M3 June 2011 Q6

EdexcelOld spec12 marksVertical Circular Motion

Figure 5: circle of radius a about O with B at the top, C at the bottom and P held at A level with O — Figure 5

A particle \(P\) 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 point \(B\) is vertically above \(O\) and the point \(C\) is vertically below \(O\), with \(OB = OC = a\), as shown in Figure 5. The particle is projected vertically upwards with speed \(3\sqrt{(ag)}\).

(a) Show that \(P\) will pass through \(B\). (6)

(b) Find the speed of \(P\) as it reaches \(C\). (2)

As \(P\) passes through \(C\) it receives an impulse. Immediately after this, the speed of \(P\) is \(\dfrac{5}{12}\sqrt{(11ag)}\) and the direction of motion of \(P\) is unchanged.

Diagram: speed V at B with force mg + T_B towards O, projection speed 3 root(ag) upwards at A — Mark scheme (a)

Diagram: string at angle theta to the downward vertical with the particle at rest, speed five-twelfths root(11ag) at the lowest point — Mark scheme (c)

Scheme	Marks
Energy to \(B\): \(\dfrac{1}{2}m\left(3\sqrt{ag}\right)^2 - \dfrac{1}{2} \times mV^2 = mag\)	M1 A1
\(9ag - V^2 = 2ag\) \(V^2 = 7ag\)
NL2 along radius at \(B\): \(T_B + mg = m\dfrac{V^2}{a}\)	M1 A1
\(T_B + mg = 7mg\)	M1
\(T_B = 6mg\) \(T_B > 0 \Rightarrow\) particle reaches \(B\)	A1
	(6)

Scheme	Marks
Energy to \(C\): \(\dfrac{1}{2} \times mU^2 - \dfrac{1}{2}m\left(3\sqrt{ag}\right)^2 = mag\)	M1
\(U^2 = 2ag + 9ag\) \(U = \sqrt{11ga}\)	A1
	(2)

Scheme	Marks
Energy from \(C\) to rest: \(\dfrac{1}{2} \times m \times \left(\dfrac{5}{12}\sqrt{11ag}\right)^2 = mga(1 - \cos\theta)\)	M1 A1
\(\dfrac{25}{144} \times 11ag = 2ga(1 - \cos\theta)\)
\(\cos\theta = \dfrac{1}{2}\left(2 - \dfrac{25 \times 11}{144}\right)\)	M1
\(\theta = 87.4\ldots\) \(\theta = 87^\circ \quad (\text{or } 1.5 \text{ rad}) \quad\) or better	A1
	(4)
	(12 marks)