AS June 2019 Q3

EdexcelCurrent spec9 marksHorizontal Circular Motion

3. A light inextensible string has length \(8a\). One end of the string is attached to a fixed point \(A\) and the other end of the string is attached to a fixed point \(B\), with \(A\) vertically above \(B\) and \(AB = 4a\). A small ball of mass \(m\) is attached to a point \(P\) on the string, where \(AP = 5a\).

The ball moves in a horizontal circle with constant speed \(v\), with both \(AP\) and \(BP\) taut.

The string will break if the tension in it exceeds \(\dfrac{3mg}{2}\)
By modelling the ball as a particle and assuming the string does not break,

(a) show that \(\dfrac{9ag}{4} \lt v^2 \leqslant \dfrac{27ag}{4}\) (7)

(b) find the least possible time needed for the ball to make one complete revolution. (2)

force diagram: A vertically above B with AB = 4a; AP = 5a at angle theta to the vertical with tension T_A; BP = 3a horizontal with tension T_B; weight mg at P; right angle at B — Mark scheme (a)

Scheme	Marks	AO

No vertical motion: \(T_A\cos\theta = mg\)	M1	1.1b
\(T_A = \dfrac{5mg}{4}\)	A1	1.1b
Circular motion: \(T_B + T_A\sin\theta = m \times \dfrac{v^2}{r}\)	M1	3.1b
\(T_B + \dfrac{3}{5}T_A = m\dfrac{v^2}{3a}\)	A1	1.1b
\(T_B \gt 0\ \left(\Rightarrow v^2 \gt \dfrac{9ag}{4}\right)\)	DM1	2.1
\(T_B \leqslant \dfrac{3mg}{2} \Rightarrow m\dfrac{v^2}{3a} - \dfrac{3}{4}mg \leqslant \dfrac{3}{2}mg,\ \left(m\dfrac{v^2}{3a} \leqslant \dfrac{9mg}{4}\right)\)	DM1	2.1
\(\Rightarrow \dfrac{9ag}{4} \lt v^2 \leqslant \dfrac{27ag}{4}\) *	A1*	2.2a
	(7)

Mark scheme (b)

Scheme	Marks	AO
Use \(v^2 = \dfrac{27ag}{4}\) and \(T = \dfrac{2\pi r}{v}\) oe	M1	3.1b
\(T = 4\pi\sqrt{\dfrac{a}{3g}}\)	A1	1.1b
	(2)
	(9 marks)

Notes

M1: Correct method to find \(T\) in terms of \(a\) and (\(g\)) only.
They may sub 9.8 for \(g\) of course

A1: Any equivalent form but no fractions within fractions.
If they use 9.8 for \(g\), the numerical part needs to be to 2 sf or 3sf. i.e \(2.3\sqrt{a}\) or \(2.32\sqrt{a}\)