CAIE M2 2003 November — Question 1 3 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2003
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeHorizontal circular track – friction only (no banking)
DifficultyModerate -0.8 This is a straightforward application of the circular motion formula F = mv²/r with all values given directly. It requires only substitution into a standard formula with no complicating factors like resolving forces, banked tracks, or multi-step reasoning—simpler than average A-level questions.
Spec6.05b Circular motion: v=r*omega and a=v^2/r

1 A railway engine of mass 50000 kg travels at a constant speed of \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) on a horizontal circular track of radius 1250 m . Find the magnitude of the horizontal force on the engine.

Question 1:
AnswerMarks Guidance
AnswerMark Guidance
For using Newton's second law with \(a = v^2/r\)M1
\(F = 50000 \frac{25^2}{1250}\)A1
Magnitude of the force is 25 000 NA1
# Question 1:

| Answer | Mark | Guidance |
|--------|------|----------|
| For using Newton's second law with $a = v^2/r$ | M1 | |
| $F = 50000 \frac{25^2}{1250}$ | A1 | |
| Magnitude of the force is 25 000 N | A1 | |

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1 A railway engine of mass 50000 kg travels at a constant speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on a horizontal circular track of radius 1250 m . Find the magnitude of the horizontal force on the engine.

\hfill \mbox{\textit{CAIE M2 2003 Q1 [3]}}