| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Horizontal road towing |
| Difficulty | Moderate -0.8 This is a straightforward M1 connected particles question requiring Newton's second law applied to a two-body system. Part (a) uses F=ma on the whole system (one equation, one unknown), part (b) applies F=ma to the trailer alone, and part (c) requires only Newton's third law. All steps are standard textbook exercises with no problem-solving insight needed. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03h Newton's third law: action-reaction pairs3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| Applying Newton's 2nd Law to whole system: \(2500 - R - 800 = (3500 + 2400)(0.2)\) | M1 | F=ma for whole system |
| \(2500 - R - 800 = 5900 \times 0.2 = 1180\) | A1 | Correct equation |
| \(R = 520 \text{ N}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| Applying Newton's 2nd Law to trailer: \(T - 800 = 2400 \times 0.2\) | M1 | F=ma for trailer alone |
| \(T - 800 = 480\) | A1 | Correct equation |
| \(T = 1280 \text{ N}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(1280 \text{ N}\) | B1 | By Newton's 3rd Law |
## Question 4:
**Part (a): Find R**
| Working/Answer | Mark | Notes |
|---|---|---|
| Applying Newton's 2nd Law to whole system: $2500 - R - 800 = (3500 + 2400)(0.2)$ | M1 | F=ma for whole system |
| $2500 - R - 800 = 5900 \times 0.2 = 1180$ | A1 | Correct equation |
| $R = 520 \text{ N}$ | A1 | cao |
**Part (b): Force of tow bar on trailer**
| Working/Answer | Mark | Notes |
|---|---|---|
| Applying Newton's 2nd Law to trailer: $T - 800 = 2400 \times 0.2$ | M1 | F=ma for trailer alone |
| $T - 800 = 480$ | A1 | Correct equation |
| $T = 1280 \text{ N}$ | A1 | cao |
**Part (c): Force tow bar exerts on tractor**
| Working/Answer | Mark | Notes |
|---|---|---|
| $1280 \text{ N}$ | B1 | By Newton's 3rd Law |
---4 A tractor, of mass 3500 kg , is used to tow a trailer, of mass 2400 kg , across a horizontal field. The trailer is connected to the tractor by a horizontal tow bar. As they move, a constant resistance force of 800 newtons acts on the trailer and a constant resistance force of $R$ newtons acts on the tractor. A forward driving force of 2500 newtons acts on the tractor. The trailer and tractor accelerate at $0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item $\quad$ Find $R$.
\item Find the magnitude of the force that the tow bar exerts on the trailer.
\item State the magnitude of the force that the tow bar exerts on the tractor.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2013 Q4 [7]}}