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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

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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 $\lambda, x, L$?

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1 Answer 1

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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 $2T\sin \theta > 3$.

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