A2 June 2025 Q4

EdexcelCurrent spec9 marksCentres of Mass

Figure 5: rectangle ACDF with AC = FD = 15a and AF = CD = 3a; B on AC with AB = 6a, BC = 9a; E on FD with FE = 9a, ED = 6a; line BE — Figure 5

The uniform rectangular lamina \(ACDF\) shown in Figure 5 has \(AC = FD = 15a\) and \(AF = CD = 3a\). The point \(B\) on \(AC\) is such that \(AB = 6a\). The point \(E\) on \(FD\) is such that \(ED = 6a\).

Figure 6: folded lamina: ABEF with AF = 3a at the top and AB = 6a vertical, and BEDC with BC horizontal, CD = 3a and ED = 6a; dashed line from F down to E — Figure 6

The rectangular lamina is folded along \(BE\) to form the folded lamina shown in Figure 6.
The folded lamina has \(AB\) perpendicular to \(BC\) and the two sections, \(ABEF\) and \(BEDC\), of the lamina lie in the same plane.

(a) Show that the distance of the centre of mass of the folded lamina from \(EF\) is \(\dfrac{2}{5}a\). (5)

(b) Explain why the centre of mass of the folded lamina lies on the perpendicular bisector of \(BE\). (1)

The folded lamina is freely suspended from \(F\) and hangs in equilibrium.

Mark scheme (a)

Scheme

Marks

	Left rectangle	2 triangles	Right rectangle	Total
Mass ratio	2 \(\left(18a^2\right)\)	1 \(\left(2 \times \dfrac{9}{2}a^2\right)\)	2 \(\left(18a^2\right)\)	5 \(\left(45a^2\right)\)
From \(EF\)	\(-1.5a\)	\(-a\)	\(3a\)	\(d\)

B1
B1

1.1b
1.1b

Moments about \(EF\)

2.1

\(2 \times -1.5a + 1 \times -a + 2 \times 3a = 5d\)

1.1b

Distance \(= \dfrac{2}{5}a\) *

A1*

2.2a

(5)

Notes

B1: Correct mass ratios

B1: Correct distances from \(EF\) or a parallel axis

M1: Moments about \(EF\) or a parallel axis. Need all terms. Dimensionally correct. Condone sign errors.

A1: Correct unsimplified moments equation (E.g. \(45d = 18 \times 1.5a + 9 \times 2a + 18 \times 6a\))

A1*: Obtain given answer from correct working.

Diagram: folded lamina hanging from F; vertical distance 6a − (2/5)a from F, horizontal distance (2/5)a, angle theta between FE and the line to the centre of mass — Mark scheme (c)

Scheme	Marks	AO
The perpendicular bisector of \(BE\) is a line of symmetry	B1	2.4
	(1)

Scheme	Marks	AO

Vertical distance of c of m from \(F = \dfrac{28a}{5}\)	B1	3.1a
\(\tan\theta = \dfrac{2}{28}\)	M1	1.1b
\(\theta = \tan^{-1}\left(\dfrac{1}{14}\right) = 4.09^\circ \quad (4.1^\circ) \quad (0.071 \text{ rads})\)	A1	1.1b
	(3)
	(9 marks)