M2 June 2011 Q5

EdexcelOld spec10 marksWork-Energy Principle

Figure 2: plane inclined at 30 degrees with A and B on a line of greatest slope, AB = 2 m — Figure 2

A particle \(P\) of mass 0.5 kg is projected from a point \(A\) up a line of greatest slope \(AB\) of a fixed plane. The plane is inclined at 30\(^\circ\) to the horizontal and \(AB = 2\) m with \(B\) above \(A\), as shown in Figure 2. The particle \(P\) passes through \(B\) with speed 5 m s\(^{-1}\). The plane is smooth from \(A\) to \(B\).

(a) Find the speed of projection. (4)

The particle \(P\) comes to instantaneous rest at the point \(C\) on the plane, where \(C\) is above \(B\) and \(BC = 1.5\) m. From \(B\) to \(C\) the plane is rough and the coefficient of friction between \(P\) and the plane is \(\mu\).

By using the work-energy principle,

(b) find the value of \(\mu\). (6)

Diagram: particle projected from A at speed u up a plane inclined at 30 degrees, passing B at 5 m/s, weight 0.5g N — Mark scheme (a)

Scheme	Marks
\(0.5g \times 2\sin 30 = \dfrac{1}{2} \times 0.5u^2 - \dfrac{1}{2} \times 0.5 \times 5^2\)	M1 A1
\(\dfrac{1}{4}u^2 = 0.5g + \dfrac{1}{2} \times 0.5 \times 5^2\)
\(u = 6.7\) m s\(^{-1}\) (accept 6.68)	DM1 A1
	(4)

Mark scheme (b)

Scheme	Marks
\(R = 0.5g\cos 30\)	B1
\(F = 0.5g\cos 30 \times \mu\)	M1
Work done by friction \(= 1.5F\)
\(\dfrac{1}{2} \times 0.5 \times 5^2 = 1.5F + 0.5g \times 1.5\sin 30\)	M1 A1 A1
\(\mu = \dfrac{\frac{1}{2} \times 0.5 \times 5^2 - 0.5g \times 1.5\sin 30}{0.5g\cos 30 \times 1.5}\)
\(\mu = 0.40\quad\) (accept 0.4 or 0.405)	A1
	(6)
	(10 marks)