| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Train with coupled trucks/carriages |
| Difficulty | Moderate -0.8 This is a straightforward connected particles question requiring repeated application of F=ma with clearly stated values. All parts involve direct substitution into Newton's second law with no geometric insight, simultaneous equations, or problem-solving required—simpler than the average A-level question which typically demands more synthesis of techniques. |
| Spec | 3.03d Newton's second law: 2D vectors3.03f Weight: W=mg |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = 14000 \times 0.25\) | M1 | Use of N2L. Allow \(F = mga\) and wrong mass. No extra forces. |
| \(= 3500\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(4000 - R = 3500\) so \(500\) N | B1 | FT \(F\) from (i). Condone negative answer. |
| Answer | Marks | Guidance |
|---|---|---|
| \(1150 - R_T = 4000 \times 0.25\) | M1 | N2L applied to truck (or engine) using all forces required. No extras. Correct mass. Do not allow use of \(F = mga\). Allow sign errors. |
| \(= 150\) N | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Component of weight down slope | M1 | Attempt to find component of *weight* (allow wrong mass). Accept \(\sin \leftrightarrow \cos\). Accept use of \(m\sin\theta\). |
| Extra driving force is component of \(mg\) down slope | M1 | May be implied. Correct mass. No extra forces. Must have resolved weight component. Allow \(\sin \leftrightarrow \cos\) |
| \(14000g\sin 3° = 14000 \times 9.8 \times 0.0523359... = 7180.49...\) so \(7180\) N (3 s.f.) | A1 | |
| or \(D - 500 - 14000g\sin 3 = 14000 \times 0.25\) | M1 | N2L with all terms present with correct signs and mass. No extras. FT 500 N. Accept their \(500 + 150\) for resistance. Must have resolved weight component. Allow \(\sin \leftrightarrow \cos\). |
| \(D = 11180.49...\) so extra is \(7180\) N (3 s.f.) | A1 | Must be the extra force. |
# Question 3:
## Part (i)
| $F = 14000 \times 0.25$ | M1 | Use of N2L. Allow $F = mga$ and wrong mass. No extra forces. |
| $= 3500$ N | A1 | |
## Part (ii)
| $4000 - R = 3500$ so $500$ N | B1 | FT $F$ from (i). Condone negative answer. |
## Part (iii)
| $1150 - R_T = 4000 \times 0.25$ | M1 | N2L applied to truck (or engine) using all forces required. No extras. Correct mass. Do not allow use of $F = mga$. Allow sign errors. |
| $= 150$ N | A1 | cao |
## Part (iv)
| Component of weight down slope | M1 | Attempt to find component of *weight* (allow wrong mass). Accept $\sin \leftrightarrow \cos$. Accept use of $m\sin\theta$. |
| Extra driving force is component of $mg$ down slope | M1 | May be implied. Correct mass. No extra forces. Must have resolved weight component. Allow $\sin \leftrightarrow \cos$ |
| $14000g\sin 3° = 14000 \times 9.8 \times 0.0523359... = 7180.49...$ so $7180$ N (3 s.f.) | A1 | |
| **or** $D - 500 - 14000g\sin 3 = 14000 \times 0.25$ | M1 | N2L with all terms present with correct signs and mass. No extras. FT 500 N. Accept **their** $500 + 150$ for resistance. Must have resolved weight component. Allow $\sin \leftrightarrow \cos$. |
| $D = 11180.49...$ so extra is $7180$ N (3 s.f.) | A1 | Must be the extra force. |
---3 A train consists of an engine of mass 10000 kg pulling one truck of mass 4000 kg . The coupling between the engine and the truck is light and parallel to the track.
The train is accelerating at $0.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ along a straight, level track.\\
(i) What is the resultant force on the train in the direction of its motion?
The driving force of the engine is 4000 N .\\
(ii) What is the resistance to the motion of the train?\\
(iii) If the tension in the coupling is 1150 N , what is the resistance to the motion of the truck?
With the same overall resistance to motion, the train now climbs a uniform slope inclined at $3 ^ { \circ }$ to the horizontal with the same acceleration of $0.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(iv) What extra driving force is being applied?
\hfill \mbox{\textit{OCR MEI M1 2006 Q3 [10]}}