M3 June 2014 (R) Q3

EdexcelOld spec11 marksHooke's Law Work-Energy Principle

3. One end \(A\) of a light elastic string \(AB\), of modulus of elasticity \(mg\) and natural length \(a\), is fixed to a point on a rough plane inclined at an angle \(\theta\) to the horizontal. The other end \(B\) of the string is attached to a particle of mass \(m\) which is held at rest on the plane. The string \(AB\) lies along a line of greatest slope of the plane, with \(B\) lower than \(A\) and \(AB = a\). The coefficient of friction between the particle and the plane is \(\mu\), where \(\mu < \tan\theta\). The particle is released from rest.

(a) Show that when the particle comes to rest it has moved a distance \(2a(\sin\theta - \mu\cos\theta)\) down the plane. (6)

(b) Given that there is no further motion, show that \(\mu \geqslant \dfrac{1}{3}\tan\theta\). (5)

Mark scheme (a)

Scheme	Marks
\(R = mg\cos\theta\)	B1
WD against friction \(= \mu xmg\cos\theta\)	B1
\(\mu xmg\cos\theta = mgx\sin\theta - \dfrac{mgx^2}{2a}\)	M1 A2
\(x = 2a(\sin\theta - \mu\cos\theta)\) **	A1
	(6)

Notes

B1 correct equation perpendicular to the plane

B1 correct expression for work done against friction

M1 work-energy equation

A2 fully correct; A1 one error;

A1 correct expression for \(x\) no errors in the working

Mark scheme (b)

Scheme	Marks
\(T = \dfrac{mg\,2a(\sin\theta - \mu\cos\theta)}{a} = 2mg(\sin\theta - \mu\cos\theta)\)	B1
No motion if \(T \leqslant mg\sin\theta + \mu mg\cos\theta\)	M1 A1
\(2mg(\sin\theta - \mu\cos\theta) \leqslant mg\sin\theta + \mu mg\cos\theta\)	DM1
\(\tfrac{1}{3}\tan\theta \leqslant \mu\) **	A1
	(5)
	(11 marks)

Notes

B1 use Hooke's law to obtain a correct expression for \(T\)

M1 using NL2 parallel to the plane to set up an inequality for situation where no motion

A1 correct inequality

DM1 solving to get an inequality for \(\mu\)

A1 correct inequality and no errors in the working

If only error is use of \(<\) instead of \(\leqslant\), deduct final A mark only