| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Applied force in addition to weights |
| Difficulty | Moderate -0.8 This is a straightforward M1 mechanics problem requiring Newton's second law applied to a two-body system. Students need to find the common acceleration using F=ma for the whole system (9N total force on 9kg), then apply F=ma to one box to find tension. It's a standard textbook exercise with clear setup and routine application of a single principle, making it easier than average. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Overall, N2L: \(135 - 9 = (5+4)a\) | M1 | Use of N2L. Allow \(F = mga\) but no extra forces. Allow 9 omitted. |
| \(a = 14\) so \(14 \text{ m s}^{-2}\) | A1 | |
| For A, N2L: \(T - 9 = 4 \times 14\) | M1 | N2L on A or B with correct mass. \(F = ma\). All relevant forces and no extras. |
| so \(65\) N | A1 | cao |
| or \(135 - T = 5a\) | M1 | 1 equation in \(T\) and \(a\). Allow sign errors. Allow \(F = mga\) |
| \(T - 9 = 4a\) | A1 | Both equations correct and consistent |
| Solving | M1 | Dependent on M* solving for \(T\) |
| \(T = 65\) so \(65\) N | A1 | cao |
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Overall, N2L: $135 - 9 = (5+4)a$ | M1 | Use of N2L. Allow $F = mga$ but no extra forces. Allow 9 omitted. |
| $a = 14$ so $14 \text{ m s}^{-2}$ | A1 | |
| For A, N2L: $T - 9 = 4 \times 14$ | M1 | N2L on A or B with correct mass. $F = ma$. All relevant forces and no extras. |
| so $65$ N | A1 | cao |
| **or** $135 - T = 5a$ | M1 | 1 equation in $T$ and $a$. Allow sign errors. Allow $F = mga$ |
| $T - 9 = 4a$ | A1 | Both equations correct and consistent |
| Solving | M1 | Dependent on M* solving for $T$ |
| $T = 65$ so $65$ N | A1 | cao |
---2 Boxes A and B slide on a smooth, horizontal plane. Box A has a mass of 4 kg and box B a mass of 5 kg . They are connected by a light, inextensible, horizontal wire. Horizontal forces of 9 N and 135 N act on A and B in the directions shown in Fig. 5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bf477f61-9f8f-418a-86d8-392bc30323b1-1_95_915_2042_650}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}
Calculate the tension in the wire joining the boxes.
\hfill \mbox{\textit{OCR MEI M1 Q2 [4]}}