Edexcel iGCSE Physics (4PH1) 1.5 Moments Exam Style Question Paper 1B - New Syllabus
Question
A pencil has a weight of \(0.16\,\mathrm{N}\).
(a) What is the mass of the pencil?
A \(1.6\,\mathrm{g}\)
B \(16\,\mathrm{g}\)
C \(160\,\mathrm{g}\)
D \(1600\,\mathrm{g}\)
(b) The diagram shows the pencil with one end resting on a small block.
A finger provides an upwards force, \(F\), to keep the pencil horizontal.
(i) The weight of the pencil is \(0.16\,\mathrm{N}\). Calculate the moment of the weight of the pencil about the pivot. Use the formula
moment = force × perpendicular distance from the pivot
(ii) State the moment of the force \(F\).
(iii) Show that force \(F\) is \(0.080\,\mathrm{N}\).
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
• 1.30P–1.33P: Moments, Principle of Moments, and Moments on a Beam — parts (b)(i)–(iii)
▶️ Answer/Explanation
(a) Mass of the pencil [1 mark]
Use the relationship:
\(W=mg\)
Taking \(g=10\,\mathrm{N\,kg^{-1}}\):
\(0.16=m\times10\)
\(m=\dfrac{0.16}{10}=0.016\,\mathrm{kg}\)
\(0.016\,\mathrm{kg}=16\,\mathrm{g}\)
Correct answer: B, \(16\,\mathrm{g}\)
(b)(i) Moment of the weight [2 marks]
Use:
\(\mathrm{moment}=\mathrm{force}\times\mathrm{perpendicular\ distance}\)
From the diagram, the perpendicular distance from the pivot to the line of action of the weight is \(3.7\,\mathrm{cm}\).
\(\mathrm{moment}=0.16\times3.7\)
\(\mathrm{moment}=0.592\,\mathrm{N\,cm}\)
Therefore:
\(\boxed{\mathrm{moment}=0.592\,\mathrm{N\,cm}}\)
(b)(ii) Moment of force \(F\) [1 mark]
The pencil remains horizontal and does not rotate, so it is in rotational equilibrium. Therefore, the clockwise and anticlockwise moments must be equal.
Hence:
\(\boxed{\mathrm{moment\ of\ }F=0.592\,\mathrm{N\,cm}}\)
(b)(iii) Force \(F\) [2 marks]
Use:
\(\mathrm{moment}=F\times\mathrm{perpendicular\ distance}\)
The perpendicular distance from the pivot to the line of action of \(F\) is \(7.4\,\mathrm{cm}\).
Therefore:
\(0.592=F\times7.4\)
Rearranging:
\(F=\dfrac{0.592}{7.4}\)
\(F=0.080\,\mathrm{N}\)
\(\boxed{F=0.080\,\mathrm{N}}\)
Question
Diagram 1 shows a wooden plank balanced horizontally on two supports, A and B. A block is suspended from the plank between the supports by a cable of negligible weight.
(a) The weight of the block is \(260\,\mathrm{N}\).
(i) State the formula linking moment, force and perpendicular distance from the pivot.
(ii) By taking moments about support A, calculate force \(F\). Assume the weight of the plank is negligible.
(iii) Explain what will happen to the magnitude of force \(F\) if the block is moved towards support B.
(b) Diagram 2 shows the block and the cable connecting the block to the plank.
(i) The centre of gravity of the block is located at point X. Draw an arrow on diagram 2 to show the weight of the block.
(ii) The block also experiences a force due to the tension in the cable. Explain why the block remains stationary when it is supported by this tension force.
(iii) Explain why the forces acting on the block are not an example of Newton’s third law of motion.
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
• 1.31P: Centre of Gravity — part (b)(i)
• 1.29P: Newton’s Third Law — part (b)(iii)
▶️ Answer/Explanation
(a)(i) Moment formula [1 mark]
The formula linking moment, force and perpendicular distance is:
\(\boxed{\mathrm{moment}=\mathrm{force}\times\mathrm{perpendicular\ distance}}\)
(a)(ii) Calculating force \(F\) [3 marks]
Take moments about support A. Since the plank is balanced, the clockwise moment equals the anticlockwise moment.
The moment of the block’s weight about A is:
\(\mathrm{moment}=260\times0.25\)
\(\mathrm{moment}=65\,\mathrm{N\,m}\)
Therefore:
\(65=F\times0.80\)
Rearranging:
\(F=\dfrac{65}{0.80}\)
\(F=81.25\,\mathrm{N}\)
\(\boxed{F\approx81\,\mathrm{N}}\)
(a)(iii) Moving the block towards support B [3 marks]
The magnitude of force \(F\) increases.
- The distance of the block from support A increases.
- Therefore, the clockwise moment produced by the block’s weight increases.
- For the plank to remain balanced, the anticlockwise moment produced by \(F\) must also increase.
Since the distance from A to the point where \(F\) acts is fixed, \(F\) must increase to produce the larger balancing moment.
(b)(i) Weight of the block [2 marks]
Draw a vertical downward arrow starting at point X, the centre of gravity of the block.
The weight acts vertically downwards through the centre of gravity.
(b)(ii) Why the block remains stationary [2 marks]
- The tension in the cable acts upwards.
- The tension is equal in magnitude to the weight of the block.
- Therefore, the forces balance and there is no resultant force.
- With no resultant force, the block has no acceleration and remains stationary.
(b)(iii) Newton’s third law [2 marks]
The tension and weight acting on the block are not a Newton’s third-law pair because both forces act on the same body, the block.
Newton’s third-law force pairs act on different bodies and are equal in magnitude and opposite in direction.
For example, the force exerted by the cable on the block has a third-law partner: the force exerted by the block on the cable.