Two particles A and B have masses 5m and km respectively, where k < 5. The particles are connected by a light inextensible string which passes over a smooth light fixed pulley. The system is held at rest with the string taut, the hanging parts of the string vertical and with A and B at the same height above a horizontal plane, as shown in Figure 4. The system is released from rest. After release, A descends with acceleration $ \frac{1}{4} g $. (a) Show that the tension in the string as A descends is $ \frac{15}{4} mg $. (b) Find the value of k. (c) State how you have used the information that the pulley is smooth. (d) After descending for 1.2 s, the particle A reaches the plane. It is immediately brought to rest by the impact with the plane. The initial distance between B and the pulley is such that, in the subsequent motion, B does not reach the pulley. (e) Find the greatest height reached by B above the plane.

A-Level Edexcel Maths Mechanics Forces: Two particles A and B have masse

Two particles A and B have masses 5m and km respectively, where k < 5 - Edexcel - A-Level Maths Mechanics - Question 6 - 2010 - Paper 1

Question 6

Two particles A and B have masses 5m and km respectively, where k < 5. The particles are connected by a light inextensible string which passes over a smooth light fi... show full transcript

Worked Solution & Example Answer:Two particles A and B have masses 5m and km respectively, where k < 5 - Edexcel - A-Level Maths Mechanics - Question 6 - 2010 - Paper 1

Step 1

Show that the tension in the string as A descends is $ \frac{15}{4} mg $.

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Answer

Utilizing Newton's second law for particle A:

For mass A:

5mg - T = 5m \cdot \frac{1}{4} g

This simplifies to:

T = 5mg - \frac{5}{4} mg = \frac{15}{4} mg

Thus, the tension in the string is indeed ( \frac{15}{4} mg ).

Step 2

Find the value of k.

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Answer

Using Newton's second law for particle B:

For mass B:

T - kmg = km \cdot \frac{1}{4} g

Substituting the value of tension from part (a):

\frac{15}{4} mg - kmg = km \cdot \frac{1}{4} g

Rearranging gives:

\frac{15}{4} mg = kmg + \frac{km}{4} g

Dividing by ( mg ):

\frac{15}{4} = k + \frac{k}{4}

Simplifying:

\frac{15}{4} = \frac{5k}{4}

Thus, ( k = 3 ).

Step 3

State how you have used the information that the pulley is smooth.

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Answer

The smoothness of the pulley indicates that there is no friction between the string and the pulley, ensuring that the tension remains constant throughout the string.

Step 4

After descending for 1.2 s, the particle A reaches the plane.

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Answer

To find the distance A descends, use the formula:

s_A = \frac{1}{2} a t^2 = \frac{1}{2} \cdot \frac{1}{4} g \cdot (1.2)^2 \approx 0.18 g \approx 1.764 \text{ m}$$ The speed just before impact is given by:

v = at = \frac{1}{4} g \cdot 1.2 \approx 2.94 \text{ m/s}$$

Since B does not reach the pulley, the distance B travels must be calculated under gravity.

Step 5

Find the greatest height reached by B above the plane.

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Answer

For particle B, we apply:

S = ut + \frac{1}{2} a t^2

where initial speed ( u = 0 ), acceleration = ( 2g ), and distance covered: After the last motion calculations, we find:

S_B = 2g \cdot (0.3)^2 / 2 = \approx 0.441 ext{ m}$$ Thus, summing the distances: \( S = s_{1.2} + 0.441 \approx 4.0 ext{ m} \).

Two particles A and B have masses 5m and km respectively, where k < 5 - Edexcel - A-Level Maths Mechanics - Question 6 - 2010 - Paper 1

Worked Solution & Example Answer:Two particles A and B have masses 5m and km respectively, where k < 5 - Edexcel - A-Level Maths Mechanics - Question 6 - 2010 - Paper 1

Show that the tension in the string as A descends is \( \frac{15}{4} mg \).

Find the value of k.

State how you have used the information that the pulley is smooth.

After descending for 1.2 s, the particle A reaches the plane.

Find the greatest height reached by B above the plane.

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