| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Train with coupled trucks/carriages |
| Difficulty | Moderate -0.3 This is a standard connected particles problem requiring force diagrams and Newton's second law applied to the system and individual components. The calculations are straightforward (finding driving force and comparing tensions) with clearly given values and no conceptual surprises, making it slightly easier than average for A-level mechanics. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Weights and normal reactions shown | B1 | 1.1a – Allow \(N_1, N_1\) and \(N_2\) instead of values to match weights for the normal reaction (condone \(N_1, N_2, N_3\)) |
| Tensions (must be two distinct) shown | B1 | 1.1a |
| Driving force (\(D\) or \(16200\)) and three resistances correctly shown | B1 [3] | 1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| N2L for the whole train: \(D - 2000 - 600 - 600 = 130\,000 \times 0.1\) | M1 | 1.1a – All forces present and no extras. Allow slip in mass. Signs must be consistent |
| \(D = 16\,200\) | A1 [2] | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| N2L for each part: \(D - 2000 - T_1 = 80\,000\times0.1\) | M1 | All forces present and no extras in each equation. Signs must be consistent. Other equations may be used instead, e.g. \(T_1 - 2\times600 = 5000\); Engine plus A: \(D-T_2-2600=10500\) |
| \(T_1 - T_2 - 600 = 25\,000\times0.1\) | ||
| \(T_2 - 600 = 25\,000\times0.1\) | ||
| Add equations to eliminate tension; \(D = 16\,200\) | M1, A1 [2] | Award M mark only when attempt made to solve equations simultaneously |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| N2L for truck A: \(T_1 - T_2 - 600 = 25\,000\times0.1\) | M1 | 3.1b – All forces present and no extras. Signs must be consistent. Values for tensions need not be found |
| Difference \(T_1 - T_2\) is \(3100\ \text{N}\) | A1 [2] | 1.1 – from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For the engine: \(D - 2000 - T_1 = 8000\); \(T_1 = 6200\ \text{N}\) | M1 | All forces present and no extras in one equation leading to a value for either \(T_1\) or \(T_2\). Signs must be consistent. The equations may have been found in part (b). Also allow \(T_1\) is twice \(T_2\) oe from correct values |
| For truck B: \(T_2 - 600 = 2500\); \(T_2 = 3100\ \text{N}\) | ||
| Difference \(T_1 - T_2\) is \(3100\ \text{N}\) | A1 [2] | FT their \(D\) |
## Question 6:
**(a)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Weights and normal reactions shown | B1 | 1.1a – Allow $N_1, N_1$ and $N_2$ instead of values to match weights for the normal reaction (condone $N_1, N_2, N_3$) |
| Tensions (must be two distinct) shown | B1 | 1.1a |
| Driving force ($D$ or $16200$) and three resistances correctly shown | B1 [3] | 1.1a |
**(b)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| N2L for the whole train: $D - 2000 - 600 - 600 = 130\,000 \times 0.1$ | M1 | 1.1a – All forces present and no extras. Allow slip in mass. Signs must be consistent |
| $D = 16\,200$ | A1 [2] | 1.1 |
**Alternative solution:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| N2L for each part: $D - 2000 - T_1 = 80\,000\times0.1$ | M1 | All forces present and no extras in each equation. Signs must be consistent. Other equations may be used instead, e.g. $T_1 - 2\times600 = 5000$; Engine plus A: $D-T_2-2600=10500$ |
| $T_1 - T_2 - 600 = 25\,000\times0.1$ | | |
| $T_2 - 600 = 25\,000\times0.1$ | | |
| Add equations to eliminate tension; $D = 16\,200$ | M1, A1 [2] | Award M mark only when attempt made to solve equations simultaneously |
**(c)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| N2L for truck A: $T_1 - T_2 - 600 = 25\,000\times0.1$ | M1 | 3.1b – All forces present and no extras. Signs must be consistent. Values for tensions need not be found |
| Difference $T_1 - T_2$ is $3100\ \text{N}$ | A1 [2] | 1.1 – from correct working |
**Alternative solution:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| For the engine: $D - 2000 - T_1 = 8000$; $T_1 = 6200\ \text{N}$ | M1 | All forces present and no extras in one equation leading to a value for either $T_1$ or $T_2$. Signs must be consistent. The equations may have been found in part (b). Also allow $T_1$ is twice $T_2$ oe from correct values |
| For truck B: $T_2 - 600 = 2500$; $T_2 = 3100\ \text{N}$ | | |
| Difference $T_1 - T_2$ is $3100\ \text{N}$ | A1 [2] | FT their $D$ |
---6 Fig. 6 shows a train consisting of an engine of mass 80 tonnes pulling two trucks each of mass 25 tonnes.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0b1c272a-f0f4-4931-be89-5d045804a7af-4_189_1262_938_246}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
The engine exerts a driving force of $D \mathrm {~N}$ and experiences a resistance to motion of 2000 N . Each truck experiences a resistance of 600 N . The train travels in a straight line on a level track with an acceleration of $0.1 \mathrm {~ms} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Complete the force diagram in the Printed Answer Booklet to show all the forces acting on the engine and each of the trucks.
\item Calculate the value of $D$.
\item The tension in the coupling between the engine and truck A is larger than that in the coupling between the trucks. Determine how much larger.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2019 Q6 [7]}}