| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Horizontal road towing |
| Difficulty | Moderate -0.8 This is a straightforward connected particles problem requiring a force diagram and application of Newton's second law to find tension. The setup is standard (tractor-trailer system), requires only basic F=ma with two equations, and involves simple arithmetic with given values. Below average difficulty as it's a routine mechanics exercise with no conceptual subtlety. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks | Guidance |
|---|---|---|
| Horizontal forces (could be mirror image): 6600 N, 1600 N, 800 N, T shown | B1 | 6600 N and resistances correctly placed and labelled. No extra horizontal forces |
| Common tension in the towbar shown | B1 | Common tension in the towbar shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(6600 - 1600 - 800 = 2800a\) | M1 | 2800 kg used. Allow a missing or extra horizontal force for the method mark |
| \(a = 1.5\ \text{ms}^{-2}\) | A1 | soi |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - 800 = 1000 \times 1.5\) | M1 | Newton's second law with correct mass. FT their acceleration. Also allow for tractor \(6600 - T - 1600 = 1800 \times 1.5\) |
| \(T = 2300\ \text{N}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - 800 = 1000a\) | M1 | Newton's second law for both parts with correct masses; allow missing or extra horizontal force |
| Both equations correct | A1 | |
| Simultaneous equations | M1 | Attempt to solve simultaneous equations leading to a value of \(T\) |
| \(T = 2300\ \text{N}\) | A1 | cao |
# Question 9:
## Part (a):
Horizontal forces (could be mirror image): 6600 N, 1600 N, 800 N, T shown | **B1** | 6600 N and resistances correctly placed and labelled. No extra horizontal forces
Common tension in the towbar shown | **B1** | Common tension in the towbar shown
## Part (b):
Tractor and trailer together:
$6600 - 1600 - 800 = 2800a$ | **M1** | 2800 kg used. Allow a missing or extra horizontal force for the method mark
$a = 1.5\ \text{ms}^{-2}$ | **A1** | soi
Newton's second law for the trailer:
$T - 800 = 1000 \times 1.5$ | **M1** | Newton's second law with correct mass. FT their acceleration. Also allow for tractor $6600 - T - 1600 = 1800 \times 1.5$
$T = 2300\ \text{N}$ | **A1** | cao
**Alternative method:**
$6600 - T - 1600 = 1800a$
$T - 800 = 1000a$ | **M1** | Newton's second law for both parts with correct masses; allow missing or extra horizontal force
Both equations correct | **A1** |
Simultaneous equations | **M1** | Attempt to solve simultaneous equations leading to a value of $T$
$T = 2300\ \text{N}$ | **A1** | cao
---9 A tractor of mass 1800 kg uses a towbar to pull a trailer of mass 1000 kg on a level field. The tractor and trailer experience resistances to motion of 1600 N and 800 N respectively. The tractor provides a driving force of 6600 N .
\begin{enumerate}[label=(\alph*)]
\item Draw a force diagram showing all the horizontal forces acting on the tractor and trailer.
\item Find the tension in the towbar.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2022 Q9 [6]}}