| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Car towing trailer, horizontal |
| Difficulty | Easy -1.2 This is a straightforward connected particles problem requiring a force diagram and application of F=ma to the system. The horizontal surface, given forces, and direct calculation make this easier than average—it's a standard textbook exercise testing basic understanding of Newton's second law with no problem-solving insight required. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03l Newton's third law: extend to situations requiring force resolution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Diagram showing: driving force 6000N (right), resistance 800N (left), resistance 300N (left), tension \(T\) labelled | B1 | Driving force and common tension correct and labelled. Allow \(T_1\) and \(T_2\) if later seen equated. Ignore vertical forces. Also allow common thrust. |
| Both resistances correct and labelled, no extra horizontal forces | B1 | Both resistances correct and labelled and no extra horizontal forces |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| N2L for system: \(6000 - 800 - 300 = 1800a\) | M1 | Attempt to find resultant horizontal force. All forces included and no extras. Allow sign errors. |
| M1 | Attempts N2L equation with total mass and resultant. Do not allow weight used instead of mass. | |
| \(a = 2.72 \text{ ms}^{-2}\) (to 3 s.f.) | A1 | cao |
| Alternative: N2L for car: \(6000 - 800 - T = 1400a\); N2L for trailer: \(T - 300 = 400a\) | M1 | Attempt N2L for at least one part. Do not allow weight instead of mass. Allow sign errors. All forces included and no extras. |
| Add equations to give \(a = 2.72 \text{ ms}^{-2}\) (to 3 s.f.) | M1, A1 | M1: Attempt to eliminate \(T\); A1: cao |
| [3] |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Diagram showing: driving force 6000N (right), resistance 800N (left), resistance 300N (left), tension $T$ labelled | B1 | Driving force and common tension correct and labelled. Allow $T_1$ and $T_2$ if later seen equated. Ignore vertical forces. Also allow common thrust. |
| Both resistances correct and labelled, no extra horizontal forces | B1 | Both resistances correct and labelled and no extra horizontal forces |
| | **[2]** | |
---
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| N2L for system: $6000 - 800 - 300 = 1800a$ | M1 | Attempt to find resultant horizontal force. All forces included and no extras. Allow sign errors. |
| | M1 | Attempts N2L equation with total mass and resultant. Do not allow weight used instead of mass. |
| $a = 2.72 \text{ ms}^{-2}$ (to 3 s.f.) | A1 | cao |
| **Alternative:** N2L for car: $6000 - 800 - T = 1400a$; N2L for trailer: $T - 300 = 400a$ | M1 | Attempt N2L for at least one part. Do not allow weight instead of mass. Allow sign errors. All forces included and no extras. |
| Add equations to give $a = 2.72 \text{ ms}^{-2}$ (to 3 s.f.) | M1, A1 | M1: Attempt to eliminate $T$; A1: cao |
| | **[3]** | |2 A car of mass 1400 kg pulls a trailer of mass 400 kg along a straight horizontal road. The engine of the car produces a driving force of 6000 N . A resistance of 800 N acts on the car. A resistance of 300 N acts on the trailer. The tow-bar between the car and the trailer is light and horizontal.
\begin{enumerate}[label=(\alph*)]
\item Draw a force diagram showing all the horizontal forces on the car and the trailer.
\item Calculate the acceleration of the car and trailer.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2024 Q2 [5]}}