A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. When P is hanging at rest vertically below O, it is projected horizontally. In the subsequent motion P completes a vertical circle. The speed of P when it is at its highest point is u. Show that the least possible value of u is $\sqrt{ag}$. [2] It is now given that u = $\sqrt{ag}$. When P passes through the lowest point of its path, it collides with, and coalesces with, a stationary particle of mass $\frac{m}{2}$. Find the speed of the combined particle immediately after the collision. [4] In the subsequent motion, when OP makes an angle $\theta$ with the upward vertical the tension in the string is T. Find an expression for T in terms of m, g, and $\theta$. [5] Find the value of $\cos \theta$ when the string becomes slack. [2]

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A particle P of mass m is attached to one end of a light inextensible string of length a - CIE - A-Level Further Maths - Question 4 - 2015 - Paper 1

Question 4

A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. When P is hangi... show full transcript

Worked Solution & Example Answer:A particle P of mass m is attached to one end of a light inextensible string of length a - CIE - A-Level Further Maths - Question 4 - 2015 - Paper 1

Step 1

Show that the least possible value of u is $\sqrt{ag}$

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Answer

To show that the least possible value of the speed $u$ at the highest point of the vertical circle is $\sqrt{ag}$ , we can apply the principles of circular motion. At the highest point, the only forces acting on the particle are its weight and the tension in the string. For the particle to complete the circle, the centripetal force must be at least equal to the weight of the particle:

$T + mg = \frac{mv^2}{a},$

If we consider the case where the particle is just about to lose contact with the string, the tension $T$ will be zero at the highest point. Hence, we have:

$mg = \frac{mu^2}{a}.$

From this, we can rearrange to find:

$u^2 = ag \Rightarrow u = \sqrt{ag}.$

Step 2

Find the speed of the combined particle immediately after the collision

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Answer

When particle P collides with the stationary particle of mass $\frac{m}{2}$ , we apply the conservation of momentum. Before the collision, the momentum of particle P is:

$p_{initial} = mv.$

After the collision, the combined mass is:

$m + \frac{m}{2} = \frac{3m}{2}.$

Using conservation of momentum:

$mv = \frac{3m}{2}v_{final}.$

We can find $v_{final}$ by rearranging:

$v_{final} = \frac{2v}{3}.$

Substituting the value of $u = \sqrt{ag}$ gives:

$v_{final} = \frac{2\sqrt{ag}}{3}.$

Step 3

Find an expression for T in terms of m, g, and $\theta$

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Answer

For the angle $\theta$ , the forces acting on the particle when the string makes an angle $\theta$ with the vertical include the tension $T$ and the weight $mg$ . The radial force equation gives:

$T - mg \cos \theta = \frac{mv^2}{a}.$

Rearranging this equation allows us to express T as:

$T = mg \cos \theta + \frac{mv^2}{a}.$

Step 4

Find the value of $\cos \theta$ when the string becomes slack

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Answer

The string becomes slack when the tension $T$ drops to zero. Thus, from our previous expression for tension:

$0 = mg \cos \theta + \frac{mv^2}{a}.$

Rearranging gives us:

$mg \cos \theta = -\frac{mv^2}{a}.$

Since $v^2$ is a positive value, this equation can be simplified to:

$\cos \theta = -\frac{v^2}{ga}.$

By considering energy conservation at the lowest point, we can find the transition conditions leading to the identification of $heta$ .

A particle P of mass m is attached to one end of a light inextensible string of length a - CIE - A-Level Further Maths - Question 4 - 2015 - Paper 1

Worked Solution & Example Answer:A particle P of mass m is attached to one end of a light inextensible string of length a - CIE - A-Level Further Maths - Question 4 - 2015 - Paper 1

Show that the least possible value of u is $\sqrt{ag}$

Find the speed of the combined particle immediately after the collision

Find an expression for T in terms of m, g, and $\theta$

Find the value of $\cos \theta$ when the string becomes slack

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