| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Two connected particles, horizontal surface |
| Difficulty | Moderate -0.3 This is a straightforward connected particles problem requiring Newton's second law applied to a system then individual blocks. Part (a) involves standard F=ma with given forces and resistances. Part (b) requires recognizing that without the string, blocks decelerate differently, creating compression. The mechanics are routine with clear diagrams and no geometric complications or novel insight required. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using N2L in direction of motion, acceleration \(a\) ms\(^{-2}\): \((40 - 10 - 5) = 10a\) | M1 | \(F = ma\) and all forces present |
| so \(a = 2.5\) | A1 | Not required, may be implied |
| Mark force in bar as tension \(T\) [N]; For A: \(T - 10 = 6a\) | M1 | Allow their \(a\) |
| or For B: \(40 - 5 - T = 4a\) | M1 | Allow their \(a\) |
| so tension is \(25\) [N] | A1 | Or \(T = 25\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Without force from string, N2L in direction of motion: \(-15 = 10a^*\) | M1 | |
| so \(a^* = -1.5\) | A1 | Not required, may be implied |
| Mark force in bar as tension \(T^*\) [N]; For A: \(T^* - 10 = 6a^* = -9\) | M1 | Allow their \(a\) |
| or For B: \(-5 - T^* = 4a^* = -6\) | M1 | Allow their \(a\) |
| so \(T^* = 1\) | A1 | |
| giving a force of \(1\) [N] which is a tension | A1 | Must be made clear in some way |
| [5] |
## Question 14:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using N2L in direction of motion, acceleration $a$ ms$^{-2}$: $(40 - 10 - 5) = 10a$ | M1 | $F = ma$ and all forces present |
| so $a = 2.5$ | A1 | Not required, may be implied |
| Mark force in bar as tension $T$ [N]; For A: $T - 10 = 6a$ | M1 | Allow their $a$ |
| **or** For B: $40 - 5 - T = 4a$ | M1 | Allow their $a$ |
| so tension is $25$ [N] | A1 | Or $T = 25$ |
| **[4]** | | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Without force from string, N2L in direction of motion: $-15 = 10a^*$ | M1 | |
| so $a^* = -1.5$ | A1 | Not required, may be implied |
| Mark force in bar as tension $T^*$ [N]; For A: $T^* - 10 = 6a^* = -9$ | M1 | Allow their $a$ |
| **or** For B: $-5 - T^* = 4a^* = -6$ | M1 | Allow their $a$ |
| so $T^* = 1$ | A1 | |
| giving a force of $1$ [N] which is a tension | A1 | Must be made clear in some way |
| **[5]** | | |
---14 Blocks A and B are connected by a light rigid horizontal bar and are sliding on a rough horizontal surface.
A light horizontal string exerts a force of 40 N on B .\\
This situation is shown in Fig. 14, which also shows the direction of motion, the mass of each of the blocks and the resistances to their motion.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ff44367e-c992-4e79-b255-5a04e0b8e21e-11_266_1283_664_255}
\captionsetup{labelformat=empty}
\caption{Fig. 14}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Calculate the tension in the bar.
The string breaks while the blocks are sliding. The resistances to motion are unchanged.
\item Determine
\begin{itemize}
\item the magnitude of the new force in the bar,
\item whether the bar is in tension or in compression.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 Q14 [9]}}