M2 January 2013 Q5

EdexcelOld spec11 marksWork-Energy Principle

5. The point \(A\) lies on a rough plane inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = \dfrac{24}{25}\). A particle \(P\) is projected from \(A\), up a line of greatest slope of the plane, with speed \(U\) m s\(^{-1}\). The mass of \(P\) is 2 kg and the coefficient of friction between \(P\) and the plane is \(\dfrac{5}{12}\). The particle comes to instantaneous rest at the point \(B\) on the plane, where \(AB = 1.5\) m. It then moves back down the plane to \(A\).

(a) Find the work done against friction as \(P\) moves from \(A\) to \(B\). (4)

(b) Use the work-energy principle to find the value of \(U\). (4)

Force diagram: particle at A on slope at angle theta, U m s^-1 up the slope, N perpendicular, F down the slope, weight 2g, AB = 1.5 m — Mark scheme (a)

Scheme	Marks

\(N = 2g\cos\theta = \dfrac{14}{25}g\)	M1
\(F = \mu N = \dfrac{5}{12} \times \dfrac{14}{25}g = \dfrac{7g}{30}\)	B1
Work done \(= \dfrac{7}{30}g \times 1.5 = 3.43\ldots = 3.4\) J	DM1 A1
	(4)

Mark scheme (b)

Scheme	Marks
\(3.43 + 2g\sin\theta \times 1.5 = \dfrac{1}{2} \times 2U^2\)	M1 A1 A1
\(U = 5.626\ldots = 5.6\)	A1
	(4)

Notes

M1 Energy equation - needs all 3 terms, but condone sign errors & trig confusion. Must have an expression for the vertical height.

A1 Correct with one slip for their WD.

A1 All correct for their WD

A1 5.6 & 5.63 only

Diagram: particle returning down the slope, 1.5 m, speed v m s^-1 — Mark scheme (c)

\(mg\sin\theta - F = ma\) and \(v^2 = (u^2) + 2as\)	M1
\(2g\sin\theta - \dfrac{7g}{30} = \dfrac{48g}{25} - \dfrac{7g}{30} = 2a\)
\(a = \dfrac{253g}{300} = 8.26\ldots\)	A1
\(v^2 = 24.794,\ v = 5.0\)	A1

Scheme	Marks

\(2g\sin\theta \times 1.5 = 3.43 + \dfrac{1}{2} \times 2v^2\) OR: \(\dfrac{1}{2} \times 2U^2 = 2 \times 3.43 + \dfrac{1}{2} \times 2v^2\)	M1 A1
\(v^2 = 3g\sin\theta - 3.43\) \(v = 4.979\ldots\) Speed \(= 5.0\) m s\(^{-1}\)	A1
	(3)
	(11 marks)

Notes

Notes

Notes

Alt (c)