| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Vehicle on slope with resistance |
| Difficulty | Standard +0.3 This is a straightforward multi-part mechanics question involving forces on slopes with small angles. It requires standard application of F=ma and resolving forces parallel to slopes, but the calculations are routine with no conceptual surprises. The small angles and given values make arithmetic manageable. Slightly above average difficulty due to multiple parts and slope work, but well within typical M1 scope. |
| Spec | 3.03d Newton's second law: 2D vectors3.03r Friction: concept and vector form3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(F - R = ma\) | M1 | Use of Newton's 2nd Law |
| \(300 - R = (750 + 50) \times 0.25\) | A1 | Correct elements present |
| \(R = 100\) | A1 | This is a given result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Carol in: Component of weight down slope | M1 | Resolving down the slope. Accept use of 750 instead of 800. For this mark only condone no \(g\) and allow sin-cos interchange |
| \(= 800g\sin 1.5° \ (= 205.2 \text{ N})\) | A1 | Give M1 A1 for \(800g\sin 15°\) seen |
| Martin has to overcome 305.2 N; \(300 < 305.2\) Martin cannot manage | A1 | This mark may be awarded for an argument based on Newton's 2nd law leading towards \(a = -0.006\) |
| Carol out: Martin has to overcome \(750g\sin 1.5° + 100 = 292.4\text{ N}\) | ||
| \(300 > 292.4\) so Martin manages | B1 | Explanation, based on correct working, that Martin can manage. This can be given retrospectively with a comment on a positive value for \(a\) |
| \(300 - 292.4 = 7.6 = 750a\) | M1 | Use of Newton's 2nd Law |
| The acceleration is \(0.010 \text{ ms}^{-2}\) | A1 | CAO. Accept 0.01 or an answer that rounds to 0.01 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Component of Carol's force parallel to the line of the car \(= 150\cos 30° \ (= 129.9)\) | M1 | For attempt at resolution in the correct direction. For this mark only, condone sin-cos interchange |
| \(= 150\cos 30° \ (= 129.9)\) | A1 | Give M1 A1 for \(150\cos 30°\) seen |
| Resultant forward force \(= 7.6 + 129.9 = 137.5\) | M1 | All forces parallel to the slope present and correct. Sign errors condoned |
| \(750a = 137.5\); The acceleration is \(0.183 \text{ ms}^{-2}\) | A1 | FT their force parallel to the slope from part (ii) (correct value 7.6 N) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Component of weight down the slope \(= (750 + 50 + 80) \times 9.8 \times \sin 3°\) | ||
| \(880a = 451.3 - 100\) | M1 | Newton's 2nd law with correct elements present. No sin-cos interchange. The same mass must be used in both places |
| \(a = 0.399\) | A1 | |
| \(v^2 - u^2 = 2as\) | M1 | Selection and use of an appropriate formula (unless with \(a = g\)) |
| When \(v = 8\), \(s = 8^2 \div (2 \times 0.399)\) | ||
| \(s = 80.1\) | A1 | FT their value of \(a\) |
| \(80.1 < 100\) so Yes they get the car started | A1 | FT their value of \(a\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(880a = 451.3 - 100\) | M1 | Newton's 2nd law with correct elements present. No sin-cos interchange |
| \(a = 0.399\) | A1 | |
| \(v^2 = (0^2) + 2 \times 0.399 \times 100\) | M1 | Selection and use of an appropriate formula (unless with \(a = g\)) |
| \(v = (\sqrt{79.8}) = 8.93...\) | A1 | FT their value of \(a\) |
| \((v > 8)\) so they get the car started | A1 | FT their value of \(a\) |
## Question 2:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $F - R = ma$ | M1 | Use of Newton's 2nd Law |
| $300 - R = (750 + 50) \times 0.25$ | A1 | Correct elements present |
| $R = 100$ | A1 | This is a given result |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| **Carol in:** Component of weight down slope | M1 | Resolving down the slope. Accept use of 750 instead of 800. For this mark only condone no $g$ and allow sin-cos interchange |
| $= 800g\sin 1.5° \ (= 205.2 \text{ N})$ | A1 | Give M1 A1 for $800g\sin 15°$ seen |
| Martin has to overcome 305.2 N; $300 < 305.2$ Martin cannot manage | A1 | This mark may be awarded for an argument based on Newton's 2nd law leading towards $a = -0.006$ |
| **Carol out:** Martin has to overcome $750g\sin 1.5° + 100 = 292.4\text{ N}$ | | |
| $300 > 292.4$ so Martin manages | B1 | Explanation, based on correct working, that Martin can manage. This can be given retrospectively with a comment on a positive value for $a$ |
| $300 - 292.4 = 7.6 = 750a$ | M1 | Use of Newton's 2nd Law |
| The acceleration is $0.010 \text{ ms}^{-2}$ | A1 | CAO. Accept 0.01 or an answer that rounds to 0.01 |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Component of Carol's force parallel to the line of the car $= 150\cos 30° \ (= 129.9)$ | M1 | For attempt at resolution in the correct direction. For this mark only, condone sin-cos interchange |
| $= 150\cos 30° \ (= 129.9)$ | A1 | Give M1 A1 for $150\cos 30°$ seen |
| Resultant forward force $= 7.6 + 129.9 = 137.5$ | M1 | All forces parallel to the slope present and correct. Sign errors condoned |
| $750a = 137.5$; The acceleration is $0.183 \text{ ms}^{-2}$ | A1 | FT their force parallel to the slope from part (ii) (correct value 7.6 N) |
### Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| Component of weight down the slope $= (750 + 50 + 80) \times 9.8 \times \sin 3°$ | | |
| $880a = 451.3 - 100$ | M1 | Newton's 2nd law with correct elements present. No sin-cos interchange. The same mass must be used in both places |
| $a = 0.399$ | A1 | |
| $v^2 - u^2 = 2as$ | M1 | Selection and use of an appropriate formula (unless with $a = g$) |
| When $v = 8$, $s = 8^2 \div (2 \times 0.399)$ | | |
| $s = 80.1$ | A1 | FT their value of $a$ |
| $80.1 < 100$ so Yes they get the car started | A1 | FT their value of $a$ |
**Alternative (iv):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $880a = 451.3 - 100$ | M1 | Newton's 2nd law with correct elements present. No sin-cos interchange |
| $a = 0.399$ | A1 | |
| $v^2 = (0^2) + 2 \times 0.399 \times 100$ | M1 | Selection and use of an appropriate formula (unless with $a = g$) |
| $v = (\sqrt{79.8}) = 8.93...$ | A1 | FT their value of $a$ |
| $(v > 8)$ so they get the car started | A1 | FT their value of $a$ |
---2 The battery on Carol and Martin's car is flat so the car will not start. They hope to be able to "bump start" the car by letting it run down a hill and engaging the engine when the car is going fast enough. Fig. 6.1 shows the road leading away from their house, which is at A . The road is straight, and at all times the car is steered directly along it.
\begin{itemize}
\item From A to B the road is horizontal.
\item Between B and C , it goes up a hill with a uniform slope of $1.5 ^ { \circ }$ to the horizontal.
\item Between C and D the road goes down a hill with a uniform slope of $3 ^ { \circ }$ to the horizontal. CD is 100 m . (This is the part of the road where they hope to get the car started.)
\item From D to E the road is again horizontal.
\end{itemize}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9bff41e0-7be0-4932-ae50-a612abb3fe19-2_239_1137_636_484}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
\end{center}
\end{figure}
The mass of the car is 750 kg , Carol's mass is 50 kg and Martin's mass is 80 kg .\\
Throughout the rest of this question, whenever Martin pushes the car, he exerts a force of 300 N along the line of the car.\\
(i) Between A and B, Martin pushes the car and Carol sits inside to steer it. The car has an acceleration of $0.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
Show that the resistance to the car's motion is 100 N .
Throughout the rest of this question you should assume that the resistance to motion is constant at 100 N .\\
(ii) They stop at B and then Martin tries to push the car up the hill BC.
Show that Martin cannot push the car up the hill with Carol inside it but can if she gets out.\\
Find the acceleration of the car when Martin is pushing it and Carol is standing outside.\\
(iii) While between B and C, Carol opens the window of the car and pushes it from outside while steering with one hand. Carol is able to exert a force of 150 N parallel to the surface of the road but at an angle of $30 ^ { \circ }$ to the line of the car. This is illustrated in Fig. 6.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9bff41e0-7be0-4932-ae50-a612abb3fe19-2_216_425_1964_870}
\captionsetup{labelformat=empty}
\caption{Fig. 6.2}
\end{center}
\end{figure}
Find the acceleration of the car.\\
(iv) At C, both Martin and Carol get in the car and, starting from rest, let it run down the hill under gravity. If the car reaches a speed of $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ they can get the engine to start.
\hfill \mbox{\textit{OCR MEI M1 Q2 [18]}}