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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 \( \frac{15}{4} mg \).
Answer
Using Newton's Second Law for mass A:
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The forces acting on A:
- Weight acting downward: ( 5mg )
- Tension acting upward: ( T )
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Setting up the equation of motion:
[ 5mg - T = 5m \cdot \frac{1}{4} g ]
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Rearranging the equation:
[ T = 5mg - \frac{5mg}{4} = \frac{20mg}{4} - \frac{5mg}{4} = \frac{15mg}{4} ]
Thus, we find that the tension ( T = \frac{15}{4} mg ).
Step 2
Find the value of k.
Answer
Applying Newton's Second Law for mass B:
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The forces acting on B:
- Tension acting upward: ( T )
- Weight acting downward: ( kmg )
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Setting up the equation of motion:
[ T - kmg = km \cdot \frac{1}{4} g ]
Substituting ( T = \frac{15}{4} mg ):
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Rearranging the equation:
[ \frac{15}{4} mg - kmg = km \cdot \frac{1}{4} g ]
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Factoring out ( mg ):
[ \frac{15}{4} - k = \frac{1}{4} k ]
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Solving for ( k ):
[ \frac{15}{4} = (1 + \frac{1}{4})k ]
[ k = 3 ]
Thus, the value of k is 3.
Step 3
State how you have used the information that the pulley is smooth.
Answer
The smoothness of the pulley implies that there is no friction between the string and the pulley. This means the tension remains constant throughout the string and does not vary from mass A to mass B. It allows us to equate the tensions directly in the equations for both masses.
Step 4
Find the greatest height reached by B above the plane.
Answer
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Calculate the distance A falls: Using the equation of motion:
[ s = ut + \frac{1}{2} a t^2 ]
For A, where ( u = 0 ), ( a = \frac{1}{4} g ), and ( t = 1.2 ) s:
[ s_{A} = 0 + \frac{1}{2} \cdot \frac{1}{4}g \cdot (1.2)^2 = 0.18g ]
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Speed of A when reaching the ground:
[ v = u + at = 0 + \left(\frac{1}{4} g \cdot 1.2\right) = 0.3g ]
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For B under gravity: Its distance fallen in the same time:
[ s_{B} = 2gs = 2g \cdot 0.3^2 = g \cdot 0.09 = 0.441 ]
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Therefore, the distance B travels is the initial height it was raised before falling:
Height reached by B = height from which it started - distance fallen.
Thus, calculating all values:
[ s_{B} = 3.969 + 0.441 \approx 4.0 \text{ m} ]