M3 June 2008 Q1

EdexcelOld spec9 marksHooke's Law Work-Energy Principle

Figure 1: spring from O at the closed end of a tube of length L, with P inside at distance one half L from O — Figure 1

A light elastic spring, of natural length \(L\) and modulus of elasticity \(\lambda\), has a particle \(P\) of mass \(m\) attached to one end. The other end of the spring is fixed to a point \(O\) on the closed end of a fixed smooth hollow tube of length \(L\).

The tube is placed horizontally and \(P\) is held inside the tube with \(OP = \tfrac{1}{2}L\), as shown in Figure 1. The particle \(P\) is released and passes through the open end of the tube with speed \(\sqrt{(2gL)}\).

(a) Show that \(\lambda = 8mg\). (4)

The tube is now fixed vertically and \(P\) is held inside the tube with \(OP = \tfrac{1}{2}L\) and \(P\) above \(O\). The particle \(P\) is released and passes through the open top of the tube with speed \(u\).

(b) Find \(u\). (5)

Sketch: horizontal spring from O to P — Mark scheme (a)

Sketch: vertical spring with P at the top — Mark scheme (b)

Scheme	Marks
EPE stored \(= \dfrac{1}{2}\dfrac{\lambda}{L}\left(\dfrac{1}{2}L\right)^2\ \ \left(= \dfrac{\lambda L}{8}\right)\)	B1
KE gained \(= \dfrac{1}{2}m\,2gL\ \ (= mgL)\)	B1
EPE = KE \(\Rightarrow \dfrac{\lambda L}{8} = mgL\) i.e. \(\lambda = 8mg\) *	M1A1cso
	(4)

Scheme	Marks
EPE = GPE + KE	M1
\(\dfrac{1}{2}\dfrac{8mg}{L}\left(\dfrac{1}{2}L\right)^2 = \dfrac{8mgL}{8} = mg\dfrac{L}{2} + \dfrac{1}{2}mu^2\)	A1A1
\(\dfrac{mgL}{2} = \dfrac{m}{2}u^2 \quad \therefore u = \sqrt{gL}\)	M1A1
	(5)
	(9 marks)