# A Mechanics Problem Involving an Elastic String

A bright sixth form pupil brought me the following problem which he said he had been struggling with for some time. The problem and my suggested solutions are recorded here.

The Problem

Suggested solution to part (a)

At point A the total energy consists of the sum of the kinetic energy of the particle $\frac{1}{2}mV^2$, the elastic potential energy which is given by

$\frac{\lambda}{2l}x^2 = \frac{6mg}{2a}\big(\frac{a}{8}\big)^2$

and the gravitational potential energy. We will take the zero level of the gravitational potential energy to be where the particle comes to instantaneous rest after being projected. The following diagram shows the situation:

Since the height of the particle at point A is $\frac{1}{4}a\sin 30^{\circ}$, the gravitational potential energy is $mg\frac{1}{4}a\sin 30^{\circ}$.

The total energy at point A after the particle has been projected is therefore

$\frac{1}{2}mV^2 + \frac{6mg}{2a}\big(\frac{a}{8}\big)^2 + mg\frac{1}{4}a\sin 30^{\circ}$

When the particle comes to instantaneous rest after being projected, the kinetic energy and the gravitational potential energy are zero. The extension of the string is $\frac{1}{8}a + \frac{1}{4}a = \frac{3}{8}a$. Therefore the elastic potential energy is

$\frac{\lambda}{2l}x^2 = \frac{6mg}{2a}\big(\frac{3a}{8}\big)^2$

By the principle of conservation of energy, we must have

$\frac{1}{2}mV^2 + \frac{6mg}{2a}\big(\frac{a}{8}\big)^2 + mg\frac{1}{4}a\sin 30^{\circ} = \frac{6mg}{2a}\big(\frac{3a}{8}\big)^2$

Solving this for $V$ gives $V = \sqrt{\frac{16}{32}ag}$. (Answer)

Suggested solution to part (b)

The extension of the string is zero when the string first becomes slack. Therefore the elastic potential energy there is zero. Let $V'$ denote the speed of $P$ there. The total energy will be the sum of the kinetic energy $\frac{1}{2}mV'^{ 2}$ and the gravitational potential energy. The following diagram shows the situation at the point where the string first becomes slack:

Since the height of the particle is $\frac{3}{8}a\sin 30^{\circ}$, the gravitational potential energy is $mg\frac{3}{8}a\sin 30^{\circ}$. The total energy at the point where the string first becomes slack is therefore

$\frac{1}{2}mV'^2 + mg\frac{3}{8}a\sin 30^{\circ}$

By the principle of conservation of energy, we must have

$\frac{6mg}{2a}\big(\frac{3a}{8}\big)^2 = \frac{1}{2}mV'^{ 2} + mg\frac{3}{8}a\sin 30^{\circ}$

Solving this for $V'$ gives $V' = \sqrt{\frac{15}{32}ag}$. (Answer)