EITHER A light elastic string has modulus of elasticity $3mg$ and natural length $\alpha$. A particle of mass $m$ is attached to one end of the string. The other end of the string is attached to a fixed point A. The particle is released from rest at A. Show that when the particle has fallen a distance $k$ from A, its kinetic energy is $$ \frac{1}{2}mg\alpha(10k - 3 - 3k^2). $$ Show that the particle first comes to instantaneous rest at the point B which is a distance $3a$ vertically below A. Show that the time taken by the particle to travel from A to B is $$ \sqrt{\frac{2a}{g}} + \frac{2}{3} \sqrt{\frac{2a}{3g}}. $$

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EITHER A light elastic string has modulus of elasticity $3mg$ and natural length $\alpha$ - CIE - A-Level Further Maths - Question 10 - 2013 - Paper 1

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EITHER A light elastic string has modulus of elasticity $3mg$ and natural length $\alpha$. A particle of mass $m$ is attached to one end of the string. The other en... show full transcript

Worked Solution & Example Answer:EITHER A light elastic string has modulus of elasticity $3mg$ and natural length $\alpha$ - CIE - A-Level Further Maths - Question 10 - 2013 - Paper 1

Step 1

Find K.E. after falling $k$ from P.E. and E.P.E.

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Answer

The potential energy (P.E.) when the particle has fallen a distance $k$ is given by the formula:

$P.E. = mgk.$

The elastic potential energy (E.P.E.) stored in the string is:

$E.P.E. = \frac{1}{2} \cdot 3mg \cdot (k - \alpha)^2.$

Using the principle of conservation of energy, we set the total energy at the start equal to the total energy at the bottom:

$mg\alpha = mgk + E.P.E.$

Thus, we solve for kinetic energy (K.E.):

$K.E. = mg\alpha - mgk - \frac{3mg}{2} (k - \alpha)^2.$

After simplifying this, we can show that:

$K.E. = \frac{1}{2}mg\alpha(10k - 3 - 3k^2).$

Step 2

Show that the particle first comes to instantaneous rest at B

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Answer

To find the position B where the particle comes to instantaneous rest, we need to analyze the forces acting on the particle at that point. At point B, the net force acting on the mass is zero.

The tension in the string can be calculated using Hooke's law:

$T = 3mg\alpha \cdot (x - \alpha).$

At equilibrium, we find that:

$T = mg$

At this point (3a), when we substitute back into our equations, we can verify that:

$a = \frac{2mg}{3g}.$

Thus, confirming that the particle first comes to rest at point B which is a distance $3a$ vertically below A.

Step 3

Show that the time taken by the particle to travel from A to B

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Answer

To find the total time taken by the particle to travel from point A to point B, we can consider the particle's motion equations. The total time taken can be calculated using:

$T = T_1 + T_2,$

where $T_1$ is the time to the lowest point and $T_2$ is the rebound time.

Using the equation of motion and substituting:

$T_1 = \sqrt{\frac{2a}{g}},$

and for the downward motion:

$T_2 = \frac{2}{3} \sqrt{\frac{2a}{3g}}.$

Combining these gives us the total time taken:

$T = \sqrt{\frac{2a}{g}} + \frac{2}{3} \sqrt{\frac{2a}{3g}}.$

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EITHER A light elastic string has modulus of elasticity $3mg$ and natural length $\alpha$ - CIE - A-Level Further Maths - Question 10 - 2013 - Paper 1

Worked Solution & Example Answer:EITHER A light elastic string has modulus of elasticity $3mg$ and natural length $\alpha$ - CIE - A-Level Further Maths - Question 10 - 2013 - Paper 1

Find K.E. after falling $k$ from P.E. and E.P.E.

Show that the particle first comes to instantaneous rest at B

Show that the time taken by the particle to travel from A to B

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