Question 1 - A-Level Maths

CAIE M2 2003 June — Question 1 4 marks

Exam Board	CAIE
Module	M2 (Mechanics 2)
Year	2003
Session	June
Marks	4
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Centre of Mass 1
Type	Frame with circular arc or semicircular arc components
Difficulty	Standard +0.3 This is a straightforward centre of mass problem requiring students to find the centroid of a circular ring (at its centre), the centroid of a rod (at its midpoint), and then combine using the standard weighted average formula. The geometry is simple (finding the perpendicular distance from centre to chord using Pythagoras), and the calculation is routine. Slightly above average difficulty only because it involves composite bodies and basic geometric reasoning.
Spec	6.04c Composite bodies: centre of mass

1 \includegraphics[max width=\textwidth, alt={}, center]{7f8646df-a7d8-4ca1-a6ee-3ceab6bb83af-2_533_497_269_824} A frame consists of a uniform circular ring of radius 25 cm and mass 1.5 kg , and a uniform rod of length 48 cm and mass 0.6 kg . The ends \(A\) and \(B\) of the rod are attached to points on the circumference of the ring, as shown in the diagram. Find the distance of the centre of mass of the frame from the centre of the ring.

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Question 1:

Answer	Marks	Guidance
Answer	Mark	Guidance
Distance from centre to rod is \(\sqrt{25^2 - 24^2}\)	B1
Taking moments about centre of ring, mid-point of rod, or C.O.M. of frame (correct number of terms required in equation)	M1
\((1.5 + 0.6)\bar{x} = 0.6 \times 7\) or \((1.5 + 0.6)(7 - \bar{x}) = 1.5 \times 7\)	A1
\(1.5\bar{x} = 0.6(7 - \bar{x})\)
Distance is 2cm	A1	SR: Allow M1 for \(48.7 = (50\pi + 48)\bar{x}\)

# Question 1:

| Answer | Mark | Guidance |
|--------|------|----------|
| Distance from centre to rod is $\sqrt{25^2 - 24^2}$ | B1 | |
| Taking moments about centre of ring, mid-point of rod, or C.O.M. of frame (correct number of terms required in equation) | M1 | |
| $(1.5 + 0.6)\bar{x} = 0.6 \times 7$ or $(1.5 + 0.6)(7 - \bar{x}) = 1.5 \times 7$ | A1 | |
| $1.5\bar{x} = 0.6(7 - \bar{x})$ | | |
| Distance is 2cm | A1 | SR: Allow M1 for $48.7 = (50\pi + 48)\bar{x}$ |

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\includegraphics[max width=\textwidth, alt={}, center]{7f8646df-a7d8-4ca1-a6ee-3ceab6bb83af-2_533_497_269_824}

A frame consists of a uniform circular ring of radius 25 cm and mass 1.5 kg , and a uniform rod of length 48 cm and mass 0.6 kg . The ends $A$ and $B$ of the rod are attached to points on the circumference of the ring, as shown in the diagram. Find the distance of the centre of mass of the frame from the centre of the ring.

\hfill \mbox{\textit{CAIE M2 2003 Q1 [4]}}

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