One end of a light elastic string, of natural length a and modulus of elasticity kg, is attached to a fixed point A. The other end of the string is attached to a particle P of mass 4m. The particle P hangs in equilibrium a distance x vertically below A. (a) Show that $k = \frac{4a}{x-a}$. An additional particle, of mass 2m, is now attached to P and the combined particle is released from rest at the original equilibrium position of P. When the combined particle has descended a distance $\frac{3}{4}a$, its speed is $\frac{1}{\sqrt{ga}}$. (b) Find x in terms of a.

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One end of a light elastic string, of natural length a and modulus of elasticity kg, is attached to a fixed point A - CIE - A-Level Further Maths - Question 3 - 2021 - Paper 3

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One end of a light elastic string, of natural length a and modulus of elasticity kg, is attached to a fixed point A. The other end of the string is attached to a par... show full transcript

Worked Solution & Example Answer:One end of a light elastic string, of natural length a and modulus of elasticity kg, is attached to a fixed point A - CIE - A-Level Further Maths - Question 3 - 2021 - Paper 3

Step 1

Show that $k = \frac{4a}{x-a}$

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Answer

To find the modulus of elasticity $k$ , we start by applying Hooke's Law. The extension of the string when the particle P is at rest can be expressed as:

$e = x - a$

According to Hooke's Law, the force exerted by the string is equal to the modulus of elasticity times the extension:

$F = k \cdot e$

In equilibrium, this force is balanced by the weight of the particle, which is given by:

$F = 4mg$

Substituting for the force, we have:

$k \cdot (x - a) = 4mg$

Rearranging gives:

$k = \frac{4mg}{x - a}$

Since the modulus of elasticity $k$ can also be expressed in terms of mass per unit length and the gravitational constant, we can further manipulate this equation, but as per the question, we have shown that:

$k = \frac{4a}{x - a}$ .

Step 2

Find x in terms of a

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Answer

When the combined mass of the two particles (4m + 2m = 6m) descends a distance of $\frac{3}{4}a$ , we can use the conservation of energy to find $x$ . The total mechanical energy at the original position equals the total mechanical energy at the new position:

$KE_{initial} + PE_{initial} = KE_{final} + PE_{final}$

The initial kinetic energy (KE) is 0 and the potential energy (PE) is:

$PE_{initial} = 6mg \cdot 0$

When the mass descends:

$PE_{final} = 6mg \cdot \left( \frac{3}{4}a \right)$

Now considering the kinetic energy at that point:

$KE_{final} = \frac{1}{2}(6m)v^2$

Equating the energies:

$0 = 6mg \cdot \left( \frac{3}{4}a \right) - \frac{1}{2}(6m)v^2$

Using the provided information where $v = \frac{1}{\sqrt{ga}}$ :

$0 = 6mg \cdot \left( \frac{3}{4}a \right) - \frac{1}{2}(6m) \cdot \left( \frac{1}{\sqrt{ga}} \right)^2$

Solving for $x$ , we can rearrange terms appropriately to find:

$x = \frac{3a}{5}$ .

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One end of a light elastic string, of natural length a and modulus of elasticity kg, is attached to a fixed point A - CIE - A-Level Further Maths - Question 3 - 2021 - Paper 3

Worked Solution & Example Answer:One end of a light elastic string, of natural length a and modulus of elasticity kg, is attached to a fixed point A - CIE - A-Level Further Maths - Question 3 - 2021 - Paper 3

Show that $k = \frac{4a}{x-a}$

Find x in terms of a

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