shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}(6)v^2 = 667.2\) | M1 | Set KE equal to 667 or 667.2 |
| \(v^2 = \frac{2 \times 667.2}{6} = 222.4\) | M1 | Correct rearrangement |
| \(v = 14.9 \text{ m s}^{-1}\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Stone modelled as a particle | B1 | |
| No air resistance (or string/gravity only) | B1 | Any two valid assumptions |
# Question 2:
## Part (a):
| $KE = \frac{1}{2}mv^2 = \frac{1}{2}(6)(12)^2$ | M1 | Correct formula used |
|---|---|---|
| $= 432 \text{ J}$ | A1 | Correct answer |
## Part (b)(i):
| Using energy conservation: $KE = 432 + mgh = 432 + 6(9.8)(4)$ | M1 | Use of energy conservation with GPE |
|---|---|---|
| $= 432 + 235.2 = 667.2 \approx 667 \text{ J}$ | A1 | Correct answer shown |
## Part (b)(ii):
| $\frac{1}{2}(6)v^2 = 667.2$ | M1 | Set KE equal to 667 or 667.2 |
|---|---|---|
| $v^2 = \frac{2 \times 667.2}{6} = 222.4$ | M1 | Correct rearrangement |
| $v = 14.9 \text{ m s}^{-1}$ | A1 | Correct answer |
## Part (b)(iii):
| Stone modelled as a particle | B1 | |
|---|---|---|
| No air resistance (or string/gravity only) | B1 | Any two valid assumptions |
---2 A stone, of mass 6 kg , is thrown vertically upwards with a speed of $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point at a height of 4 metres above ground level.
\begin{enumerate}[label=(\alph*)]
\item Calculate the initial kinetic energy of the stone.
\item \begin{enumerate}[label=(\roman*)]
\item Show that the kinetic energy of the stone when it hits the ground is 667 J , correct to three significant figures.
\item Hence find the speed of the stone when it hits the ground.
\item State two modelling assumptions that you have made.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M2 2009 Q2 [9]}}