M3 January 2011 Q6

EdexcelOld spec13 marksHooke's Law Work-Energy Principle

Figure 4: ball P of mass 3m hanging on strings from A and B, AB = 2l, P a distance three-quarters l below the midpoint C — Figure 4

A small ball of mass \(3m\) is attached to the ends of two light elastic strings \(AP\) and \(BP\), each of natural length \(l\) and modulus of elasticity \(kmg\). The ends \(A\) and \(B\) of the strings are attached to fixed points on the same horizontal level, with \(AB = 2l\). The mid-point of \(AB\) is \(C\). The ball hangs in equilibrium at a distance \(\tfrac{3}{4}l\) vertically below \(C\) as shown in Figure 4.

(a) Show that \(k = 10\) (7)

The ball is now pulled vertically downwards until it is at a distance \(\tfrac{12}{5}l\) below \(C\). The ball is released from rest.

(b) Find the speed of the ball as it reaches \(C\). (6)

Force diagram: tensions T_a and T_b along PA and PB at angle theta to the vertical, weight 3mg, P three-quarters l below C — Mark scheme (a)

Diagram: ball pulled down to twelve-fifths l below C with both strings stretched — Mark scheme (b)

Scheme	Marks
length \(AP =\) length \(BP = \dfrac{5}{4}l\)	B1
\(T_a = T_b = \dfrac{kmg\left(\frac{1}{4}l\right)}{l} = \dfrac{1}{4}kmg \qquad (\text{or } T = \ldots)\)	M1A1
R\((\uparrow)\quad T_a\cos\theta + T_b\cos\theta = 3mg \qquad (\text{or } 2T\cos\theta = 3mg)\)	M1A1
\(\dfrac{1}{4}kmg \times \dfrac{3}{5} + \dfrac{1}{4}kmg \times \dfrac{3}{5} = 3mg \qquad \left(\text{or } \dfrac{1}{2}kmg \times \dfrac{3}{5} = 3mg\right)\) \(\dfrac{3}{10}kmg = 3mg\)	A1
\(k = 10\) *	A1
	(7)

Scheme	Marks
initial extn \(= \dfrac{13}{5}l - l = \dfrac{8}{5}l\)	B1
E.P.E. lost \(= 2 \times \dfrac{\lambda x^2}{2l} = 2 \times \dfrac{10mg}{2l}\left(\dfrac{8l}{5}\right)^2 = \dfrac{128mgl}{5}\) P.E. gained \(= 3mg \times \dfrac{12l}{5} = \dfrac{36mgl}{5}\)	M1A1
\(\dfrac{1}{2} \times 3mv^2 + \dfrac{36mgl}{5} = \dfrac{128mgl}{5}\) \(v^2 = \dfrac{256gl}{15} - \dfrac{72gl}{15}\)	M1A1
\(v = \sqrt{\left(\dfrac{184}{15}gl\right)}\)	A1
	(6)
	(13 marks)