| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - smooth inclined plane (no resistance) |
| Difficulty | Moderate -0.8 This is a straightforward application of energy conservation on an inclined plane with clear step-by-step guidance ('hence' structure). It requires only standard formulas (PE = mgh, KE = ½mv²) and basic trigonometry to find the height drop, making it easier than average with no problem-solving insight needed. |
| Spec | 6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\Delta PE = mgh = 4 \times 9.8 \times 5\sin 20°\) | M1 | Correct use of \(mgs\sin\theta\) |
| \(= 67.0\) J (67.08…) | A1 | Accept \(g = 9.81\) giving \(67.4\) J |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(KE = 67.0\) J (equal to PE lost, from (a)) | B1 | Follow through from (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2}mv^2 = 67.0\) | M1 | Use of \(KE = \frac{1}{2}mv^2\) with their (b) |
| \(v = \sqrt{\frac{2 \times 67.0}{4}} = 5.79\) m s\(^{-1}\) | A1 | Correct final answer |
# Question 2:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\Delta PE = mgh = 4 \times 9.8 \times 5\sin 20°$ | M1 | Correct use of $mgs\sin\theta$ |
| $= 67.0$ J (67.08…) | A1 | Accept $g = 9.81$ giving $67.4$ J |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $KE = 67.0$ J (equal to PE lost, from (a)) | B1 | Follow through from (a) |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2}mv^2 = 67.0$ | M1 | Use of $KE = \frac{1}{2}mv^2$ with their (b) |
| $v = \sqrt{\frac{2 \times 67.0}{4}} = 5.79$ m s$^{-1}$ | A1 | Correct final answer |
---2 A particle is placed on a smooth plane which is inclined at an angle of $20 ^ { \circ }$ to the horizontal. The particle, of mass 4 kg , is released from rest at a point $A$ and travels down the plane, passing through a point $B$. The distance $A B$ is 5 m .\\
\includegraphics[max width=\textwidth, alt={}, center]{9d039ec3-fd0a-40ae-9afe-7627439081df-04_371_693_500_680}
\begin{enumerate}[label=(\alph*)]
\item Find the potential energy lost as the particle moves from point $A$ to point $B$.
\item Hence write down the kinetic energy of the particle when it reaches point $B$.
\item Hence find the speed of the particle when it reaches point $B$.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2011 Q2 [5]}}