M2 June 2005 Q2

Figure 1: trapezium ABCD with AB = AD = 4 cm, CD = 5 cm, right angles at A and B — Figure 1

A thin uniform wire, of total length 20 cm, is bent to form a frame. The frame is in the shape of a trapezium \(ABCD\), where \(AB = AD = 4\) cm, \(CD = 5\) cm, and \(AB\) is perpendicular to \(BC\) and \(AD\), as shown in Figure 1.

(a) Find the distance of the centre of mass of the frame from \(AB\). (5)

The frame has mass \(M\). A particle of mass \(kM\) is attached to the frame at \(C\). When the frame is freely suspended from the mid-point of \(BC\), the frame hangs in equilibrium with \(BC\) horizontal.

(b) Find the value of \(k\). (3)

Diagram: the trapezium with BC = 4 + x, AD = 4 and CD = 5 — Mark scheme (a)

Scheme	Marks
\((x = 3)\)
\(M(AB)\): \(7 \times 3.5 + 5 \times 5.5 + 4 \times 2 = 20 \times \bar{x}\)	M1 A2,1,0
\(\Rightarrow\ 20\bar{x} = 24.5 + 27.5 + 8 = 60\ \Rightarrow\ \bar{x} = 3\) cm dep	M1 A1
	(5)

Scheme	Marks
\(M(XY)\):
\(M \times (3.5 - 3) = kM \times 3.5\)	M1 A1ft
\(\Rightarrow k = \tfrac{1}{7}\).	A1
	(3)
	(8 marks)