| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work done against friction/resistance - inclined plane or slope |
| Difficulty | Standard +0.3 This is a straightforward multi-part mechanics question requiring standard application of work-energy principles and Newton's second law. Part (a) uses energy conservation with one equation to solve, part (b)(i) applies F=ma on an incline (routine calculation), and part (b)(ii) requires resolving forces with friction—all standard M2 techniques with no novel problem-solving required. Slightly above average due to the multi-step nature and need to carefully track energy terms, but well within typical A-level mechanics scope. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle6.02k Power: rate of doing work |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Energy equation: \(\frac{1}{2}(80)V^2 + 80(9.8)(1600) = \frac{1}{2}(80)(6)^2 + 1.3\times10^6\) | M1 | Use work-energy theorem |
| \(40V^2 + 1254400 = 1440 + 1300000\) | A1 | Correct terms |
| \(40V^2 = 1301440 - 1254400 = 47040\) | M1 | Rearrange |
| Wait: \(\frac{1}{2}(80)V^2 = \frac{1}{2}(80)(36) + 1.3\times10^6 - 80(9.8)(1600)\) | ||
| \(40V^2 = 1440 + 1300000 - 1254400 = 47040\) | A1 | |
| \(V^2 = 1176\), \(V \approx 34.3 \text{ ms}^{-1}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Newton's 2nd law for whole system (vehicle only, mass 800 kg): \(F - 1150 - 800g\sin\theta = 800 \times 0.25\) | M1 | Apply N2L |
| \(F = 200 + 1150 + 800(9.8)(0.1)\) | A1 | |
| \(F = 200 + 1150 + 784 = 2134 \text{ N}\) | A1 | shown |
| Power \(= F \times v = 2134 \times 8\) | M1 | |
| Power \(= 17072 \text{ W} \approx 17100 \text{ W}\) | A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| For sledge (mass 300 kg): \(900 - F_{friction} - 300g\sin\theta = 300 \times 0.25\) | M1 A1 | Apply N2L to sledge |
| \(900 - F_f - 300(9.8)(0.1) = 75\) | A1 | |
| \(F_f = 900 - 294 - 75 = 531 \text{ N}\) | A1 | |
| Normal reaction: \(N = 300g\cos\theta = 300(9.8)\sqrt{1-0.01} \approx 2940\cos\theta\) | M1 | |
| \(\mu = \frac{531}{2940\cos\theta}\); \(\cos\theta = \sqrt{1-0.01} \approx 0.995\) | M1 | |
| \(\mu \approx \frac{531}{2925} \approx 0.181\) | A1 |
# Question 4:
## Part (a)
| Working/Answer | Marks | Guidance |
|---|---|---|
| Energy equation: $\frac{1}{2}(80)V^2 + 80(9.8)(1600) = \frac{1}{2}(80)(6)^2 + 1.3\times10^6$ | M1 | Use work-energy theorem |
| $40V^2 + 1254400 = 1440 + 1300000$ | A1 | Correct terms |
| $40V^2 = 1301440 - 1254400 = 47040$ | M1 | Rearrange |
| Wait: $\frac{1}{2}(80)V^2 = \frac{1}{2}(80)(36) + 1.3\times10^6 - 80(9.8)(1600)$ | | |
| $40V^2 = 1440 + 1300000 - 1254400 = 47040$ | A1 | |
| $V^2 = 1176$, $V \approx 34.3 \text{ ms}^{-1}$ | A1 | cao |
## Part (b)(i)
| Working/Answer | Marks | Guidance |
|---|---|---|
| Newton's 2nd law for whole system (vehicle only, mass 800 kg): $F - 1150 - 800g\sin\theta = 800 \times 0.25$ | M1 | Apply N2L |
| $F = 200 + 1150 + 800(9.8)(0.1)$ | A1 | |
| $F = 200 + 1150 + 784 = 2134 \text{ N}$ | A1 | shown |
| Power $= F \times v = 2134 \times 8$ | M1 | |
| Power $= 17072 \text{ W} \approx 17100 \text{ W}$ | A1 A1 | |
## Part (b)(ii)
| Working/Answer | Marks | Guidance |
|---|---|---|
| For sledge (mass 300 kg): $900 - F_{friction} - 300g\sin\theta = 300 \times 0.25$ | M1 A1 | Apply N2L to sledge |
| $900 - F_f - 300(9.8)(0.1) = 75$ | A1 | |
| $F_f = 900 - 294 - 75 = 531 \text{ N}$ | A1 | |
| Normal reaction: $N = 300g\cos\theta = 300(9.8)\sqrt{1-0.01} \approx 2940\cos\theta$ | M1 | |
| $\mu = \frac{531}{2940\cos\theta}$; $\cos\theta = \sqrt{1-0.01} \approx 0.995$ | M1 | |
| $\mu \approx \frac{531}{2925} \approx 0.181$ | A1 | |
The images you've shared show only **blank pages** (page 7 and page 8) from an OCR 2011 examination paper (reference 4762 Jun11). These pages contain no mark scheme content — page 7 is explicitly labelled "BLANK PAGE" and page 8 contains only the OCR copyright notice.
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\begin{enumerate}[label=(\alph*)]
\item A parachutist and her equipment have a combined mass of 80 kg . During a descent where the parachutist loses 1600 m in height, her speed reduces from $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and she does $1.3 \times 10 ^ { 6 } \mathrm {~J}$ of work against resistances.
Use an energy method to calculate the value of $V$.
\item A vehicle of mass 800 kg is climbing a hill inclined at $\theta$ to the horizontal, where $\sin \theta = 0.1$. At one time the vehicle has a speed of $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and is accelerating up the hill at $0.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ against a resistance of 1150 N .
\begin{enumerate}[label=(\roman*)]
\item Show that the driving force on the vehicle is 2134 N and calculate its power at this time.
The vehicle is pulling a sledge, of mass 300 kg , which is sliding up the hill. The sledge is attached to the vehicle by a light, rigid coupling parallel to the slope. The force in the coupling is 900 N .
\item Assuming that the only resistance to the motion of the sledge is due to friction, calculate the coefficient of friction between the sledge and the ground.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2011 Q4 [18]}}