M3 June 2008 Q5

EdexcelOld spec15 marksVertical Circular Motion

5. A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point \(O\). The particle is released from rest with the string taut and \(OP\) horizontal.

(a) Find the tension in the string when \(OP\) makes an angle of 60\(^\circ\) with the downward vertical. (6)

A particle \(Q\) of mass \(3m\) is at rest at a distance \(a\) vertically below \(O\). When \(P\) strikes \(Q\) the particles join together and the combined particle of mass \(4m\) starts to move in a vertical circle with initial speed \(u\).

(b) Show that \(u = \sqrt{\left(\dfrac{ga}{8}\right)}\). (3)

The combined particle comes to instantaneous rest at \(A\).

(i) the angle that the string makes with the downward vertical when the combined particle is at \(A\),

(ii) the tension in the string when the combined particle is at \(A\). (6)

Sketch: string at angle theta to the vertical, tension T, weight mg, acceleration v squared over a towards O, u = 0 at the horizontal — Mark scheme (a)

Sketch: string at angle theta to the vertical at A, where v = 0 — Mark scheme (c)

Scheme	Marks
Energy \(\dfrac{1}{2}mv^2 = mga\cos\theta\) \(v^2 = 2ga\cos\theta\)	M1A1
\(F = ma\) \(\nwarrow\) \(T - mg\cos\theta = \dfrac{mv^2}{a}\)	M1A1
Sub for \(\dfrac{v^2}{a}\): \(T = mg\cos\theta + 2mg\cos\theta\): \(\theta = 60\) \(\therefore T = \dfrac{3}{2}mg\)	M1A1
	(6)

Scheme	Marks
Speed of \(P\) before impact \(= \sqrt{2ga}\)	B1
PCLM : \(\therefore u = \dfrac{\sqrt{2ga}}{4} = \sqrt{\dfrac{ga}{8}}\) *	M1A1cso
	(3)

Scheme	Marks
(i) At \(A\) \(v = 0\) so conservation of energy gives: \(\dfrac{1}{2}\cancel{4m}u^2 = \cancel{4m}ga(1 - \cos\theta)\)	M1A1
\(\dfrac{ga}{16} = ga(1 - \cos\theta)\)	M1
\(\cos\theta = \dfrac{15}{16}\), \(\theta = 20^\circ\)	A1
(ii) \(\nearrow\) At \(A\) \(T = 4mg\cos\theta = \dfrac{15mg}{4}\) (accept 3.75mg)	M1A1
	(6)
	(15 marks)