M3 January 2013 Q4

EdexcelOld spec11 marksHorizontal Circular Motion

Figure 2: particle P of mass m on a string from A moving in a horizontal circle with centre O, OA = 2a — Figure 2

A particle \(P\) of mass \(m\) is attached to one end of a light elastic string, of natural length \(2a\) and modulus of elasticity \(6mg\). The other end of the string is attached to a fixed point \(A\). The particle moves with constant speed \(v\) in a horizontal circle with centre \(O\), where \(O\) is vertically below \(A\) and \(OA = 2a\), as shown in Figure 2.

(a) Show that the extension in the string is \(\dfrac{2}{5}a\). (6)

(b) Find \(v^2\) in terms of \(a\) and \(g\). (5)

Diagram: string at angle theta to the vertical AO of length 2a, tension T along the string, weight mg at P — Mark scheme (a)

Scheme	Marks
R\((\uparrow)\) \(T\cos\theta = mg\)	M1
\(T \times \dfrac{2a}{(2a + x)} = mg\)	A1
Hooke’s Law: \(T = \dfrac{6mgx}{2a} = \dfrac{3mgx}{a}\)	M1A1
\(\dfrac{3mgx}{a} \times \dfrac{2a}{(2a + x)} = mg\)	M1dep
\(6x = 2a + x\)
\(x = \dfrac{2}{5}a\) *	A1

Mark scheme (b)

Scheme	Marks
\(T\sin\theta = \dfrac{mv^2}{r}\)	M1A1
\(3mg \times \dfrac{2}{5}\sin\theta = \dfrac{mv^2}{\left(\dfrac{12a}{5}\right)\sin\theta}\)	M1dep
\(v^2 = \dfrac{6}{5}g \times \dfrac{12a}{5}\sin^2\theta\)
\(\sin^2\theta = 1 - \left(\dfrac{4a^2}{\left(\dfrac{12a}{5}\right)^2}\right) = \dfrac{11}{36}\)
\(v^2 = \dfrac{72ag}{25} \times \dfrac{11}{36} = \dfrac{22ag}{25}\)	M1depA1