Question 5 - A-Level Maths

OCR MEI M1 — Question 5 8 marks

Exam Board	OCR MEI
Module	M1 (Mechanics 1)
Marks	8
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 and resolving forces. All parts follow standard procedures with clear numerical values given, requiring only basic substitution and arithmetic. The inclined plane extension in part (iv) is routine for M1 level.
Spec	3.03c Newton's second law: F=ma one dimension 3.03d Newton's second law: 2D vectors 3.03k Connected particles: pulleys and equilibrium 3.03o Advanced connected particles: and pulleys 3.03v Motion on rough surface: including inclined planes

5 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.

What is the resultant force on the train in the direction of its motion? The driving force of the engine is 4000 N .
What is the resistance to the motion of the train?
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 \(0.25 \mathrm {~m} \mathrm {~s} ^ { 2 }\).
What extra driving force is being applied?

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Question 5:

(i)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(F = 14000 \times 0.25\)	M1	Use of N2L. Allow \(F = mga\) and wrong mass. No extra forces.
so \(3500\) N	A1

(ii)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(4000 - R = 3500\) so \(500\) N	B1	FT \(F\) from (i). Condone negative answer.

(iii)

Answer	Marks	Guidance
Answer/Working	Mark	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.
so \(150\) N	A1	cao

(iv)

Answer	Marks	Guidance
Answer/Working	Mark	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 M1	Attempt component of weight. 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 5:

**(i)**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = 14000 \times 0.25$ | M1 | Use of N2L. Allow $F = mga$ and wrong mass. No extra forces. |
| so $3500$ N | A1 | |

**(ii)**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4000 - R = 3500$ so $500$ N | B1 | FT $F$ from (i). Condone negative answer. |

**(iii)**

| Answer/Working | Mark | 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. |
| so $150$ N | A1 | cao |

**(iv)**

| Answer/Working | Mark | 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 M1 | Attempt component of weight. 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 |

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Show LaTeX source

5 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  Q5 [8]}}

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