A2 October 2021 Q4

EdexcelCurrent spec10 marksHorizontal Circular Motion

Figure 4: A vertically above B with AB = l; the string runs from A down to the ring R and from R horizontally to B; R moves on a dashed horizontal circle with centre B — Figure 4

One end of a light inextensible string of length \(2l\) is attached to a fixed point \(A\). A small smooth ring \(R\) of mass \(m\) is threaded on the string and the other end of the string is attached to a fixed point \(B\). The point \(B\) is vertically below \(A\), with \(AB = l\). The ring is then made to move with constant speed \(V\) in a horizontal circle with centre \(B\). The string is taut and \(BR\) is horizontal, as shown in Figure 4.

(a) Show that \(BR = \dfrac{3l}{4}\) (2)

Given that air resistance is negligible,

(b) find, in terms of \(m\) and \(g\), the tension in the string, (4)

Scheme	Marks	AO
\(l^2 + r^2 = (2l - r)^2\), using Pythagoras	M1	1.1b
\(BR = \dfrac{3l}{4}\) *	A1*	1.1b
	(2)

Mark scheme (b)

Scheme	Marks	AO
Resolve vertically	M1	2.1
\(T\cos\alpha = mg\)	A1	1.1b
Overall strategy to solve problem: substitute for \(\cos\alpha\) and solve for \(T\)	M1	3.1b
\(T = \dfrac{5mg}{4}\)	A1	1.1b
	(4)

Notes

M1: Allow sin/cos confusion

A1: Correct equation

M1: Substituting for their trig ratio and solving for \(T\)

A1: cao

Mark scheme (c)

Scheme	Marks	AO
Equation of motion horizontally	M1	2.1
\(T + T\sin\alpha = \dfrac{mV^2}{r}\)	A1	1.1b
Overall strategy to solve problem: substitute for \(T\), \(\sin\alpha\) and \(r\) and solve for \(V\)	M1	3.1b
\(V = \sqrt{\dfrac{3gl}{2}}\)	A1	1.1b
	(4)
	(10 marks)

Notes

M1: Correct no. of terms, dimensionally correct

A1: Correct equation

M1: Substitute for \(T\), \(\sin\alpha\) and \(r\) and solve for \(V\)

A1: cao. Accept other equivalent forms