# Question

The ends of a light elastic string of natural length 0.8m and modulus of elasticity $$\lambda$$ N are attached to fixed points A and B which are 1.2m apart at the same horizontal level. A particle of mass 0.3kg is attached to the centre of the string, and released from rest at the mid-point of AB. The particle descends 0.32 m vertically before coming to instantaneous rest. Calculate $$\lambda$$.

Source : Cambridge A level Mathematics 9709 Paper 53 June 11 Q4

# My work

Let C be the equillibrium position of the particle and $$\angle ABC = \theta$$

Consider half string BC,

Using Pythagoras, BC = $$0.68$$ m

$$\sin \theta = 0.32/0.68 =8/17$$

Since the original length of this half-string was $$0.4$$ m, extension, $$x=0.68-0.4=0.28$$ m.

Tension in BC = $$T$$

$$2T\sin \theta= 3$$

$$T =3.1875$$

$$T=\frac{\lambda x}{L}=\frac{\lambda (0.28)}{0.4}$$

$$\lambda=\frac{255}{56}$$

# Mark scheme

Why is my method incorrect? Why is the principle of conservation of energy required? Isn't the formula $$T=\frac{\lambda x}{L}$$ enough since we already know the values of $$\lambda, x, L$$?

You are only told that the particle comes instantaneously to rest at C. It is not in equilibrium there, so you cannot conclude that $$2T\sin \theta = 3$$. Indeed, it is clear that there must be a net upwards force on the particle at C, so $$2T\sin \theta > 3$$.