(a) A light inextensible string passes over a small fixed smooth pulley. A particle A of mass 10 kg is attached to one end of the string and a particle B of mass 5 kg is attached to the other end. The system is released from rest when B touches the ground and A is 1 m above the ground. Find (i) the speed of A as it hits the ground (ii) the height that B rises above the horizontal ground. (b) A mass $m_k$ kg is at rest on a smooth horizontal table. It is attached to a light inextensible string. The string, after passing over a small fixed pulley at the edge of the table, passes under a small movable pulley C, of mass $m_2$ kg. The string then passes over a smooth fixed pulley and supports a mass of 1 kg. The system is released from rest. (i) Find, in terms of $m_1$ and $m_2$, the tension in the string. (ii) The pulley C will remain at rest if $rac{2}{m_2} = rac{1}{m_1}k$. Find the value of k.

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(a) A light inextensible string passes over a small fixed smooth pulley - Leaving Cert Applied Maths - Question 4 - 2009

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(a) A light inextensible string passes over a small fixed smooth pulley. A particle A of mass 10 kg is attached to one end of the string and a particle B of mass 5 k... show full transcript

Worked Solution & Example Answer:(a) A light inextensible string passes over a small fixed smooth pulley - Leaving Cert Applied Maths - Question 4 - 2009

Step 1

(i) the speed of A as it hits the ground

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Answer

To find the speed of A as it strikes the ground, we utilize the principle of conservation of energy or kinematic equations. Since A descends a height of 1 m, we can apply the equation of motion:

$v^2 = u^2 + 2as$

Here, the initial velocity $u = 0$ , $a = g = 9.81 ext{ m/s}^2$ (acceleration due to gravity), and $s = 1 ext{ m}$ . Thus:

$v^2 = 0 + 2(9.81)(1)$ $v^2 = 19.62$ $v = rac{2}{ ext{3}} ext{ m/s}$

Therefore, the speed of A as it hits the ground is approximately $2.56 ext{ m/s}$ .

Step 2

(ii) the height that B rises above the horizontal ground

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Answer

When B touches the ground, it reveals that the mass A has fallen 1 m. Given the constraints of the system, as B rises, A falls the same distance. Therefore:

Using the kinematic equation for B: $v^2 = u^2 + 2as$ With $u = 0$ , the equation simplifies to: $0 = 2h - 2gs$

From the equation, we can deduce the height B rises: $h = rac{1}{3} ext{ m} + 1 ext{ m}$ Thus, the height B rises above the ground is approximately $$rac{4}{3} ext{ m}.$

Step 3

(i) Find, in terms of m1 and m2, the tension in the string

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Answer

For the mass $m_k$ on the table with mass attached to a pulley, we can apply Newton’s second law.

Let T be the tension in the string:

For $m_1$ (1 kg):
- The net force equation is :
- $T - g = m_1 a$
- Therefore, $T = m_1 g + m_1 a$ .
For $m_k$ :
- The mass is at rest:
- $m_k g - 2T = 0$
- Substitute back into the equation from $m_1$ : $m_k g = 2T$ Thus, $T = rac{m_k g}{2}$

Step 4

(ii) Find the value of k.

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Answer

To determine the value of k, we can utilize the given relation where the pulley will remain at rest: $rac{2}{m_2} = rac{1}{m_1}k$

Rearranging gives: k = $rac{2 m_1}{m_2}$

From our previous relationships, it identifies that: k = 1 when both terms are equal.

(a) A light inextensible string passes over a small fixed smooth pulley - Leaving Cert Applied Maths - Question 4 - 2009

Worked Solution & Example Answer:(a) A light inextensible string passes over a small fixed smooth pulley - Leaving Cert Applied Maths - Question 4 - 2009

(i) the speed of A as it hits the ground

(ii) the height that B rises above the horizontal ground

(i) Find, in terms of m1 and m2, the tension in the string

(ii) Find the value of k.

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