M3 June 2011 Q1

EdexcelOld spec6 marksVariable Force & Kinematics

1. A particle \(P\) of mass 0.5 kg moves on the positive \(x\)-axis under the action of a single force directed towards the origin \(O\). At time \(t\) seconds the distance of \(P\) from \(O\) is \(x\) metres, the magnitude of the force is \(0.375x^2\) N and the speed of \(P\) is \(v\) m s\(^{-1}\).

When \(t = 0\), \(OP = 8\) m and \(P\) is moving towards \(O\) with speed 2 m s\(^{-1}\).

(a) Show that \(v^2 = 260 - \tfrac{1}{2}x^3\). (4)

(b) Find the distance of \(P\) from \(O\) at the instant when \(v = 5\). (2)

Diagram: P of mass 0.5 kg on the x-axis with force 0.375x squared towards O and acceleration x-double-dot in the positive x-direction — Mark scheme (a)

Scheme	Marks
\(0.5v\dfrac{\mathrm{d}v}{\mathrm{d}x} = -0.375x^2\)	M1
\(\dfrac{1}{2}v^2 = -0.25x^3 + c\)	M1 A1
\(t = 0,\ v = 2,\ x = 8\) \(\dfrac{1}{2} \times 2^2 = -0.25 \times 8^3 + c\) \(c = 130\)
\(\therefore v^2 = -\dfrac{1}{2}x^3 + 260\) *	A1
	(4)

Scheme	Marks
\(v = 5 \quad x^3 = 520 - 50\)	M1
\(x = 7.77\)	A1
	(2)
	(6 marks)