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P3 07 Moments
224 minutes
222 marks
Q1. The diagram shows two buses. Bus A is empty. Bus B contains bags of sand upstairs to represent passengers.
Each bus has been tilted as far as it can without falling over.
(a) Each bus will topple over if it is tilted any further.
Explain, in as much detail as you can, why this will happen.
(You can draw on one of the diagrams as part of your answer if you want to.)
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(b) What difference does it make to the stability of the bus when the upper deck is full of “passengers”? Explain your answer as fully as you can.
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(c) Why are the bags of sand in bus B only put upstairs?
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(Total 6 marks)
Q2. A spanner makes it a lot easier to loosen a bolt.
You cannot usually loosen It is easier a bolt with your fingers. with a spanner.
Choose words from this list to complete the sentences below.
lever piston pivot pulley turning effect
The spanner is a simple ............................................................
You use it to produce a bigger ........................................................... on the bolt.
A longer spanner works better.
This is because there is a bigger distance between your force and the ...............................(Total 3 marks)
Q3. The diagram shows a simple machine for lifting water from a river.
(a) Calculate the turning force (moment) of the bucket of water.
(Show your working.)
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Answer .......................................................... Nm (newton metre)(2)
(b) What can you say about the size of downwards force the operator must use to balance the moment of the bucket of water?
(Explain your answer, using numbers if you can.)
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(Total 6 marks)
Q4. A child stands a wooden brick on its end as shown in the diagram.
The child then pushes the brick to make it tilt.
How far must the brick be tilted to make it fall over?
Explain your answer.
(You may draw a labelled diagram if you wish.)
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(Total 2 marks)
Q5. The diagram below shows an outline of a balance. The balance is used to weigh lorries. A fraction of the weight of a lorry is used as the load on the right side of the pivot.
A standard weight W is moved along the arm until the weight of the load is balanced.
(a) As the weight W is moved away from the pivot it can support a heavier load.Why is this?
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(b) (i) The weight W is 100 N. When it is 0.2 m from the pivot it balances the load.Calculate the moment of the weight W about the pivot.
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Answer ..................................... Nm(2)
(ii) The load is one hundredth of the weight of the lorry and is 0.02 m from the pivot.Calculate the weight of the lorry.
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Answer ..................................... N(2)
(Total 6 marks)
Q6. (a) The diagram shows three similar toys. Each toy should be able to balance on a narrow rod. The arrows show the direction in which the weight of the toy acts.
Only one of the toys balances on the rod, the other two fall over. Which one of the toys is balanced? Explain the reason for your choice.
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(3)
(b) The diagram shows a simple toy. Different animal shapes can be positioned so that the 50 cm rod balances horizontally.
(i) Use the following equation to calculate the moment exerted by the elephant shape of weight 2N about the pivot P. Show clearly how you work out your answer and give the unit.
moment = force × perpendicular distance from pivot
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Moment = ...................................(3)
(ii) Use the following relationship to calculate the weight of the monkey shape.
total clockwise moment = total anticlockwise moment
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Weight = ................................ N(2)
(c) The graph shows how the length of the spring changes as the total weight of the different animal shapes change.
Use the graph to find how much the spring extends when the elephant shape and the monkey shape are hung from the rod. Show how you get your answer.
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Extension of spring = ................... cm(2)
(Total 10 marks)
Q7. For part of the ride the cyclist pushed on the pedals with a constant vertical force of 300 N. The simplified diagrams show the pedals in three different positions.
(i) Which position, A, B, or C, gives the largest moment on the pedal?
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(ii) Use the following equation to calculate, in Newton metres, the size of the largest moment on the pedal.
moment = force × perpendicular distance from pivot
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Moment = ................................... Nm(2)
(Total 3 marks)
Q8. The diagram shows a spanner being used to undo a tight nut.
The nut was tightened using a moment of 120 newton metres.
Use the following equation to calculate the force needed to undo the nut. Show clearly how you work out your answer.
moment = force × perpendicular distance from pivot
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Force = ............................................. N(Total 2 marks)
Q9. (a) The diagrams show a windsurfer pulling up the sail of a sailboard. The mast pivots at point P.
In which position, A, B or C must the windsurfer pull with the largest force? Give a reason for your answer.
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(b) Once the mast is upright, the windsurfer and the sailboard are in equilibrium.
(i) What does in equilibrium mean?
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(ii) The weight of the windsurfer is 700 newtons. Use the equation below to calculate the moment exerted by the windsurfer on the sailboard. Show clearly how you work out your answer.
moment = force × perpendicular distance from pivot
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Moment = ......................................... Nm(2)
(iii) Use the relationship below to calculate the horizontal force of the wind on the sail. Show clearly how you work out your answer.
total clockwise moment = total anticlockwise moment
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Force = ........................................ N(2)
(c) As the wind speed increases the windsurfer leans further out from the sailboard.
This position allows the windsurfer and sailboard to stay in equilibrium. Explain why.
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(Total 10 marks)
Q10. (a) The diagram shows a lifebelt. It is hanging freely from hook Y.
(i) On the diagram, mark with an X the point where you think the centre of mass of the lifebelt will be.
(1)
(ii) Explain why you have chosen this point.
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(b) The drawing shows Susan on a diving board. She is 1.5 metres from point P and she weighs 500 N.
Calculate her moment (turning effect) about point P.Show clearly how you work out your answer and give the unit.
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Moment about P = ................................................(3)
(c) Susan has a case with wheels.
When she packs this case, she puts the heaviest items at the end where the wheels are.This means that the heaviest items are less likely to crush the other contents and it helps her to find things when she opens the case.
Explain another advantage of packing her case in this way.
To gain full marks in this question you should write your ideas in good English. Put them into a sensible order and use the correct scientific words.
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(Total 10 marks)
Q11. (a) Every object has a centre of mass. What is meant by the centre of mass?
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(b) The drawing shows a thin sheet of plastic. The sheet is 250 mm wide. Two holes, each with a radius of 2 mm, have been drilled through the sheet.
Describe how you could use:
• a clamp and stand• a steel rod 100 mm long and with a radius of I mm• a weight on a thin piece of string (= a plumb line)• a ruler• a pen which will write on the plastic sheet
to find the centre of mass of the plastic sheet.
To gain full marks in this question you should write your ideas in good English. Put them into a sensible order and use the correct scientific words.
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(c) There is a trapdoor in the ceiling of a house.The trapdoor weighs 44 N.The drawing shows a side view of the trapdoor.
(i) Complete the three spaces to give the equation which is used to calculate the turning effect of a force.
......................... = ......................... × perpendicular between ......................... line of action and pivot
(1)
(ii) Calculate the turning effect, about the hinge, due to the weight of the trapdoor.
Show clearly how you work out your final answer and give the unit.
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Turning effect = ..............................................(3)
(Total 10 marks)
Q12. The diagram shows a crane which is loading containers onto a ship.
(a) Use the equation in the box to calculate the moment of the container which is being loaded.
moment = force ×
perpendicular distance from the line of
action of the force to the axis of rotation
Show clearly how you work out your answer and give the unit.
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Moment of the container = .......................................................(3)
(b) Suggest and explain the purpose of the large concrete blocks.
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(Total 6 marks)
Q13. A girl and her father visit a children’s playground.
(a) The diagram shows the girl holding on to a roundabout which is turning.
A centripetal force must act because the girl moves in a circular path.
(i) In which direction, P, Q, R or S, does the centripetal force act?
Direction ..........(1)
(ii) What provides this centripetal force?
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(iii) Her father pushes the roundabout so that it turns faster. The girl continues to stand on the same part of the roundabout.
Complete the following sentence by drawing a ring around the correct line in the box.
The centripetal force on the girl
decreases
does not change
increases
(1)
(b) The diagram shows the girl and her father on a see-saw.
(i) Use the equation in the box to calculate the moment of the girl.
moment = force × perpendicular distance from the line of action of the force to the axis of rotation
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Moment of the girl = ....................................... Nm(2)
(ii) What must her father do to increase his moment?
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(c) The diagram shows part of a level road that they take when they drive home. They drive at a steady speed.
(i) At which point, A, B, C or D, will the centripetal force on the car be greatest?
Centripetal force is greatest at .................... .(1)
(ii) What provides the centripetal force when the car goes round a bend?
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(Total 8 marks)
Q14. Tractors are often used on sloping fields, so stability is important in their design.
On the diagram, the centre of the X marks the centre of mass of the tractor.
(a) Explain why the tractor has not toppled over. You may add to the diagram to help you to explain.
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(b) Give two features of the tractor which affect its stability and state how each feature could be changed to increase the tractor’s stability.
Feature 1 .....................................................................................................................
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Feature 2 .....................................................................................................................
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(Total 5 marks)
Q15. A spanner gives a turning effect to undo a nut.
(a) Complete the sentence.
The turning effect of a force is called the .................................................... of the force.(1)
(b) The diagram shows a spanner being used.
Use the equation in the box to calculate the spanner’s turning effect in newton metres.
turning effect = perpendicular distance from the line of action of the force to the axis of rotation
Show clearly how you work out your answer.
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Turning effect = .............................. Nm(2)
(c) Give two ways in which you can increase the spanner’s turning effect.
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(Total 5 marks)
Q16. The drawing shows a sign which hangs outside a shop.
(a) Draw an X on the sign so that the centre of your X is at the centre of mass of the sign.(1)
(b) Use a ruler to draw one axis of symmetry on the sign.(1)
(c) One force which acts on the sign is its weight.
Complete the following sentence by drawing a ring around the correct line in the box.
The moment of the weight produces
an accelerating
a balancing
a turning
effect.
(1)(Total 3 marks)
Q17. The diagram shows a small mobile crane. It is used on a building site.
The distance, d, is measured to the front of the cab.
The table shows information from the crane driver’s handbook.
Load in kilonewtons (kN) Maximum safe distance, d, in metres (m)
10 6.0
15 4.0
24 2.5
40 1.5
60 1.0
(a) What is the relationship between the load and the maximum safe distance?
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(b) The crane driver studies the handbook and comes to the conclusion that a load of 30 kN would be safe at a distance, d, of 2.0 metres.
Is the driver correct?
Explain your answer.
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(c) What is the danger if the driver does not follow the safety instructions?
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(d) How should the data in the table have been obtained?
Put a tick ( ) in the box next to your answer.
average results from an opinion poll of mobile crane drivers
copied from a handbook for a similar crane
results of experiments on a model mobile crane
results of experiments on this mobile crane
(1)(Total 6 marks)
Q18. A student wants to weigh himself but the only balance available is a newtonmeter that measures up to 200 newtons.The diagram shows how the student solved the problem using moments.
(a) Use the information in the diagram to calculate the weight of the student given by this method.
Write down the equation you use, and then show clearly how you work out your answer and give the unit.
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Weight = ....................................(5)
(c) Even though all the measurements are accurate the student’s weight obtained by this method is inaccurate.
Explain why.
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(Total 7 marks)
Q19. Tractors are often used on sloping fields, so stability is important in their design.
On the diagram, the centre of the X marks the centre of mass of the tractor.
(a) What is meant by the term centre of mass?
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(b) Explain how the design of the tractor could be changed in order to increase the tractor’s stability.
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(c) Explain why the tractor does not topple over. You may add to the diagram to help your explanation.
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(Total 6 marks)
Q20. (a) The diagram shows a gardener using a steel bar to lever a tree stump out of the ground.
When the gardener pushes with a force of 300 N the tree stump just begins to move.
Calculate the moment produced by the gardener on the steel bar.
Write down the equation you use, and then show clearly how you work out your answer and give the unit.
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Moment = ........................................(4)
(b) Using a longer steel bar would have made it easier for the gardener to lever the tree stump out of the ground.
Explain why.
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(Total 7 marks)
Q21. The diagram shows a design for a crane. The crane is controlled by a computer.
The purpose of the motors and gears is to change the pulling force in the steel cable.This is done so that the jib stays horizontal whatever the size of the load or the position of the load.
Use the equation in the box to answer questions (a) and b).
moment = force × perpendicular distance from the line of
actionof the force to the axis of rotation
(a) Calculate the moment caused by the load in the position shown in the diagram.
Show clearly how you work out your answer and give the unit.
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Moment = ..............................................................(3)
(b) Calculate the pulling force that is needed in the steel cable to keep the jib horizontal.
Show clearly how you work out your answer.
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Pulling force = .............................................................. N(2)
(Total 5 marks)
Q22. The diagrams show two concrete mixers.
Concrete mixer A Concrete mixer B
On each diagram, the centre of the white X marks the centre of mass of the concrete mixer and its contents.
(a) Complete the sentence to explain what the term centre of mass means.
The centre of mass of a concrete mixer and its contents is ........................
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(b) Both diagrams are drawn to the same scale.
Concrete mixer B is more stable than concrete mixer A.
The two features which make concrete mixer B more stable are:
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(c) Use the terms ‘line of action of the weight’ and ‘resultant moment’ to explain why a stable concrete mixer does not fall over when it is given a small push.
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(Total 5 marks)
Q23. Two children visit a playground.
(a) The diagram shows them on a see-saw. The see-saw is balanced.
Complete the following sentences by drawing a ring around the correct word or line in the box.
force.
(i)
The turning effect of the girl’s weight is called her load.
moment.
(1)
balance
(ii)
Point P is the axis of rotation of the see-saw.
turning
(1)
the boy moves nearer to point P.
(iii) To make end A of the see-saw go up, the girl moves nearer to point P.
the girl moves nearer to end A.
(1)
(b) In another part of the playground, a tyre has been suspended from a bar.
(i) Draw an X on the diagram so that the centre of the X marks the centre of mass of the tyre.
(1)
(ii) Complete the sentence by using the correct word or phrase from the box.
above below to the left of to the right of
If the suspended tyre is pushed, it will come to rest with its centre of mass
directly .................................................. the point of suspension.(1)
(Total 5 marks)
Q24. The diagram shows a back view of a computer monitor.
(a) In normal use, the monitor is stable.
(i) Explain the meaning, in the above sentence, of the word stable.
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(ii) State the relationship between the total clockwise moment and the total anticlockwise moment about any axis of the monitor when it is stable.
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(b) The instruction booklet explains that the screen can be tilted.It also includes a warning.
Caution
The monitor can tip over if the screen is tilted too far back.
Explain why the monitor will tip over if the screen is tilted too far back.
Include the words centre of mass, weight and moment in your explanation.
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(Total 6 marks)
Q25. (a) A student investigates the moment of a force.
(i) What does the word moment mean in this sentence?
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(ii) The diagram shows how she sets up her apparatus.
Suggest the purpose of the G-clamp.
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(iii) A horizontal rod fits into a hole at the centre of the metre ruler. This is the axis of rotation. The student changes the load Y and adjusts the distance X until the metre ruler is horizontal. She takes six pairs of measurements which are shown in the table.
Load Yin newtons
Distance Xin centimetres
1 7
2 14
3 21
4 28
5 35
6 42
Explain fully how distance X varies with load Y.
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(iv) The weight of the ruler can be ignored in this experiment.
Which statement gives the reason why?
Put a tick ( ) in the box next to your answer.
The weight of the ruler is so small it is negligible.
The centre of mass of the ruler is at the axis of rotation.
The ruler is a symmetrical object.
(1)
(b) In the summer, a town council fits hanging baskets to some of its lamp posts.
Use the information in the diagram and the equation in the box to calculate the moment produced by the weight of the hanging basket about an axis through point A.
moment = force × perpendicular distance from the line ofaction of the force to the axis of rotation
Show clearly how you work out your answer and give the unit.
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Moment =................................................................................(3)
(Total 8 marks)
Q26. The diagram shows a fork-lift truck with a load of 2.4 kN. The clockwise moment caused by this load is 2880 Nm.
(a) Use the equation in the box to calculate the distance d.
moment = force × perpendicular distance from the line ofaction of the force to the axis of rotation
Show clearly how you work out the answer and give the unit.
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Distance d = .......................................................(3)
(b) This warning notice is in the driver’s cab.
Explain in terms of moments why the maximum load must not be exceeded.
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(Total 5 marks)
Q27. (a) A student holds a ruler at one end and slides a weight along the ruler.
At which point, A, B or C, will the turning effect of the weight feel greatest?
Write your answer, A, B or C, in the box. Point
(1)
(b) Complete the following sentence by drawing a ring around the correct word in the box.
axis
The turning effect of a force is called the equilibrium of the force.
moment
(1)
(c) In a human arm, the biceps muscle provides the force needed to hold the arm horizontal.A student uses a model in which a rubber band represents the biceps muscle.
Complete the following sentence by drawing a ring around the correct line in the box.
To hold the model arm horizontal, the pull from the rubber band will be
bigger than
smaller than
the force caused by the weight.
the same as
(1)
(d) The diagram shows a long spanner.
Use the equation in the box to calculate the moment, in N cm, being produced.
moment = force × perpendicular distance from the line ofaction of the force to the axis of rotation
Show clearly how you work out your answer.
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Moment = .................................. N cm(2)
(Total 5 marks)
Q28.The diagram shows a device called a current balance.
(a) (i) When the switch is closed, the part of the wire labelled XY moves upwards.
Explain why.
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(ii) What is the name of the effect that causes the wire XY to move?
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(iii) An alternating current (a.c.) is a current which reverses direction. How many times the current reverses direction in one second depends on the frequency of the alternating supply.
Describe the effect on the wire XY if the battery is replaced by an a.c. supply having
a frequency of 5 hertz.
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(b) The diagram shows how a small weight can be used to make the wire XY balance horizontally.
Use the data in the diagram and the equation in the box to calculate the force, F, acting on the wire XY.
moment = force × perpendicular distance from the line ofaction of the force to the axis of
rotation
Show clearly how you work out your answer.
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Force = ............................................................ N(3)
(Total 8 marks)
Q29.The diagram shows someone starting to lift the end of a heavy wooden pole.
(a) Use the equation in the box to calculate the moment produced by the weight of the pole.
moment = force × perpendicular distance from the line ofaction of the force to the axis of rotation
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Moment = .................................................. Nm(2)
(b) (i) Complete the following sentence by drawing a ring around the correct line in the box.
bigger than
The smallest force needed to lift the end of the pole will be the same as
smaller than
the weight of the pole.(1)
(ii) Give a reason for your answer to part (b)(i).
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(c) How could the person lifting the end of the pole increase the moment?
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(Total 5 marks)
Q30.The diagram shows a father and his two children sitting on a playground see-saw.The see-saw is not moving.
(a) What is the total clockwise moment of the two children about the axis of rotation?
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Explain the reason for your answer.
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(b) (i) What is the clockwise moment of the boy, B, about the axis of rotation?
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Moment = .................................................. Nm(1)
(ii) Use the information in the diagram and the equation in the box to calculate the weight, W, of the boy, B.
moment = force × perpendicular distance from the line ofaction of the force to the axis of rotation
Show clearly how you work out your answer.
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Weight of boy B = .................................................. N(2)
(Total 6 marks)
Q31. A company makes a wheel wrench with an extending handle. The company claims that the extending handle makes it easier to loosen the wheel nuts on a car.
The diagram shows the wheel wrench being used without the handle extended.
(a) (i) Use the equation in the box to calculate the moment produced by the force on the wrench.
moment = force × perpendicular distance from the line ofaction of the force to the axis of rotation
Show clearly how you work out your answer.
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Moment = ............................................................. newton metres(2)
(ii) Units can be written in words or symbols.
Which of the following is the unit for a moment written using symbols?
Draw a ring around your answer.
nm Nm nM NM(1)
(b) The wheel nut will not move and so the handle of the wrench is extended.
It is now easy to loosen the wheel nut using the same force as before.
Explain why.
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(Total 5 marks)
Q32.The diagram shows a design for a crane. The crane is controlled by a computer.
The purpose of the motors and gears is to change the pulling force in the steel cable. This is done so that the jib stays horizontal whatever the size of the load or the position of the load.
Use the equation in the box to answer questions (a) and (b).
moment = force x perpendicular distance from the line of
action of the force to the axis of rotation
(a) Calculate the moment caused by the load in the position shown in the diagram.
Show clearly how you work out your answer and give the unit.
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........................................................................................................................
Moment = ........................................................................(3)
(b) Calculate the pulling force that is needed in the steel cable to keep the jib horizontal.
Show clearly how you work out your answer.
........................................................................................................................
........................................................................................................................
Pulling force = .................................................. N
(2)(Total 5 marks)
Q33.The drawing shows a plastic toy which can stand on its feet.
(a) (i) Draw an X on the diagram so that the centre of the X marks the likely position of the centre of mass of the toy.
(1)
(ii) Explain the reason for your choice in part (a)(i).
...............................................................................................................
...............................................................................................................(1)
(b) Suggest two ways in which the design of the toy could be altered to make the toy more stable.
1 .....................................................................................................................
........................................................................................................................
2 .....................................................................................................................
.........................................................................................................................(2)
(Total 4 marks)
Q34.The diagram shows a gardener using a steel bar to lift a tree stump out of the ground.
When the gardener pushes with a force of 300 N, the tree stump just begins to move.
(a) Use the equation in the box to calculate the moment produced by the 300 N force.
moment = force x
perpendicular distance from the line of action of the force to the axis of rotation
Show clearly how you work out your answer.
........................................................................................................................
........................................................................................................................
Moment = .................................................. newton metres(2)
(b) Using a longer steel bar would have made it easier for the gardener to lift the tree stump out of the ground.
Explain why.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................(2)
(Total 4 marks)
Q35.The diagram shows a gardener using a steel bar to lift a tree stump out of the ground.
When the gardener pushes with a force of 300 N, the tree stump just begins to move.
(a) Use the equation in the box to calculate the moment produced by the 300 N force.
moment = force x
perpendicular distance from the line of action of the force to the axis of rotation
Show clearly how you work out your answer.
........................................................................................................................
........................................................................................................................
Moment = .................................................. newton metres(2)
(b) Using a longer steel bar would have made it easier for the gardener to lift the tree stump out of the ground.
Explain why.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................(2)
(Total 4 marks)
M1. (a) idea
• line of action of weight/force/gravity(if drawn: a vertical line through the centre of mass)
• falls outside the (wheel) base (mark NOT from diagram)for 1 mark each
2
(b) ideas that
• less stable/topples more easily
• centre of mass at a higher level
• so need small angle to make line of action of weight fall outside(wheel) base
for 1 mark each3
(c) idea that
this is the most unstable condition (when bus used)orthis makes c. of m. as high as it is likely to be
for 1 mark1
[6]
M2. lever
turning effect
pivotfor 1 mark each
[3]
M3. (a) evidence of moment = force × distance or 200 × 1.5
gains 1 mark
but 300gains 2 marks
2
(b) ideas that smaller than loadgains 1 mark
but 100 N or half the loadgains 2 marks
because applied further from pivotgains 1 mark
but applied 2 × distance from pivot or evidence of balancing momentsgains 2 marks(working for (b) shown in (a) gains credit – transfer mark)
4[6]
M4. any evidence of idea that weight acts through/near centre of mass/gravity/brickgains 1 mark
but clear indication that brick topples ifvertical line through centre of mass is outside base line of brickor line of action of weight is outside base line of brick
gains 2 marks[2]
M5. (a) moment/torque increases as moves awaygains 2 marks
leverage/force increases as moves awaygains 1 mark
2
(b) (i) 20gains 2 marks
else workinggains 1 mark
2
(ii) 100 000 ecfgains 2 marks
else workinggains 1 mark
2[6]
M6. (a) Z1
weight or mass acts through pivotaccept rod or base for pivotaccept centre of gravity in line with pivot
1
no (resultant) (turning) momentaccept clockwise moment equals anticlockwise momentdo not accept same weight on each side of rod
1
(b) (i) 30allow 1 mark for 2 15or 2 0.15
2
N cm
orfor full credit the unit must be consistent with the numerical answer
0.3
Nmdo not accept joules
1
(ii) 1.5 (N)allow 1 mark for correct transformationallow 2 marks ecf their part (b)(i)/20 (ecf only if correct physics)
2
(c) 5 (cm)allow 1 mark for 6.0 (cm)allow 1 mark for a subtraction of 1 from a value clearly obtained from the graphallow 2 marks for correct ecf using an incorrect value for (b)(i) 0.2cmallow 1 mark for clearly showing correct use of graph using an incorrect value for (b)(ii)
2[10]
M7. (i) C1
(ii) 48an answer of 4 800 gains 1 markif answer (b)(i) is given as A then 42 scores 1 mark4200 scores 0 marks substitution of correct figures = 1 mark
2[3]
M8. 300allow 1 mark for rearranging equation or correct substitution
[2]
M9. (a) Amust be correct for reason to score
moment (due to weight) of sail is the largest1
or
(perpendicular) distance from pivot to rope the smallestdo not accept sail is low or sail is too heavy
1
(b) (i) no resultant turning moment or in a state of balance or balancedallow clockwise moments = anticlockwise momentsallow no resultant forceallow (forces are) balancedallow no accelerationdo not allow forces are equal
1
(ii) moment = 420allow 1 mark for moment = 700 × 0.6or700 × a distance from diagram (1.5, 2.1, 0.9)
2
(iii) force = 280420 = F × 1.5or
F = 1 mark only
if (b)(ii) obtained by a correct method (1470, 630, 1050)
2
(c) (as wind speed increases) the force on the sail increasesaccept pressure
1
aniticlockwise moment increases or moment on sail increases1
so clockwise moment (or opposite moment) needs to increase (by increasingthe distance from the pivot)
1[10]
M10. (a) (i) X at the centre of the lifebeltmeasuring from the centre of X, allow 2 mm tolerance in any direction
1
(ii) any two from:if X is on vertical line below the hanger (but not at centre) can gain the first point only
below the point of suspensionaccept ‘(vertically) below Y’
at the centre (of the lifebelt)accept ‘in the middle’
(because) the lifebelt / it is symmetricalor (because) the mass / weight is evenly distributed
2
(b) Nm or newton metre(s)accept Newton metre(s)do not accept any ambiguity in the symbol ie NM, nM or nm
1
750(moment) = force (perpendicular) distance (between line of action and pivot)or (moment) = 500 1.5 gains 1 mark
2
(c) Quality of written communication:for 2 of the underlined terms used in the correct context
1
any three connected points from:
low(er) centre of mass / gravityor centre of mass / gravity will be close(r) to the wheels / axle / ground
(more) stableor less unstable
less likely to fall overaccept ‘less likely to overturn’do not accept ‘will not fall over’
the turning effect / moment (of the weight of case) is lessor so less effort is needed to hold the caseignore references to pulling the case
so the pull on her arm is less3
[10]
M11. (a) point at which its mass (seems to) act or point at which gravity (seems to) actaccept ... its weight actsaccept correct statements if the intent is clear e.g.. .. if suspended, the centre of gravity will be directly under the point of suspensione.g.... (if the object is symmetrical), the centre of gravity is on theor an axis (of symmetry)do not credit just 'it is a point'
1
(b) The answer to this question requires good English in a sensible order with correct use of scientific terms. Quality of written communication should be considered in crediting points in the mark scheme
maximum of 4 marks if ideas not well expressed
any five from:
clamp (steel) rod (horizontally)no marks if method quite unworkable
hang plastic / sheet by rod through (one) hole
hang plumb line from rod
mark ends of plumb line on the sheet anduse the ruler to draw a straight line
repeat with other hole
centre of mass is where the lines cross
check by balancing at this pointmaximum of 3 marks if no 'repeat with other hole'
5
(c) (i) (turning) effect or momentforcedistance
all three correctaccept weightaccept length
1
(ii) 17.6allow 44 x 0.4 or 0.4 x 44 for 1 mark
2
Nm or newton metre(s)do not accept N/m or N/cm1760 Ncm gains all 3 marks
1[10]
M12. (a) 810 000allow 45 000 × 18 for 1 mark
2
newton-metres / Nm1
(b) any three from:ignore references to force throughout
• their weight / mass can be altered / adjusted
• so that the crane remains stableallow does not topple
• so that the (total) clockwise moment equals the (total)anticlockwise moment
do not allow just ‘moments are equal’
• because not all containers are the same weight / massdo not allow ‘not all containers are the same size / volume’
• because not all containers will be / need to move the samedistance (from the crane)
• to keep the centre of mass (of the upper crane and container) in/above the base of the tower
• so that the crane remains in equilibrium/balanced3
[6]
M13. (a) (i) P1
(ii) the child’s grip / hold / pull (on the roundabout / bar / rail)or ‘the tension in the child’s arms’accept ‘the child’s muscles’accept ‘friction between the child and the roundabout’do not accept just ‘friction’
1
(iii) increasesaccept any unambiguous indication that this ending has been selected
1
(b) (i) 360 (Nm)credit either 240 × 1 ½or 240 × 1.5 with 1 mark
2
(ii) move to(wards) the left / to(wards) the / his endor move away from the centre / pivot / axis (of rotation)or move away from the girl / the child / his daughter
1
(c) (i) C1
(ii) friction / grip of the car / tyres / wheels (on the road)do not accept just ‘friction / grip’
1[8]
M14. (a) (line of action of) its weight1
falls inside its wheel baseaccept ‘falls between the wheels’the first two points may be credited by adding a vertical line from the centre of the X on the diagram (1)and labelling it weight / force / with a downwards arrow (1)provided there is no contradiction between what is added to the diagram and anything which may be written
1
(so there is) no (resultant / clockwise) moment / turning effect1
(b) centre of mass should be loweraccept ‘… centre of gravity’accept ‘weight / mass low down’not just ‘lower the roof’
1
wheel base should be wideraccept ‘long axle(s)’ for ‘wide wheel base’allow bigger / larger wheel basedo not credit ‘long wheel base’responses in either order
1[5]
M15. (a) momentor torque do not credit ‘leverage’
1
(b) 4 (2)either 0.20 × 20 (1) or allow ‘400’ (1)
2
(c) use a longer spanneror increases the perpendicular distance / length
or ‘fit a pipe over the (end of the) spanner (to lengthen it)’note ‘lever’ refers to ‘spanner’note change the . . . (0)ignore references to wider / larger nut
1
use a greater force / pulleither order
1[5]
M16. (a) centre of X at the centre of the concentric circlesjudge by eye that the intention is correct
1
(b) drawn from any corner to the diagonally opposite cornerjudge by eye that the intention is correct
or from the mid-point of any side to the mid-point of the opposite sideif more than one axis of symmetry has been drawn,accept only if both / all are correct
1
(c) a turningaccept any unambiguous indication
1[3]
M17. (a) any two from:
• inversely proportional
• as the load gets biggerthe (maximum safe) distance gets lessallow ‘as the mass increases the distance decreases’accept an unspecified response e.g. ‘big load at a short distance’ for (1)
• load × distance = 60 (kNm)2
(b) yes, because 30 × 2 = 60 (2)accept for (1) a correct but insufficiently explained responsee.g. ‘yes because it’s safe’accept for (2) a correct response which is sufficiently explainede.g. ‘yes, because 60 (kNm) at 1 metre is safe and 30 (kNm) is half the load at twice the distancedo not accept ‘no’ and do not accept just ‘yes’do not accept ‘yes, because 30 is between 24 and 40 and 2 is between 2.5 and 1.5’do not accept ‘the crane/ cable may break’ or other dangers
2
(c) the crane may/will topple over/fall over/forward
1
(d) results of experiments on this mobile craneaccept any unambiguous indication
1[6]
M18. (a) 560allow 1 mark forclockwise (moments) = anticlockwise (moments)allow 1 mark for correct substitutionie 160 × 1.75 = W × 0.5allow 1 mark for correct transformation
ie 4
newtons, N1
(c) the weight of plank which has been ignored1
causes an anticlockwise moment which has not beenconsidered / included in the calculation
1[7]
M19. (a) where the mass of the object can be thought to be concentrated1
(b) lower the C of M1
and make the wheelbase wider1
accept a practical description of how these changes could be achieved
(c) the line of action of its weightaccept a vertical arrow drawn from X
2
falls inside its wheel base
accept falls between the wheels
therefore there is no resultant / clockwise moment1
[6]
M20. (a) (i) 360allow 1 mark for correct length used ie 1.2 mallow 2 marks for substitution into correct equation - ie 300 × 1.2allow 1 mark only for an answer 240
3
(ii) Newton-metre or Nm1
(b) the force is applied further from the pivot1
which causes an increased moment to act on the steel bar1
and therefore an increased force acts on the tree stump1
[7]
M21. (a) 38 400allow 6.4 × 6000 for 1 mark
2
Nm or newton metresdo not credit ‘nm’, ‘mN’ or ‘metre newtons’
1
(b) 16 000 (N) or 16 kNallow 1 mark for 38 400 ÷ 2.4accept their (a) ÷ 2.4 correctly calculated for 2 marksaccept their (a) ÷ 2.4 for 1 mark
2[5]
M22. (a) the point at which the (total) mass seems to act / appears to be concentratedaccept ‘weight’ for ‘mass’accept the point at which gravity seems to actdo not accept a definitive statement eg where (all) the mass is
1
(b) wider / larger basemarks are for a correct comparison
1
lower centre of massaccept lower centre of gravity / c of g
1
(c) line of action (of the weight) lies / falls inside the basein each case the underlined term must be used correctly to gain the mark
1
the resultant moment returns mixer to its original positionaccept there is no resultant moment / resultant moment is zeroaccept resulting moment for resultant momentdo not accept converse argument
1[5]
M23. (a) (i) moment1
(ii) rotation1
(iii) the girl moves nearer to point P1
(b) (i) X drawn in the centre of the space enclosed by the tyrejudge by eye
1
(ii) below1
[5]
M24. (a) (i) will not fall over (1)
accept will not easily fall over (2)
orcentre of mass will remain above the base (1)
(line of action of the) weight will remain above within the baseaccept centre of gravity / c of g / c of m / c m
if the monitor is given a small push (1)depends on mark above
2
(ii) (total) clockwise moment = (total) anticlockwise momentor they are equal / balanced
1
(b) the position of the centre of mass has changed (1)the line of action of the weight is outside the base (1)producing a (resultant) moment (1)
points may be expressed in any order3
[6]
M25. (a) (i) turning effectaccept turning forceaccept force x distance(accept symbols only if correctly defined)do not accept newtons x metres
1
(ii) stop apparatus falling overaccept holds the stand in placeaccept make it safer / stablereferences to balanced / equilibrium are insufficient
1
(iii) as x increases y increases1
in same proportion / ratiosallow both marks for they are directly proportionalora specific example eg doubling y, doubles xallow both marks for a correct answer giving figureseg they increase in the ratio of 1 to 7allow for 1 mark positive correlation
1
(iv) the centre of mass of the ruler is at the axis of rotation1
(b) 108allow 1 mark for correct substitution ie 240 x 0.45
2
newton metres / Nmsymbols must be correctfor full credit the unit must be consistent with the numerical answer
1[8]
M26. (a) 1.2allow 1 mark for conversion of 2.4 kN to 2400 Nor for correct transformation without conversionie d = 2880 ÷ 2.4
2
metre(s)/m1
(b) any two from:
• as the load increases the (total) clockwise moment increases
• danger is that the fork lift truck / the load will topple / tip forward
• (this will happen) when the total clockwise moment isequal to (or greater than) the anticlockwise moment
accept moments will not be balanced
• (load above 10.0 kN) moves line of action (from C of M)outside base (area)
2[5]
M27. (a) C1
(b) momentaccept any unambiguous correct indication
1
(c) bigger than
accept any unambiguous correct indication1
(d) 120 (Ncm)allow 1 mark for correct substitutionie 12 × 10
2[5]
M28. (a) (i) current produces a magnetic field (around XY)accept current (in XY) is perpendicular to the (permanent) magnetic field
1
(creating) a force (acting) on XY / wire / upwardsreference to Fleming's left hand rule is insufficient
1
(ii) motor (effect)1
(iii) vibrate / move up and down1
5 times a secondonly scores if first mark point scoresallow for 1 mark only an answer ‘changes direction 5 times a second’
1
(b) 0.005allow 1 mark for calculating moment of the weight as 0.04 (Ncm)andallow 1 mark for correctly stating principle of momentsorallow 2 marks for correct substitutionie F × 8 = 2 × 0.02 or F × 8 = 0.04
3[8]
M29. (a) 1250allow 1 mark for correct substitutionie 500 × 2.5 provided there is no subsequent calculation
2
(b) (i) smaller than1
(ii) force (exerted) further from axis of rotation (than the weight)accept pivot for axis of rotation
1
(c) increase the force (exerted)do not accept increase distance of force from axis of rotation
1[5]
M30. (a) 960 (Nm)1
see-saw is in equilibriumaccept see-saw is balancedsee-saw is stationary is insufficient
1
(total) clockwise moments = anticlockwise momentaccept no resultant momentforces are balanced is insufficientan answer clockwise moments balance the anticlockwise moments gains 2 marks
1
(b) (i) 600 (Nm)1
(ii) 375 (N) or their (b)(i) ÷ 1.6 correctly calculateddo not credit if (b)(i) is larger than 960allow 1 mark for correct substitution and transformation ie
2[6]
M31. (a) (i) 75allow 1 mark for correct substitution ie 250 × 0.3do not credit if subsequent step shownallow 1 mark for an answer 7500
2
(ii) Nm1
(b) force is (applied) further from the nut / pivot / axis of rotationhandle is longer is insufficientdo not accept less force needed
1
moment (on wrench) is larger1
[5]
M32.(a) 38 400allow 6.4 × 6000 for 1 mark
2
Nm or newton metresdo not credit ‘nm’, ‘mN’ or ‘metre newtons’
1
(b) 16 000 (N) or 16 kNallow 1 mark for 38 400 ÷ 2.4accept their (a) ÷ 2.4 correctly calculated for 2 marksaccept their (a) ÷ 2.4 for 1 mark
2[5]
M33.(a) (i) centre of X above the feet and in the bodya vertical line from their X falls between two lines in diagram – judged by eye
1
(ii) where the mass seems to be concentratedaccept it’s above the base (area)
accept because otherwise it would toppleaccept line of action (of weight) passes through the basedo not accept where the mass is concentrated
1
(b) any two from:
• make (the area of) feet / base bigger
• make feet wider apart
• make legs shorter / heavier
• make head smaller / lighter
• make tail touch the ground / make the tail longeraccept ‘make centre of mass / gravity lower’
2[4]
M34.(a) 360allow 1 mark for correct substitution ie 300 × 1.2 provided no subsequent step shown
2
(b) the force is applied further from the axis of rotationaccept pivot / (tree) stump for ‘axis of rotation’
1
or
this increases the moment of the force
increases the force on the (tree) stump1
[4]
M35.(a) 360allow 1 mark for correct substitution ie 300 × 1.2 provided no subsequent step shown
2
(b) the force is applied further from the axis of rotationaccept pivot / (tree) stump for ‘axis of rotation’
1
or
this increases the moment of the force
increases the force on the (tree) stump1
[4]
E1. Paper I3
This question was generally poorly answered.
(a) Nearly all candidates compared Bus A and B rather than explaining that the vertical line through the centre of mass must be within the wheel base for the bus to remain upright.
(b) Many candidates thought the greater weight of the extra passengers upstairs would make the bus more stable. Good answers noted that the centre of mass would be higher, making it less stable, so the bus could not be tilted at such a big angle.
(c) Few candidates realised that it was to make the centre of mass as high as possible or make the bus as unstable as possible.
Paper H5
(a) Many candidates failed to indicate, either in words or - equally acceptable - on a diagram, that it is the line of action of the weight, vertically through the centre of mass, which must not fall outside of the wheel-base.
(b) This item was generally answered as well as candidates’ answers to (a) permitted.
(c) This was badly answered by the great majority of candidates, very few candidates referring to the need to test the most extreme situation likely to be met when the bus is in operation. Providing candidates with information at the start of the question about the point of the test would probably have resulted in better answers to this item.
E2. There were many accurate answers to this recognition question.
E3. In part (a) good answers multiplied 1.5m by 200N to get 300Nm.
Where candidates showed evidence of multiplying force by distance credit was given. Weaker candidates multiplied or divided 200N by various combinations of 1.5/3.0/4.5.
In part (b) weaker candidates guessed as to whether more or less force was needed. Stronger candidates compared the distances 1.5:3.0 and thus compared forces in the same way i.e. 100N:200N.
E4. Paper 3 Option Q
This question elicited a lot of angles with little mention of the centre of gravity being beyond the base of the brick.
Paper 5 Option R
This question was generally well answered.
E5. Part (a) was badly answered. In part (b) many candidates scored at least part marks.
E6. Foundation Tier
In part (a) most candidates could identify the balanced toy but found difficulty explaining the reason for their choice, even though the concept of moments was in the stem of part (b)(i). Although many candidates calculated the moment correctly in part (b)(i), the unit was often incorrect, N/cm was a favourite error. Many candidates were unable to correctly complete the subsequent calculation of the weight. In part (c) the original length of the spring was often not subtracted to produce a correct value for the extension. However, the better candidates were able to correctly complete all the calculations demonstrating a sound understanding of the topic.
Higher Tier
In part (a) most candidates could identify the balanced toy but found difficulty explaining the reason for their choice, even though the concept of moments was in the stem of part (b)(i). Although many candidates calculated the moment correctly in part (b)(i), the unit was often incorrect, N/cm was a favourite error. Most candidates were able to complete the subsequent calculation of the weight. In part (c) the original length of the spring was often not subtracted to produce a correct value for the extension.
E7. Foundation Tier
In part (i) few candidates identified the position C as being the situation to produce the largest moment on the pedal. Of those giving either B or C, the majority gave the numerical answer of 4 800 in part (ii), though the correct scientific unit was given in both the stem of the question and on the answer line.
Higher Tier
In part (i) only the more able candidates correctly used the idea of perpendicular distance to identify position C as being the situation to produce the largest moment on the pedal. Of those giving either B or C, the majority gave the numerical answer of 4 800 in part (ii), though the correct scientific unit was given in both the stem of the question and on the answer line.
E8. The majority of candidates found the idea of moments difficult to understand and subsequently very few completely correct answers to this question were seen. Most candidates were able to rearrange the equation but very few converted distance to metres and even fewer could identify the correct distance to use.
E9. Foundation Tier
Answers to all parts of this question were poor, with many candidates scoring few marks. Many identified ‘A’ as the correct response in part (a), but were unable to explain why. The word ‘equilibrium’ was often interpreted as ‘equal forces’. The equations given in parts (b)(ii) and (b)(iii) allowed some candidates to score marks, however, there was a general lack of understanding as to which distance to use. In part (c) answers often mentioned balanced forces rather than balanced moments. Few candidates mentioned that the wind force had increased. Clearly many candidates understood the mechanics of windsurfing, but were unable to articulate the explanation necessary to score any marks.
Higher Tier
Many candidates identified ‘A’ to be the correct response in part (a), but were unable to explain why. The word ‘equilibrium’ was not well understood and it was often interpreted as meaning ‘equal forces’. The equations given in parts (b)(ii) and (b)(iii) allowed most candidates to score marks, however, there was a general lack of understanding as to which distance to use. In part (c) answers often mentioned balanced forces rather than balanced moments. Few candidates mentioned that the wind force had increased. Clearly many candidates understood the mechanics of windsurfing, but were unable to articulate the explanation necessary to score any marks.
E10. Only a small minority were able to give the correct position of the centre of mass. Usually positions in the body of the lifebelt either above or below the hook were suggested and very few earned any marks in part (ii). Incorrect responses to part (i) made a rational explanation more difficult in part (ii) but responses which could still gain credit, such as ‘below the point of suspension’ and ‘the lifebelt is symmetrical’, were only given by a minority of candidates.
Susan’s moment was usually numerically correct but the unit was often omitted or incorrect.
In part (c) it was rare for candidates to gain more than one mark. Where credit was gained it was usually for ‘low centre of mass’, ‘less likely to fall over’ or ‘easier to hold’. Very few made correct statements in relation to the moments involved and the communications mark, which was given for the correct use of terms, was rarely awarded.
E11. Foundation Tier
(a) In this part the examiners were looking for a correct statement which would explain, or help to explain, the term centre of mass. The candidate did not have to account for cases in which the centre of mass lies outside of the object but, even so, appropriate responses, for example ‘the object will balance if it is supported at its centre of mass’, were rare. ‘The centre of the object’ and ‘the mass will be the same on both sides’ were fairly popular, but incorrect, responses.
(b) The specification states that ‘candidates should be able to describe how to find the centre of mass of a thin sheet of material’. Some invalidated their responses either by clamping the sheet or by suspending it through one hole and hanging the plumb line from the other hole. Others ignored any sensible consideration of the size of the sheet and bent it so that the 100 mm long rod could go through both holes. However, there was a minority of candidates who seemed to understand what was required.
(c) Only a minority of candidates were able to recollect the equation to calculate the turning effect of a force. A large proportion injected the words mass and/or height into their erroneous responses. Some were able to find the product of 44 and 0.4 but some spoilt their efforts by increasing or decreasing by a factor of 10 or by doubling the distance to 0.8 m. If there was an attempt at a unit it was usually to suggest N/m.
Higher Tier
(a) In this part the examiners were looking for a correct statement which would explain, or help to explain, the term centre of mass. The candidate did not have to account for cases in which the centre of mass lies outside of the object but, even so, appropriate responses, for example ‘the object will balance if it is supported at its centre of mass’, were not too frequent. ‘The centre of the object’ and ‘the mass will be the same on both sides’ were fairly popular, but incorrect, responses.
(b) The Specification states that ‘candidates should be able to describe how to find the centre of mass of a thin sheet of material’. Many were able to do this, and to express themselves clearly, so gained all five marks. A small minority invalidated their responses by clamping the sheet or by suspending it through one hole and hanging the plumb line from the other hole. Some others ignored any sensible consideration of the size of the sheet and bent it so that the 100 mm long rod could go through both holes.
(c) Most candidates were able to recollect the equation to calculate the turning effect of a force. Some injected the words mass and/or height into their erroneous responses. Most candidates were able to find the product of 44 and 0.4 but some spoilt their efforts by increasing or decreasing by a factor of 10 or by doubling the distance to 0.8 m. A significant minority gave the unit as N/m. It could be that some correctly knew that their answer was in newton-metres but incorrectly thought that this was the way to write its
symbol.
E12. (a) The calculation was generally correct, although some candidates multiplied the 45 000 N by 10. However, a significant proportion of candidates either failed to give the unit, or gave an incorrect unit or could not write newton metres correctly either in words or as the symbol (Nm).
(b) Almost all candidates showed some understanding of the principle involved and scored at least one mark. A large proportion failed to use scientific terms, such as clockwise/anticlockwise moments, correctly, or wrote about centre of mass but failed to show understanding of where it is or should be.
E13. (a) (i) Slightly fewer than half the candidates could identify P as the direction of the centripetal force.
(ii) Very few candidates realised that the centripetal force was provided by the girl’s grip.
(iii) Three quarters of candidates identified that if the speed of the roundabout increases that the centripetal force increases.
(b) (i) This question was well answered with more than 80 % of candidates gaining the 2 marks available.
(ii) Two thirds of candidates were able to identify the direction in which the father moves correctly.
(c) (i) Over 80 % of candidates gave the correct answer and therefore gained a mark.
(ii) Very few candidates could identify the friction between the tyres and the road as the source of centripetal force.
E14. (a) Many candidates were able to explain, either on the diagram or in writing, that the line of action from the centre of mass falls vertically and is within the wheel base.
However, very few indicated that the weight of the tractor results in a moment which keeps it turned to the slope or alternatively that if the line of action were to fall to the right of the right wheel in the diagram the resulting moment would topple the tractor.
(b) The great majority of candidates gained two marks because they knew that having a wider wheel base and a lower centre of mass would increase the tractor’s stability.
E15. Foundation tier
(a) Only about a quarter of candidates knew that a turning effect is a ‘moment’. There were many responses which appeared to be guesses involving other terms used in Physics.
(b) Many candidates obtained one mark for ‘400’ but few took account of the units and so the correct response ‘4 (Nm)’ was infrequently seen.
(c) Most candidates knew that increasing the force and/or using a longer spanner could increase the turning effect.
Higher tier
(a) Nearly all candidates knew that a turning effect is a moment.
(b) Many candidates gained one mark because they offered ‘400’. However, less than half took account of the units and gave the correct response of ‘4 (Nm)’.
(c) Almost all candidates knew that increasing the force and using a longer spanner could increase the turning effect and so obtained full marks. However, only a small minority expressed themselves in practical terms, for example, ‘fit a pipe over the handle (of the spanner)’.
E16. (a) Nearly all candidates were able to mark the centre of mass of the square sign.
(b) Nearly all candidates were able to draw one axis of symmetry.
(c) Only a small minority of candidates associated a moment with a turning effect.
E17. Foundation Tier
(a)(b) Many candidates realised that the data shows that as the load increases the maximum safe distance decreases but only a minority of these could explain why the crane driver’s conclusion is correct.
(c) Several hazards were mentioned with a minority of candidates stating that the mobile
crane may topple over, or words to that effect.
(d) Most candidates were able to select ‘results of experiments on this crane’ as the appropriate source for the data in the table.
Higher Tier
(a)(b) A large majority realised that the data shows that as the load increases the maximum safe distance decreases but fewer could explain why the crane driver’s conclusion is correct.
(c) Several hazards were mentioned with the majority of candidates stating that the mobile crane may topple over, or words to that effect.
(d) A large majority of candidates correctly selected ‘results of experiments on this crane’ as the appropriate source for the data in the table.
E21. (a) Most candidates were able to complete the calculation and give the correct unit. However, a significant minority (about a third) were unable to give the correct unit or simply missed it out.
(b) Just under three quarters of the candidates were able to complete this calculation correctly.A few candidates scored one mark for showing a correct method but incorrect numerical answer.
E22. (a) Many candidates failed to gain marks here as their responses were definite ie it is where all the mass is.
(b) Just over half of the candidates were able to make two clear comparative statements with a further quarter able to make one correct clear statement.
(c) Many responses were not worth a mark because candidates had not used the terms as instructed in the stem of the question. It was encouraging to note that the fuller answers often used the phrase ‘resultant moment restores equilibrium’ or words to that effect.
E23. (a) Over three fifths of candidates answered parts (i) and (iii) correctly but only just over a quarter correctly named the axis of rotation, part (ii), most opting for axis of balance.
(b) (i) Only just under a half of candidates were able to mark the centre of mass of the tyre correctly, most wrongly putting it somewhere on the tyre itself.
(ii) A large majority of candidates gave the correct answer.
E24. (a) (i) Just over two-thirds of the candidates could relate being stable to not falling over, or that the centre of mass remains above the base. Very few candidates gained the second mark, the effect of a small push.
(ii) Most candidates correctly stated that the moments are equal.
(b) The few candidates who gained all 3 marks generally started their responses with the statement that the centre of mass of the monitor is not in the same place now the screen has been tilted. Many candidates realised that ‘something’ went outside the base and /or caused a (resultant) moment.
E25. (a) (i) Candidates continue to be confused between moments and momentum. Under a quarter of candidates gave an acceptable meaning of the word ‘moment’.Very few candidates did not attempt to give an answer.
(ii) Just over three quarters of candidates correctly gave the purpose of the G-clamp.
(iii) More than half of candidates described fully the relationship between X and Y as detailed in the table. Over a third of candidates gave a more general relationship which scored one mark.
(iv) Just over three quarters of candidates knew the reason for ignoring the weight of the ruler.
(b) Just over a quarter of candidates obtained full marks for the calculation and were able to give the correct unit. Nearly two thirds of candidates obtained two out of the three marks. It was evident that quite a lot of candidates were unable to do the calculation because they did not have a calculator. Some candidates tried to work it out long hand with varying success.
E26. (a) The calculation was generally well done with nearly two-thirds of candidates gaining full marks, including that given for the unit.
(b) Many candidates appreciated the possibility of the fork lift truck ‘toppling’, however fewer were able to give a correctly worded explanation as to why this should happen. Of those who attempted this, some candidates only referred to a single moment which was not identified as the resultant; others referred to more than one moment but did not identify the direction of action.
A significant number of candidates did not appreciate that part (b) was linked to part (a) and did not read the instruction to explain in terms of moments. This resulted in many responses linked to other vehicles, most often a taxi. Answers in terms of momentum and the problems in cornering and the weight of the passengers were then common.
E27. (a) Over four-fifths of candidates identified the correct point.
(b) Over three-quarters of candidates identified the turning effect as a moment.
(c) Half of the candidates understood the model arm and realised that the rubber band would exert a greater force than the weight. However a significant number of candidates thought that the forces were equal because the moments were equal.
(d) Nearly all candidates were able to put the correct numbers into the equation and produce the correct numerical answer.
E28. (a) (i) Students rarely included the idea of the current producing a magnetic field around the wire XY. Some students were clearly confused with the induction process.
(ii) Just over half of students scored this mark.
(iii) Just over half of students scored at least one mark. Incorrect responses included: back and forth; spinning; faster; repelled instead of ‘up and down’. The frequency of 5 hertz was sometimes interpreted as once every five seconds.
(b) There was a clear distinction in this question between the students who could do the calculation perfectly and those who struggled. Some answers were left as fractions by students without calculators; there were a number of 1 mark responses for calculating the left hand moment. A number of students did not gain marks because of inconsistent units
and moving between cm and m.
E29. (a) Most students completed the calculation correctly to score both marks.
(b) (i) Less than a quarter of students correctly answered this question and it became obvious from part (b)(ii) that often their reasoning was incorrect.
(ii) Many students did not consider moments in this question but merely considered the lifting force and the weight of the pole. Other students thought that the person was only lifting part of the weight. A few students left this question blank.
(c) Most students gave an answer in terms of changing the length of the pole or varying the distance of the lifting force from the pivot rather than simply increasing the force.
E30. (a) Most candidates correctly identified 960 Nm but fewer followed this with correct reasoning. Some candidates failed to score a second or third mark because the moments were not directional i.e. ‘clockwise moments’.
(b) (i) This question was well answered with 80% of candidates scoring this mark.
(ii) This part question proved to be a good end to the paper for the majority of candidates. Just over 71% of candidates gained both marks, few candidates scored just one mark.
E31. (a) (i) The calculation was well done with 92% of candidates achieving full marks.
(ii) Just over 85% of candidates were able to identify the correct units for a moment from the list.
(b) Not many candidates gained both marks for this question, mainly because they did not give a full explanation. A significant number of candidates lost marks by using careless terminology, with a considerable number of references to ‘centrifugal force’, ‘pressure’, ‘leverage’, and a few of ‘momentum’ and/or ‘weight’.
E32.Generally well answered.
E33.(a) (i) This was well answered with the majority of the students gaining the mark.
(ii) This was very poorly done: Three quarters of the students scored zero and of those that did score a mark, credit was usually given for the “toppling” mark rather than a correct definition regarding centre of mass.
(b) The majority of students scored at least one mark.
E34.(a) The majority of students achieved full marks for the moments calculation.
(b) A very low number of students achieved the full 2 marks for this part of the question although just over half managed to gain one mark.
E35.(a) The majority of students achieved full marks for the moments calculation.
(b) A very low number of students achieved the full 2 marks for this part of the question although just over half managed to gain one mark.
