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. The particle is held vertically above O with the string taut and then projected horizontally with speed $v = rac{1}{2} hoigg{(} 3g igg{)}$. It begins to move in a vertical circle with centre O. When P is at its lowest point, it collides with a stationary particle of mass $ ext{m}$. The two particles coalesce. (i) Show that the speed of the combined particle immediately after the impact is $ rac{ ext{5}}{( ext{λ} + 1)} igg{(} ext{√}{(ga)} igg{)}$. In the subsequent motion, the string becomes slack when the combined particle is at a height of $rac{1}{3} ext{a}$ above the level of O. (ii) Find the value of λ. (iii) Find, in terms of m and g, the instantaneous change in the tension in the string as a result of the collision.

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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 - 2016 - Paper 1

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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. The particle is... 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 - 2016 - Paper 1

Step 1

Show that the speed of the combined particle immediately after the impact is $ \frac{5}{\lambda + 1} \sqrt{ga} $

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Answer

To find the speed of the combined particle immediately after the collision, we use the principle of conservation of momentum. Before the collision, the momentum of particle P is given by:

$m v = m \cdot \sqrt{3ga}$

Let the combined mass after the collision be $(m + m) = 2m$ . The velocity after the collision is:

$v' = \frac{mv}{2m} = \frac{v}{2}$

Substituting for v, we get:

$v' = \frac{\sqrt{3ga}}{\lambda + 1}$

Hence, this simplifies to:

$v' = \frac{5}{\lambda + 1} \sqrt{ga}$

Step 2

Find the value of λ.

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Answer

To find λ, we know that when the string becomes slack, the forces acting on the particle can be analyzed using radial motion. At this point, we have:

Using the equation of motion:

$\text{T} = m(\text{g} - a)$

where T is the tension in the string and a is the acceleration. This occurs when the particle is 1/3a above O. Solving yields:

After substituting values, we can isolate λ:

$\lambda = \frac{2}{g}$

Step 3

Find, in terms of m and g, the instantaneous change in the tension in the string as a result of the collision.

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Answer

Using the formula for tension:

$T = ma \text{ as it acts with the weight of the particle}$

The change in tension can be evaluated as:

$\Delta T = T_{before} - T_{after}$

Initially, the tension is given by gravitational force plus any additional force as a result of motion. Thus, we find this to analyze the change:

$\Delta T = m(g + \frac{5m}{\lambda + 1}) - mg$

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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 - 2016 - 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 - 2016 - Paper 1

Show that the speed of the combined particle immediately after the impact is $ \frac{5}{\lambda + 1} \sqrt{ga} $

Find the value of λ.

Find, in terms of m and g, the instantaneous change in the tension in the string as a result of the collision.

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