M3 June 2018 Q6

EdexcelOld spec15 marksVertical Circular Motion

Figure 2: particle P on a string of length a from O, projected vertically upwards with speed u from A where OA is horizontal; theta is the angle of the string above the horizontal — Figure 2

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 held at the point \(A\), where \(OA = a\) and \(OA\) is horizontal. The particle is projected vertically upwards with speed \(u\), as shown in Figure 2. When the string makes an angle \(\theta\) with the horizontal through \(O\) and the string is still taut, the tension in the string is \(T\).

(a) Show that \(T = \dfrac{m}{a}\left(u^2 - 3ag\sin\theta\right)\) (8)

The particle moves in complete circles.

(b) Find, in terms of \(a\) and \(g\), the minimum value of \(u\). (2)

Given that the least tension in the string is \(S\) and the greatest tension in the string is \(4S\),

Mark scheme (a)

Scheme	Marks
\(\dfrac{1}{2} \times m \times u^2 - \dfrac{1}{2} \times mv^2 = mga\sin\theta\)	M1A1A1
\(T + mg\sin\theta = m\dfrac{v^2}{a}\)	M1A1A1
\(T + mg\sin\theta = \dfrac{mu^2}{a} - 2mg\sin\theta\)
\(T = \dfrac{m}{a}\left(u^2 - 3ga\sin\theta\right)\) *	dM1A1cso
	(8)

Notes

M1 Attempt an energy equation from \(A\) to the general position (as shown in the diagram).
Must have a difference of two KE terms and a gain of PE (one or two terms)

A1A1 - 1 each error

M1 Attempt an equation of motion along the radius at the general position. Weight must be resolved (sin or cos). Acceleration can be in either form

A1 Resultant force correct

A1 Correct acceleration (as shown)

dM1 Eliminate \(v^2\) between their two equations and solve to \(T = \ldots\) Depends on both previous M marks.

A1cso Obtain the given expression for \(T\) with no errors seen

Mark scheme (b)

Scheme	Marks
At top \(T \geqslant 0 \ \Rightarrow u^2 \geqslant 3ag\)	M1
\(u_{\min} = \sqrt{3ag}\)	A1
	(2)

Notes

M1 Use \(\sin\theta = 1\) so \(T \geqslant 0\) at the top. Allow with \(\geqslant\) or \(>\) OR State min \(u\) when \(T = 0\) at the top

A1 \(u_{\min} = \sqrt{3ag}\) \(u_{\min} > 3ag\) or \(u_{\min} \geqslant 3ag\) scores A0

Mark scheme (c)

Scheme	Marks
Least at top: \(T_{least} = \dfrac{m}{a}\left(u^2 - 3ag\right)\ \ (= S)\)	M1A1
Greatest at bottom: \(T_{greatest} = \dfrac{m}{a}\left(u^2 + 3ag\right)\ \ (= 4S)\) (M1A1 for either A1 for second one)	A1
\(4 \times \dfrac{m}{a}\left(u^2 - 3ag\right) = \dfrac{m}{a}\left(u^2 + 3ag\right)\)	dM1
\(4u^2 - 12ag = u^2 + 3ag\)
\(3u^2 = 15ag, \quad u = \sqrt{5ag}\)	A1
	(5)
	(15 marks)

Notes

M1 Use the result given in (a) to obtain the tension at the top \((\theta = 90^\circ)\) or the tension at the bottom \((\theta = 270^\circ)\) Alt: Energy equation from \(A\) to either or both top and bottom.

A1 A1 One mark for each correct (Enter A1A1, A1A0 or A0A0)

dM1 Form an equation with 4 x their least = their greatest Both tensions to be of the form \(\dfrac{m}{a}\left(u^2 \pm kag\right)\) \(k \neq 0\). Depends on the previous M mark

A1 Correct expression for \(u\).