AS June 2024 Q4

EdexcelCurrent spec12 marksCentres of Mass

Figure 3: right-angled triangle ABC with the right angle at A, AB horizontal along the top and AC = 6a vertical; rectangle DEFG with AD = a, DE = 2a and EF = 3a removed from the top edge, EB = 6a; the template is shaded — Figure 3

The uniform triangular lamina \(ABC\) has \(AB\) perpendicular to \(AC\), \(AB = 9a\) and \(AC = 6a\). The point \(D\) on \(AB\) is such that \(AD = a\).

The rectangle \(DEFG\), with \(DE = 2a\) and \(EF = 3a\), is removed from the lamina to form the template shown shaded in Figure 3.

The distance of the centre of mass of the template from \(AC\) is \(d\).

(a) Show that \(d = \dfrac{23}{7}a\) (3)

The template is freely suspended from \(A\) and hangs in equilibrium with \(AB\) at an angle \(\theta^\circ\) to the downward vertical through \(A\).

(b) Find the value of \(\theta\) (5)

A new piece, of exactly the same size and shape as the template, is cut from a lamina of a different uniform material. The template and the new piece are joined together to form the model shown in Figure 4. Both parts of the model lie in the same plane.

Figure 4: the shaded template ADGFEBC with AC vertical, joined along AC to the new piece CPQRSTA, its mirror image on the left with TS = 6a along the bottom, SR = 3a, RQ = 2a, PC = a; a horizontal force X is applied at T — Figure 4

The weight of \(CPQRSTA\) is \(W\)

The weight of \(ADGFEBC\) is \(4W\)

The model is freely suspended from \(A\).

A horizontal force of magnitude \(X\), acting in the same vertical plane as the model, is now applied to the model at \(T\) so that \(AC\) is vertical, as shown in Figure 4.

Mark scheme (a)

Scheme	Marks	AO
M(\(AC\))	M1	2.1
\((27-6)a^2 d = 27a^2 \times 3a - 6a^2 \times 2a\ \left(= 69a^3\right)\)	A1	1.1b
\(\Rightarrow d = \dfrac{69}{21}a = \dfrac{23}{7}a\) *	A1*	2.2a
	(3)

Notes

M1: Take moments about \(AC\) or a parallel axis. Dimensionally correct terms. Correct number of terms.

A1: Correct unsimplified equation

A1*: Obtain the given answer including “\(d =\)” from correct working

Mark scheme (b)

Scheme	Marks	AO
M(\(AB\))	M1	3.1a
\((27-6)a^2\bar{y} = 27a^2 \times 2a - 6a^2 \times 1.5a\ \left(= 45a^3\right)\)	A1	1.1b
\(\bar{y} = \dfrac{45}{21}a = \dfrac{15}{7}a\)	A1	1.1b
\(\tan\theta^\circ = \dfrac{15}{23}\)	M1	3.1a
\(\theta = 33\) \((\theta = 33.111..)\)	A1	1.1b
	(5)

Notes

M1: Take moments about \(AB\) or a parallel axis. Dimensionally correct terms. Correct number of terms.

A1: Correct unsimplified equation

A1: Correct vertical distance. Any equivalent form

M1: Correct use of trig to find a relevant angle

A1: 2 sf or better (33.111…)

Mark scheme (c)

Scheme	Marks	AO
M(\(A\))	M1	3.1a
\(6aX + \dfrac{23}{7}a \times W = \dfrac{23}{7}a \times 4W\)	A1 A1	1.1b 1.1b
\(X = \dfrac{23}{14}W\)	A1	1.1b
	(4)
	(12 marks)

Notes

M1: Complete method to form an equation in \(X\) e.g by taking moments about \(A\).
\(5W\) must be split correctly and equation must be dimensionally correct.

A1: Unsimplified equation with at most one error

A1: Correct unsimplified equation

A1: \(1.6W\) or better \((1.6428\ldots W)\) (corrected from the printed mark scheme: it prints “\(3.1W\) or better \((3.1428\ldots W)\)”, which does not match \(X = \dfrac{23}{14}W\))