Forces — Paper 3 · IGCSE Physics 0625

1 A bus travels from one bus stop to the next. The journey has three distinct parts. Stated in For order they are Examiner’s uniform acceleration from rest for 8.0 s, Use uniform speed for 12 s, non-uniform deceleration for 5.0 s. Fig. 1.1 shows only the deceleration of the bus. 15 speed m/s 10 5 0 0 5 10 15 20 25 time/s Fig. 1.1 (a) On Fig. 1.1, complete the graph to show the first two parts of the journey. [3] (b) Calculate the acceleration of the bus 4.0 s after leaving the first bus stop. acceleration = … [2] (c) Use the graph to estimate the distance the bus travels between 20 s and 25 s. estimated distance = … [2] (d) On leaving the second bus stop, the uniform acceleration of the bus is 1.2 m / s2. The mass of the bus and passengers is 4000 kg. Calculate the accelerating force that acts on the bus. force = … [2] (e) The acceleration of the bus from the second bus stop is less than that from the first bus stop. Suggest two reasons for this. 1. … … 2. … … [2]

11 marks

Mark scheme: 1 (a) point 8,12 identified B1 straight line joining 0,0 and 8,12 B1 straight line joining 8,12 and 20,12 B1 3 (b) acceleration = change in v/change in t or 12/8 etc C1 = 1.5 m/s2 A1 2 (c) distance = area under graph between t = 20 and t = 25 C1 = 24 m to 28 m A1 2 (d) F = ma or 4000 x 1.2 C1 = 4800 N A1 2 (e) more passengers got on (so mass increased) driver pressed accelerator less (so force decreased) more traffic or going uphill any two lines B2 2 [11]

This question in 0625/31 May/June 2006

2 A student sets up the apparatus shown in Fig. 2.1 in order to find the resultant of the two For Examiner’s tensions T1 and T2 acting at P. When the tensions T1, T2 and T3 are balanced, the angles between T1 and the vertical and T2 and the vertical are as marked on Fig. 2.1. Use pulley vertical pulley T1 = 6.0 N T2 = 8.0 N board 44° 69° P T3 Fig. 2.1 In the space below, draw a scale diagram of the forces T1 and T2. Use the diagram to find the resultant of the two forces. State (a) the scale used, scale = … (b) the value of the resultant, value = … (c) the direction of the resultant. direction = … [6]

6 marks

Mark scheme: 2 any closed triangle or parallelogram C1 forces in correct directions relative to each other C1 correct resultant indicated C1 resultant 7.7 N to 8.1 N A1 4 scale stated B1 resultant vertically upwards B1 2 [6]

This question in 0625/31 May/June 2006

2 In an experiment, forces are applied to a spring as shown in Fig. 2.1a. The results of this For experiment are shown in Fig. 2.1b. Examiner’s Use 16 RR QQ 12 ruler force/N spring 8.0 PP 4.0 weights 0 0 2.0 4.0 6.0 extension/mm Fig. 2.1a Fig. 2.1b (a) What is the name given to the point marked Q on Fig. 2.1b? … [1] (b) For the part OP of the graph, the spring obeys Hooke’s Law. State what this means. … … [1] (c) The spring is stretched until the force and extension are shown by the point R on the graph. Compare how the spring stretches, as shown by the part of the graph OQ, with that shown by QR. … … [1] (d) The part OP of the graph shows the spring stretching according to the expression F = kx. Use values from the graph to calculate the value of k. k = … [2]

5 marks

Mark scheme: 2 (a) limit of proportionality (allow elastic limit) B1 [1] (b) force is proportional to extension or in terms of doubling B1 [1] (c) (up to Q extension proportional to force applied) Q to R extension/unit force more however expressed B1 [1] (d) k = force/extension or 8/2 or other correct ratio C1 = 4.0 N/mm A1 [2] [Total: 5]

This question in 0625/31 Oct/Nov 2006

2 Fig. 2.1 shows a steam safety valve. When the pressure gets too high, the steam lifts the weight W and allows steam to escape. 0.2 m pivot W force of steam Fig. 2.1 (a) Explain, in terms of moments of forces, how the valve works. … … … … [2] (b) The moment of weight W about the pivot is 12 N m. The perpendicular distance of the line of action of the force of the steam on the valve from the pivot is 0.2 m. The area of the piston is 0.0003 m2. Calculate (i) the minimum steam force needed for the steam to escape, force = … [2] (ii) the minimum steam pressure for the steam to escape. pressure = … [2] [Total: 6]

6 marks

Mark scheme: 2 (a) moment of W down/anticlockwise, moment of steam opposite C1 when moment of steam > moment of W, steam escapes OR when clockwise moment > anticlockwise moment, steam escapes A1 [2] (b) (i) 12 = 0.2 F C1 F = 60 N c.a.o. allow 60–61 for ans if working for 60 N shown A1 [2] (ii) (P =) F/A or 60/0.0003 e.c.f. C1 2 × 105 Pa or 200 000 Pa e.c.f. (accept N/m2) OR 20 N/cm2 A1 [2] [Total: 6] IGCSE – May/June 2007 0625 03

This question in 0625/31 May/June 2007

3 (a) A spring of original length 3.0 cm is extended to a total length of 5.0 cm by a force of 8.0 N. Assuming the limit of proportionality of the spring has not been reached, calculate the force needed to extend it to a total length of 6.0 cm. force = … [3] (b) Fig. 3.1 shows the arrangement for an experiment on moments. spring pivot F metre rule Fig. 3.1 The spring exerts a force F on the metre rule. (i) On Fig. 3.1, mark another quantity which must be measured to find the moment of the force F. [1] (ii) State how the moment of the force F is calculated. … … [1] [Total: 5]

5 marks

Mark scheme: 3 (a) any logical method e.g. extension is 2 cm for 8 N or 1 cm for 4 N C1 final extension is 3 cm C1 need 12 N to extend to 6 cm A1 (b) (i) shown on diagram: distance from pivot to F OR value of weights OR dist from weights to pivot B1 (ii) force/weight of load × distance from pivot to force (accept symbols if clear) B1 [Total: 5]

This question in 0625/31 Oct/Nov 2007

4 (a) In an accident, a truck goes off the road and into a ditch. Two breakdown vehicles A and For B are used to pull the truck out of the ditch, as shown in Fig. 4.1. Examiner’s Use road 45° A breakdown B vehicles ditch Fig. 4.1 At one point in the rescue operation, breakdown vehicle A is exerting a force of 4000 N For and breakdown vehicle B is exerting a force of 2000 N. Examiner’s Use (i) Using a scale of 1 cm = 500 N, make a scale drawing to show the resultant force on the truck. [4] (ii) Use your diagram to find the magnitude and direction of the resultant force on the truck. magnitude of resultant force = … direction of resultant force = … to direction of road [2] (b) (i) State why the resultant force is an example of a vector quantity. … [1] (ii) Give an example of a vector quantity that is not a force. … [1] [Total: 8]

8 marks

Mark scheme: 4 (a) (i) (note: diagram may be drawn in any orientation) sides correct length, by eye B1 forces drawn at 45°, by eye B1 parallelogram completed B1 correct diagonal drawn / correct resultant if intersecting arcs shown B1 (ii) magnitude: between 5500 N and 5700 B1 direction: between 28° and 32° B1 (b) (i) it has direction (as well as magnitude) B1 (ii) any example which is clearly a vector B1 [8] IGCSE – May/June 2009 0625 31 2

This question in 0625/31 May/June 2009

6 (a) A man squeezes a pin between his thumb and finger, as shown in Fig. 6.1. For Examiner’s Use finger pinhead pin thumb Fig. 6.1 The finger exerts a force of 84 N on the pinhead. The pinhead has an area of 6.0 × 10–5 m2. (i) Calculate the pressure exerted by the finger on the pinhead. pressure = … [2] (ii) State the value of the force exerted by the pin on the thumb. … [1] (iii) Explain why the pin causes more pain in the man’s thumb than in his finger. … … [2] (b) The density of the water in a swimming pool is 1000 kg / m3. The pool is 3 m deep. For Examiner’s (i) Calculate the pressure of the water at the bottom of the pool. Use pressure = … [2] (ii) Another pool has the same depth of water, but has twice the area. State the pressure of the water at the bottom of this pool. pressure = … [1] [Total: 8]

8 marks

Mark scheme: 6 (a) (i) P = F /A in any form, letters, words or numbers C1 1.4 × 106 Pa accept N/m2 A1 (ii) 84 N OR 84.0 N B1 (iii) same force over (much) smaller area B1 (much) bigger pressure B1 (b) (i) P = hdg in any form, letters, words or numbers C1 3 × 104 Pa OR 30 000 Pa OR 30 kPa accept N/m2 A1 (ii) his (i) B1 [8]

This question in 0625/31 May/June 2009

3 A student investigated the stretching of a spring by hanging various weights from it and For measuring the corresponding extensions. The results are shown below. Examiner’s Use weight / N 0 1 2 3 4 5 extension / mm 0 21 40 51 82 103 (a) On Fig. 3.1, plot the points from these results. Do not draw a line through the points yet. [2] 120 100 extension / mm 80 60 40 20 0 0 1 2 3 4 5 6 weight / N Fig. 3.1 (b) The student appears to have made an error in recording one of the results. For Examiner’s Which result is this? Use … [1] (c) Ignoring the incorrect result, draw the best straight line through the remaining points. [1] (d) State and explain whether this spring is obeying Hooke’s Law. … … … … … [2] (e) Describe how the graph might be shaped if the student continued to add several more weights to the spring. … … … … [1] (f) The student estimates that if he hangs a 45 N load on the spring, the extension will be 920 mm. Explain why this estimate may be unrealistic. … … … … [1] [Total: 8]

8 marks

Mark scheme: 3 (a) 5 points correctly plotted ±½ small square –1 e.e.o.o. (ignore 0,0) B2 (b) 3 N one, however identified OR 3rd value OR 4th value B1 (c) good straight line through origin and candidate’s remaining points B1 (d) straight line / constant gradient M1 does obey Hooke’s Law A1 OR special case: obeys Hooke’s law because force ú extension or wtte B1 IGCSE – October/November 2009 0625 31 (e) graph becomes non-linear / curves / bends B1 Ignore reference to direction of curve or bend. (f) will have exceeded / reached proportional / elastic limit OR permanently deformed or equiv OR staightened OR will have broken OR no longer elastic or wtte B1 [8]

This question in 0625/31 Oct/Nov 2009

4 (a) A force acting on an object causes the object to accelerate. For Examiner’s In which direction is the acceleration? Use … [1] (b) Any object moving in a circle has a force acting on it towards the centre of the circle. What does this force do to the object? … [1] (c) A woman of mass 60 kg is standing in a lift at a shopping centre. (i) The lift is at rest. 1. State the value of the weight of the woman. … [1] 2. State the value of the force exerted on the woman by the floor of the lift. … [1] (ii) Calculate the force required to accelerate a mass of 60 kg at 2.5 m / s2. force = … [2] (iii) The lift accelerates upwards at 2.5 m / s2. Calculate the force exerted on the woman by the floor when the lift is accelerating. force = … [1] (iv) The lift reaches a steady upward speed. State the value of the force exerted on the woman by the floor at this steady speed. … [1] [Total: 8]

8 marks

Mark scheme: 4 (a) in direction of the force Do not accept forward on is own. B1 (b) changes direction / causes acceleration / stops straight line motion / keeps object from leaving circle / keeps path circular / pulls object into circle B1 (c) (i) 1. 600 N B1 2. same as his 1. accept 600 N if no value given in (c) (i) 1. B1 (ii) ma OR 60 × 2.5 C1 150 N A1 (iii) 750 N e.c.f. from (c) (i) 2 and/or (c) (ii) B1 (iv) same as his (c) (i) 2 accept 600 N if no value given in (c) (i) 2. B1 [8]

This question in 0625/31 Oct/Nov 2009

3 A student investigated the stretching of a spring by hanging various weights from it and For measuring the corresponding extensions. The results are shown below. Examiner’s Use weight / N 0 1 2 3 4 5 extension / mm 0 21 40 51 82 103 (a) On Fig. 3.1, plot the points from these results. Do not draw a line through the points yet. [2] 120 100 extension / mm 80 60 40 20 0 0 1 2 3 4 5 6 weight / N Fig. 3.1 (b) The student appears to have made an error in recording one of the results. For Examiner’s Which result is this? Use … [1] (c) Ignoring the incorrect result, draw the best straight line through the remaining points. [1] (d) State and explain whether this spring is obeying Hooke’s Law. … … … … … [2] (e) Describe how the graph might be shaped if the student continued to add several more weights to the spring. … … … … [1] (f) The student estimates that if he hangs a 45 N load on the spring, the extension will be 920 mm. Explain why this estimate may be unrealistic. … … … … [1] [Total: 8]

8 marks

Mark scheme: 3 (a) 5 points correctly plotted ±½ small square -1 e.e.o.o. (ignore 0,0) B2 (b) 3 N one, however identified OR 3rd value OR 4th value B1 (c) good straight line through origin and candidate’s remaining points B1 (d) straight line / constant gradient M1 does obey Hooke’s Law A1 OR special case: obeys Hooke’s law because force ú extension or wtte B1 (e) graph becomes non-linear / curves / bends B1 Ignore reference to direction of curve or bend. IGCSE – October/November 2009 0625 32 (f) will have exceeded / reached proportional / elastic limit OR permanently deformed or equiv OR straightened OR will have broken OR no longer elastic or wtte B1 [8]

This question in 0625/32 Oct/Nov 2009

4 Fig. 4.1 illustrates an object on a string being whirled anticlockwise in a vertical circle. For Examiner’s A string Use B ground Fig. 4.1 The lowest point of the circle is a small distance above the ground. The diagram shows the object at the top A of the circle, and at B, when it is at the same height as the centre of the circle. (a) On Fig. 4.1, mark clearly (i) the force of the string on the object 1. at A, 2. at B. [2] (ii) the path the object would take until it hit the ground, if the string broke 1. at A, 2. at B. [3] (b) The mass of the object is 0.05 kg. At A, the tension in the string is 3.6 N. (i) Calculate the weight of the object. weight = … [1] (ii) Calculate the total force on the object at A. total force = … [2] [Total: 8]

8 marks

Mark scheme: 4 (a) (i) 1. force marked towards centre B1 2. force marked towards centre B1 (ii) 1. clearly horizontal at start to left or right M1 horizontal to the left curving down to reach ground to left of A B1 2. vertically down, not necessarily to reach ground B1 (b) Allow use of g = 9.81 or 9.8 throughout (i) 0.5 N B1 (ii) 4.1 N or 3.1 N e.c.f. from (i) C1 4.1 N e.c.f. from (i) A1 [8]

This question in 0625/32 Oct/Nov 2009

3 Two students make the statements about acceleration that are given below. For Examiner’s Student A: For a given mass the acceleration of an object is proportional to the resultant Use force applied to the object. Student B: For a given force the acceleration of an object is proportional to the mass of the object. (a) One statement is correct and one is incorrect. Re-write the incorrect statement, making changes so that it is now correct. For a given … the acceleration of an object is … … [1] (b) State the equation which links acceleration a, resultant force F and mass m. [1] (c) Describe what happens to the motion of a moving object when (i) there is no resultant force acting on it, … [1] (ii) a resultant force is applied to it in the opposite direction to the motion, … [1] (iii) a resultant force is applied to it in a perpendicular direction to the motion. … [1] [Total: 5]

5 marks

Mark scheme: 3 (a) 2nd statement re-written to include force in first gap and inversely proportional to mass in second gap. NOT indirectly proportional B1 (b) F = ma OR in words in any correct arrangement B1 (c) (i) nothing OR continues as before OR same / constant velocity OR same / constant speed & direction OR no acceleration B1 (ii) idea of retardation. Ignore stop. Ignore brakes. Ignore goes in opposite direction B1 (iii) moves in (arc of a) circle or curve OR deflected OR turns OR changes direction B1 [5] IGCSE – May/June 2010 0625 31

This question in 0625/31 May/June 2010

2 Four students, A, B, C and D, each have a spring. They measure the lengths of their springs For when the springs are stretched by different loads. Examiner’s Use Their results are shown in Fig. 2.1. student A student B student C student D load / N spring length / cm spring length / cm spring length / cm spring length / cm 0.5 6.7 9.2 9.1 10.0 1.0 7.7 10.0 9.9 11.1 1.5 8.7 10.8 10.7 12.2 2.0 9.7 11.6 11.5 13.3 2.5 10.7 12.6 12.3 14.4 3.0 11.7 13.8 13.1 15.5 3.5 12.7 15.2 13.9 16.6 4.0 13.7 16.8 14.7 17.7 Fig. 2.1 (a) (i) State which student had loaded the spring beyond the limit of proportionality. … [1] (ii) Explain how you obtained your answer to (a)(i). … … … [2] (b) For the spring used by student A, calculate (i) the extra extension caused by each additional 0.5 N, extra extension = … [1] (ii) the unloaded length of the spring. unloaded length = … [1] (c) Student A obtains a second spring that is identical to his first spring. He hangs the two For springs side by side, as shown in Fig. 2.2. Examiner’s Use identical springs load Fig. 2.2 Use the table to calculate the length of each of the springs when a load of 2.5 N is hung as shown in Fig. 2.2. Show your working. length = … [2] [Total: 7]

7 marks

Mark scheme: 2 (a) Mark (i) and (ii) together. Note both M1s required to score the A1 mark (i) B M1 (ii) idea of greater / different (NOT less) increase in length for each additional load accept load not proportional to extension or reverse argument M1 at 4th or 5th reading / value between 2.0 – 2.5 N / 11.6 – 12.6 cm A1 (b) (i) 1.0 cm B1 (ii) 5.7 cm B1 (c) 2.5 (cm) OR 1.25 (N) OR 5.0(cm) ignore 2.5N e.c.f. from (b) if clear C1 8.2 cm e.c.f. from (b) if clear A1 e.g. 10.7/2 (= 5.35) scores 0/2 [7] IGCSE – May/June 2010 0625 32 3 3

This question in 0625/32 May/June 2010

5 The apparatus shown in Fig. 5.1 is used to demonstrate how a coin and a piece of paper fall For when they are released from rest. Examiner’s Use piece of paper coin tube containing air A Fig. 5.1 (a) At the positions shown in Fig. 5.1, the paper is descending at constant speed but the coin still accelerates. In terms of the forces acting, explain these observations. paper … … … coin … … … [4] (b) A vacuum pump is now connected at A and the air in the tube is pumped out. The paper and coin are again made to fall from rest. State one difference that would be observed, compared with what was observed when air was present. … … [1] [Total: 5]

5 marks

Mark scheme: 5 Ignore upthrust throughout this question (a) paper: drag / air resistance / friction (upwards) (seen anywhere in (a)) B1 drag /air resistance / friction = weight / force of gravity B1 no resultant (force) / forces balance / upwards force = downwards force AND no acceleration B1 coin: weight / force of gravity (always) bigger than air resistance OR force down bigger than force up OR air resistance hasn’t time / distance to equal weight B1 (b) fall at same speed / acceleration / rate, ignore fall at same time ) hit bottom at same time/together ) paper now accelerates (all the way) ) any 1 B1 paper no longer flutters side-side ) they/paper NOT coin fall(s) faster ) the paper (ignore coin) hits sooner ) NOT constant speed/rate [5]

This question in 0625/32 May/June 2010

2 Four students, A, B, C and D, each have a spring. They measure the lengths of their springs For when the springs are stretched by different loads. Examiner’s Use Their results are shown in Fig. 2.1. student A student B student C student D load / N spring length / cm spring length / cm spring length / cm spring length / cm 0.5 6.7 9.2 9.1 10.0 1.0 7.7 10.0 9.9 11.1 1.5 8.7 10.8 10.7 12.2 2.0 9.7 11.6 11.5 13.3 2.5 10.7 12.6 12.3 14.4 3.0 11.7 13.8 13.1 15.5 3.5 12.7 15.2 13.9 16.6 4.0 13.7 16.8 14.7 17.7 Fig. 2.1 (a) (i) State which student had loaded the spring beyond the limit of proportionality. … [1] (ii) Explain how you obtained your answer to (a)(i). … … … [2] (b) For the spring used by student A, calculate (i) the extra extension caused by each additional 0.5 N, extra extension = … [1] (ii) the unloaded length of the spring. unloaded length = … [1] (c) Student A obtains a second spring that is identical to his first spring. He hangs the two For springs side by side, as shown in Fig. 2.2. Examiner’s Use identical springs load Fig. 2.2 Use the table to calculate the length of each of the springs when a load of 2.5 N is hung as shown in Fig. 2.2. Show your working. length = … [2] [Total: 7]

7 marks

Mark scheme: 2 (a) Mark (i) and (ii) together. Note both M1s required to score the A1 mark (i) B M1 (ii) idea of greater / different (NOT less) increase in length for each additional load accept load not proportional to extension or reverse argument M1 at 4th or 5th reading / value between 2.0 – 2.5 N / 11.6 – 12.6 cm A1 (b) (i) 1.0 cm B1 (ii) 5.7 cm B1 (c) 2.5 (cm) OR 1.25 (N) OR 5.0(cm) ignore 2.5N e.c.f. from (b) if clear C1 8.2 cm e.c.f. from (b) if clear A1 e.g. 10.7/2 (= 5.35) scores 0/2 [7] IGCSE – May/June 2010 0625 33 3 3

This question in 0625/33 May/June 2010

5 The apparatus shown in Fig. 5.1 is used to demonstrate how a coin and a piece of paper fall For when they are released from rest. Examiner’s Use piece of paper coin tube containing air A Fig. 5.1 (a) At the positions shown in Fig. 5.1, the paper is descending at constant speed but the coin still accelerates. In terms of the forces acting, explain these observations. paper … … … coin … … … [4] (b) A vacuum pump is now connected at A and the air in the tube is pumped out. The paper and coin are again made to fall from rest. State one difference that would be observed, compared with what was observed when air was present. … … [1] [Total: 5]

5 marks

Mark scheme: 5 Ignore upthrust throughout this question (a) paper: drag / air resistance / friction (upwards) (seen anywhere in (a)) B1 drag /air resistance / friction = weight / force of gravity B1 no resultant (force) / forces balance / upwards force = downwards force AND no acceleration B1 coin: weight / force of gravity (always) bigger than air resistance OR force down bigger than force up OR air resistance hasn’t time / distance to equal weight B1 (b) fall at same speed / acceleration / rate, ignore fall at same time ) hit bottom at same time/together ) paper now accelerates (all the way) ) any 1 B1 paper no longer flutters side-side ) they/paper NOT coin fall(s) faster ) the paper (ignore coin) hits sooner ) NOT constant speed/rate [5]

This question in 0625/33 May/June 2010

1 An object of weight W is suspended by two ropes from a beam, as shown in Fig. 1.1. 86.6 N 30° 50.0 N 60° W Fig. 1.1 The tensions in the ropes are 50.0 N and 86.6 N, as shown. (a) In the space below, draw a scale diagram to find the resultant of the two tensions. Use a scale of 1.0 cm = 10 N. Clearly label the resultant. [3] (b) From your diagram, find the value of the resultant. resultant = … [1] (c) State the direction in which the resultant is acting. … [1] (d) State the value of W. W = … [1] [Total: 6]

6 marks

Mark scheme: 1 (a) (parallelogram or triangle may have any orientation) NOT a copy of Fig. 1.1 two sides at right angles, by eye B1 one side longer than the other B1 diagonal or completion of triangle drawn and labelled “resultant” OR R Ignore numerical values. Condone arrows in wrong direction B1 (b) 98 N – 102 N B1 (accept value found by calculation) (c) (vertically) up/opposite to W NOT North B1 (d) his (b) OR correct value calculated B1 ignore mass [Total: 6]

This question in 0625/31 Oct/Nov 2010

2 A car travels around a circular track at constant speed. (a) Why is it incorrect to describe the circular motion as having constant velocity? … [1] (b) A force is required to maintain the circular motion. (i) Explain why a force is required. … … … [2] (ii) In which direction does this force act? … [1] (iii) Suggest what provides this force. … [1] [Total: 5]

5 marks

Mark scheme: 2 (a) constant velocity must be in a straight line/direction of motion is changing B1 (b) (i) if no force, then constant velocity in straight line OR force is needed to change direction B1 body moving in circle is changing direction/velocity/accelerating so force is needed B1 (ii) towards centre (of circle)/at right angles to motion/inwards B1 (iii) friction between tyres and road/reaction from banking of track B1 [Total: 5]

This question in 0625/31 Oct/Nov 2010

3 Fig. 3.1 shows a hydraulic lift in a car repair workshop. car support hydraulic fluid 4 pistons, each of area 0.02 m2 piston A, area 0.01 m2 Fig. 3.1 The hydraulic fluid transmits the pressure, caused by piston A, equally to each of the four pistons holding up the car supports. The pressure throughout the fluid is the same. A force of 1000 N on piston A is just enough to raise the car. (a) Using values from Fig. 3.1, find (i) the pressure caused by piston A on the fluid, pressure = … [2] (ii) the total upward force caused by the fluid. force = … [3] (b) The weight of each of the two car supports is 1000 N. Calculate the mass of the car. mass = … [2] [Total: 7]

7 marks

Mark scheme: 3 (a) (i) (P =) F/A in any form OR 1000/0.01 C1 100 000 Pa accept N/m2 A1 (ii) multiplication of either force or area by 4 C1 0.08 × his (i) OR 0.02 × his (i) C1 8000 N e.c.f. from (i) A1 (2000 N gets C0, C1, A1) (b) his (ii) – 2000 correctly evaluated C1 600 kg e.c.f. A1 [Total: 7] IGCSE – October/November 2010 0625 31

This question in 0625/31 Oct/Nov 2010

1 A hillside is covered with snow. A skier is travelling down the hill. Fig. 1.1 The table below gives the values of the acceleration of the skier at various heights above the bottom of the hill. height / m 350 250 150 50 acceleration 7.4 3.6 1.2 0 m / s2 (a) On Fig. 1.2, plot the values given in the table, using dots in circles. Draw the best curve for these points. [2] 8 7 acceleration m / s2 6 5 4 3 2 1 0 0 50 100 150 200 250 300 350 height / m Fig. 1.2 (b) Describe what is happening, during the descent, to (i) the acceleration of the skier, … … … [1] (ii) the speed of the skier. … … … [1] (c) The acceleration becomes zero before the skier reaches the bottom of the hill. Use ideas about forces to suggest why this happens. … … [1] (d) Below a height of 50 m, further measurements show that the acceleration of the skier has a negative value. What does this mean is happening to the speed of the skier in the last 50 m? … … [1] (e) The skier has a mass of 60 kg. Calculate the resultant force on the skier at a height of 250 m. resultant force = … [3] [Total: 9]

9 marks

Mark scheme: 1 (a) all points plotted correctly ±½ small square B1 smooth curve through points, by eye B1 (b) (i) decreasing OR idea of greater at greater heights NOT decelerating B1 (ii) increasing OR idea of slower at greater heights NOT accelerating B1 (c) idea of resultant force becomes zero B1 (d) decreasing/slowing down, ignore deceleration NOT accelerating B1 (e) F = ma in any form, letters, words, numbers C1 (a =) 3.6 (m/s2) c.a.o. C1 (F =) 216 N / 220 N A1 [Total: 9]

This question in 0625/32 Oct/Nov 2010

3 (a) A uniform metre rule is pivoted at its centre, which is also the position of its centre of mass. Three loads, 2.0 N, F and 3.0 N are positioned on the rule at the 20 cm, 30 cm and 90 cm marks respectively, as shown in Fig. 3.1. 0 cm 20 cm 30 cm 50 cm 90 cm 100 cm pivot 2.0 N F 3.0 N Fig. 3.1 (i) Calculate the moment of the 3.0 N load about the pivot. moment = … [1] (ii) Calculate the moment of the 2.0 N load about the pivot. moment = … [1] (iii) The force F maintains the metre rule in equilibrium on the pivot. Calculate the value of F. F = … [3] (b) The weight of the metre rule is 1.2 N and can be considered to act at the 50 cm mark. All the weights in (a) are removed. The pivot is positioned under the 30 cm mark and the 2.0 N load is placed on the rule as shown in Fig. 3.2. 30 cm 50 cm pivot 2.0 N 1.2 N Fig. 3.2 The position of the 2.0 N load is adjusted until the metre rule is again in equilibrium. Determine the position of the 2.0 N load. 2.0 N load is at the … cm mark [3] [Total: 8]

8 marks

Mark scheme: 3 (a) (i) 120 Ncm OR 1.2 Nm B1 (ii) 60 Ncm OR 0.6 Nm B1 (iii) idea of CW moments = ACW moments C1 60 + 20F = 120 OR 0.6 + 0.2F = 1.2 e.c.f. C1 3.0 N OR 3 N e.c.f. A1 (b) 1.2 × 20 = 2.0 × d OR 1.2 × 0.2 = 2.0 × d C1 (d =) 12 OR 0.12 C1 18 c.a.o. OR special case (30 – his 12) correctly evaluated B1 A1 [Total: 8]

This question in 0625/32 Oct/Nov 2010

5 The front views of two cars are shown in Fig. 5.1, to the same scale. family car racing car Fig. 5.1 (a) Suggest which car has the greater stability, and give two reasons. car … reason 1 … … reason 2 … … [2] (b) The cars have the same weight. Study Fig. 5.1 and suggest why the stationary racing car exerts less pressure on the ground. … … [1] (c) The family car’s tyres each have an area of 0.012 m2 in contact with the ground. The weight of the car and its contents is 9600 N. Calculate the pressure exerted by the car on the ground. pressure = … [2] [Total: 5]

5 marks

Mark scheme: 5 (a) racing car + 1 correct reason M1 2nd correct reason A1 correct reasons: • wider (car) • lower (centre of mass/gravity) NOT wider tyre/surfaces o.w.t.t.e. (b) larger/wider tyres/area (of contact) ignore base area B1 (c) F/A OR 9600/0.012 OR 9600/0.048 OR 9600/(4 × 0.012) OR 800,000 C1 2 x 105 Pa OR 200 000 Pa (accept N/m2) c.a.o. A1 [Total: 5] IGCSE – October/November 2010 0625 32

This question in 0625/32 Oct/Nov 2010

1 A young athlete has a mass of 42 kg. On a day when there is no wind, she runs a 100 m race in 14.2 s. A sketch graph (not to scale) showing her speed during the race is given in Fig. 1.1. speed m / s 8.0 0 0 3.0 14.2 time / s Fig. 1.1 (a) Calculate (i) the acceleration of the athlete during the first 3.0 s of the race, acceleration = … [2] (ii) the accelerating force on the athlete during the first 3.0 s of the race, force = … [2] (iii) the speed with which she crosses the finishing line. speed = … [3] (b) Suggest two differences that might be seen in the graph if there had been a strong wind opposing the runners in the race. 1. … … 2. … … [2] [Total: 9]

9 marks

Mark scheme: 1 (a) (i) (v – u)/t OR v/t OR 8/3 C1 2.7 m/s2 A1 (ii) ma OR 42 × answer from (i) OR 42 × 8/3 C1 110/112 N e.c.f. A1 (iii) (distance in 1st 3 secs =) 12 m OR (dist in last 3 secs =) 88 m C1 use of area of trapezium OR area of “top” triangle C1 7.7 m/s A1 (b) longer time to top speed ) longer total time ) lower top speed ) lower finishing speed ) any 2 B1+B1 specific/all speeds lower (not speed decreases) ) less slope/less acceleration (in first section) ) greater slope/greater deceleration in 2nd section ) [Total: 9]

This question in 0625/33 Oct/Nov 2010

2 (a) A loose uniform wooden floorboard weighs 160 N and rests symmetrically on four supports P, Q, R and S. The supports are 0.50 m apart, as shown in Fig. 2.1. floorboard P Q R S support 0.20 m 0.50 m 0.25 m 160 N Fig. 2.1 Calculate the force exerted on the floorboard by each of the supports, and state the direction of these forces. One value is already given for you. force exerted by P = … force exerted by Q = … 40 N force exerted by R = … force exerted by S = … direction = … [2] (b) A workman of weight W stands on the end of the floorboard described in (a). This just causes the floorboard to tip up, as shown in Fig. 2.2. The supports are each 0.060 m thick. P Q R S 0.060 m 160 N W Fig. 2.2 (i) Calculate the weight W of the workman. weight W = … [3] (ii) Calculate the force that each of the supports now exerts on the floorboard. force exerted by P = … force exerted by Q = … force exerted by R = … force exerted by S = … [2] [Total: 7]

7 marks

Mark scheme: 2 (a) all four = 40 N OR all four add up to 160 N B1 upwards B1 (b) (i) W × 0.17/0.20/0.23 = 160 × 0.72/0.75/0.78 C1 W × 0.17 = 160 × 0.78 or 600 N C1 730/734 N A1 (ii) force by P = 160 + answer to (i) correctly evaluated B1 all others = 0 B1 [Total: 7]

This question in 0625/33 Oct/Nov 2010

1 In a laboratory, an experiment is carried out to measure the acceleration of a trolley on a horizontal table, when pulled by a horizontal force. trolley force Fig. 1.1 The measurements are repeated for a series of different forces, with the results shown in the table below. force / N 4.0 6.0 10.0 14.0 acceleration 0.50 0.85 1.55 2.25 m / s2 (a) On Fig. 1.2, plot these points and draw the best straight line for your points. [2] 16 force / N 12 8 4 0 0 0.5 1.0 1.5 2.0 2.5 acceleration m / s2 Fig. 1.2 (b) The graph shows that below a certain force there is no acceleration. (i) Find the value of this force. … [1] (ii) A force smaller than that in (b)(i) is applied to the stationary trolley. Suggest what happens to the trolley, if anything. … [1] (c) Show that the gradient of your graph is about 5.7. gradient = … [1] (d) (i) State the equation that links resultant force F, mass m and acceleration a. [1] (ii) Use your gradient from (c) to find the mass of the trolley. mass = … [2] (e) On Fig. 1.3, sketch a speed / time graph for a trolley with constant acceleration. speed 0 0 time Fig. 1.3 [1] [Total: 9]

9 marks

Mark scheme: 1 (a) all points correctly plotted ±½ small square B1 straight line of best fit for candidate’s points B1 (b) (i) candidate’s correct value with unit (± 0.2), (expect 1.2 N) B1 (ii) remains stationary / nothing happens / no acceleration NOT constant speed B1 (c) Correct data from candidates graph for ∆F and ∆m, used in ∆F/∆m B1 (d) (i) F = ma in any form, letters, words B1 (ii) gradient = F/a OR gradient = m ignore m=F/a C1 candidate’s (c) with correct unit A1 (e) straight line of positive gradient B1 [9]

This question in 0625/31 May/June 2011

3 (a) Complete the following statement: The moment of a force about a point is … multiplied by … [1] (b) Fig. 3.1 shows a uniform iron bar B of weight 30 N and length 1.40 m. The bar is being used to lift one edge of a concrete slab S. A stone, placed 0.20 m from one end of B, acts as a pivot. A force of 40 N pushing down at the other end of B is just enough to lift the slab and hold it as shown. 1.40 m 0.20 m force 40 N concrete slab iron bar B S stone Fig. 3.1 (i) On Fig. 3.1, draw an arrow to show the weight of bar B acting from its centre of mass. [1] (ii) State the distance d of the centre of mass of bar B from the pivot. d = … [1] (iii) Calculate the total clockwise moment, about the pivot, of the forces acting on bar B. total clockwise moment = … [3] (iv) Calculate the downward force which the slab S exerts on the end of bar B. force = … [2] (v) Suggest a change to the arrangement in Fig. 3.1 that would reduce the force required to lift the slab. … … [1] [Total: 9]

9 marks

Mark scheme: 3 (a) force AND perpendicular distance (of force) from the point. B1 (b) (i) downward arrow at centre of bar B1 (ii) 0.5(0) m / 50 cm B1 (iii) 40 × 1.2 OR 48 seen anywhere C1 (+) 30 × 0.5 0R 15 seen anywhere C1 = 63 N m A1 (iv) F × 0.2 = 63 C1 F = 63/0.2 = 315 N A1 (v) make bar / B longer OR move pivot / stone to the left OR increase distance between force and pivot (by moving pivot to left) OR increase mass of the bar / B B1 [9] IGCSE – May/June 2011 0625 32

This question in 0625/32 May/June 2011

3 (a) Complete the following statement: The moment of a force about a point is … multiplied by … [1] (b) Fig. 3.1 shows a uniform iron bar B of weight 30 N and length 1.40 m. The bar is being used to lift one edge of a concrete slab S. A stone, placed 0.20 m from one end of B, acts as a pivot. A force of 40 N pushing down at the other end of B is just enough to lift the slab and hold it as shown. 1.40 m 0.20 m force 40 N concrete slab iron bar B S stone Fig. 3.1 (i) On Fig. 3.1, draw an arrow to show the weight of bar B acting from its centre of mass. [1] (ii) State the distance d of the centre of mass of bar B from the pivot. d = … [1] (iii) Calculate the total clockwise moment, about the pivot, of the forces acting on bar B. total clockwise moment = … [3] (iv) Calculate the downward force which the slab S exerts on the end of bar B. force = … [2] (v) Suggest a change to the arrangement in Fig. 3.1 that would reduce the force required to lift the slab. … … [1] [Total: 9]

9 marks

Mark scheme: 3 (a) force AND perpendicular distance (of force) from the point. B1 (b) (i) downward arrow at centre of bar B1 (ii) 0.5(0) m / 50 cm B1 (iii) 40 × 1.2 OR 48 seen anywhere C1 (+) 30 × 0.5 0R 15 seen anywhere C1 = 63 N m A1 (iv) F × 0.2 = 63 C1 F = 63/0.2 = 315 N A1 (v) make bar / B longer OR move pivot / stone to the left OR increase distance between force and pivot (by moving pivot to left) OR increase mass of the bar / B B1 [9] IGCSE – May/June 2011 0625 33

This question in 0625/33 May/June 2011

3 (a) State the two conditions required for the equilibrium of a body acted upon by a number of forces. 1. … … 2. … … [2] (b) Fig. 3.1 shows a diagram of an arm with the hand holding a weight of 120 N. F 2 cm P 15 cm 20 N 120 N 33 cm Fig. 3.1 The 20 N force is the weight of the forearm, acting at its centre of mass. F is the force in the muscle of the upper arm. P is the point in the elbow about which the arm pivots. The distances of the forces from point P are shown. (i) By taking moments about point P, calculate the force F. force F = … [3] (ii) A force acts on the forearm at point P. Calculate this force and state its direction. force = … direction = … [2] [Total: 7]

7 marks

Mark scheme: 3 (a) 1. no resultant force acts / no net force acts OR total force up / in any direction = total force down / in opposite direction B1 allow sum of forces or resultant force for total force 2. no resultant moment / couple / torque acts OR (sum of) clockwise moments and (sum of) anti-clockwise moments (about any point / axis) balance B1 (b) (i) (anti-clockwise moment =) F × 2 C1 (total clockwise moment =) (120 × 33) + (20 × 15) = 4260 (N cm) C1 2130 N A1 (ii) 1990 N OR candidate’s (b)(i) – 140 N B1 force is downwards B1 [7]

This question in 0625/31 Oct/Nov 2011

3 Fig. 3.1 shows a house brick of dimensions 21.0 cm × 10.0 cm × 7.00 cm. 21.0 cm 10.0 cm 7.00 cm Fig. 3.1 The brick is held under water with its largest surfaces horizontal. The density of water is 1000 kg / m3. (a) Calculate the difference in pressure between the top and the bottom surfaces of the brick. pressure difference = … [2] (b) Use your value from (a) to calculate the upward force exerted on the brick by the water. upward force = … [2] (c) The mass of the brick is 3.09 kg. Calculate the acceleration of the brick when it is released. acceleration = … [3] [Total: 7]

7 marks

Mark scheme: 3 (a) ρgh in symbols, words or numbers C1 700 Pa or N/m2 A1 [2] (b) use of F = pA C1 14.7 N ecf from (a) A1 [2] (c) (30.9 – 14.7 = )16.2 N OR evidence of calculation of resultant C1 use of a = F/m C1 5.24 m/s2 A1 [3] IGCSE – October/November 2011 0625 32

This question in 0625/32 Oct/Nov 2011

2 A person is standing on the top of a cliff, throwing stones into the sea below. cliff sea Fig. 2.1 (a) The person throws a stone horizontally. (i) On Fig. 2.1, draw a line to show the path which the stone might take between leaving the person’s hand and hitting the sea. (ii) On the line you have drawn, at a point halfway to the sea, mark the stone and the direction of the force on the stone. [3] (b) Later, the person drops a small stone and a large stone vertically from the edge of the cliff. Comment on the times taken for the two stones to hit the water. … … … … [2] (c) 800 m from the point where the person is standing, a navy ship is having target practice. The person finds that if a stone is dropped vertically at the same time as the spurt of smoke from the ship’s gun is seen, the stone hits the water at the same time as the sound from the gun is heard. Sound travels at 320 m / s in that region. Calculate the velocity with which the stone hits the water. velocity = … [4] [Total: 9]

9 marks

Mark scheme: 2 (a) (i) downward curve B1 initially horizontal at top and not vertical at bottom B1 (ii) force shown vertically down (accept leaning back a small amount) B1 (b) any two from: same (times) / air resistance negligible / same acceleration B2 OR times different B1 one has (more) air resistance B1 (c) (time =) 800/320 C1 2.5 (s) C1 (v =) at OR 10 × candidate’s t value C1 25 m/s A1 [9]

This question in 0625/33 Oct/Nov 2011

2 (a) Underline the vectors in the following list of quantities. For Examiner’s density energy force mass velocity volume [2] Use (b) A small metal ball is projected into the air with a velocity of 40 m / s vertically upwards. The graph in Fig. 2.1 shows how the velocity changes with time until the ball reaches its maximum height. 40 velocity m / s 20 0 0 1.0 2.0 3.0 4.0 5.0 6.0 time / s –20 –40 Fig. 2.1 Use the graph to find, (i) the time at which the ball reaches its maximum height, time = … [1] (ii) the deceleration of the ball, deceleration = … [2] (iii) the maximum height reached by the ball. For Examiner’s Use maximum height = … [2] (c) On Fig. 2.1, add a line to the graph to show how the velocity of the ball changes after it reaches its maximum height. Your line should extend to time 6.0 s. [1] [Total: 8]

8 marks

Mark scheme: 2 (a) underline or circle force B1 underline or circle velocity B1 (b) (i) 4.07 – 4.1 (s) B1 (ii) (v – u)/t OR ∆v/t OR in words OR use of 40 ÷ (ans. to (b)(i)) OR other correct values from graph C1 answer between 9.7 and 10 m/s2 or m/s/s A1 (iii) area under graph OR ½ (u + v)t OR ½ × 40 × (ans. to (b)(i)) C1 OR s = ut + ½at2 OR v2 = u2 + 2as OR numbers substituted 82 m A1 (c) graph continues in straight line to 6 s B1 [Total 8] IGCSE – May/June 2013 0625 31

This question in 0625/31 May/June 2013

2 Fig. 2.1 shows the extension-load graph for a spring. For Examiner’s Use P extension 0 0 load Fig. 2.1 Point P is the limit of proportionality. (a) (i) Name the law obeyed by the spring from the origin to P. … [1] (ii) Describe two features of the graph which show that the law is obeyed. 1. … 2. … [2] (b) On Fig. 2.1, sketch a possible continuation of the graph when the spring is loaded beyond the limit of proportionality. [1] [Total: 4]

4 marks

Mark scheme: 2 (a) (i) Hooke’s Law B1 [1] (ii) straight line (graph) / constant gradient B1 through origin/(0,0) B1 [2] ignore through zero ignore extension proportional to load (b) curved extension to graph with increasing gradient, condone decreasing NOT if any part of curve is vertical/horizontal or has negative gradient B1 [1] [Total: 4] IGCSE – May/June 2013 0625 32

This question in 0625/32 May/June 2013

1 (a) State Hooke’s law. For Examiner’s … Use … [1] (b) Fig. 1.1 shows a graph of the stretching force F acting on a spring against the extension x of the spring. 250 200 F / N 150 100 50 0 0 10 20 30 40 50 60 70 80 x / mm Fig. 1.1 (i) State the features of the graph that show that the spring obeys Hooke’s law. … … [1] (ii) Calculate k, the force per unit extension of the spring. k = … [3] (iii) The limit of proportionality of the spring is reached at an extension of 50 mm. For Examiner’s Continue the graph in Fig. 1.1 to suggest how the spring behaves when the Use stretching force is increased to values above 125 N. [1] (iv) Another spring has a smaller value of k. This spring obeys Hooke’s law for extensions up to 80 mm. On the grid of Fig. 1.1, draw a possible line of the variation of F with x for this spring. [1] [Total: 7]

7 marks

Mark scheme: 1 (a) extension (of spring) proportional to load / force (applied) OR load / force (applied) proportional to extension OR force = constant × extension OR extension = constant × force OR F = kx in any form with symbols explained B1 (b) (i) graph is through the origin AND is a straight line / has a constant gradient B1 (ii) F = kx in any form OR (k =) F/x C1 use of a point anywhere on graph e.g. 50 / 20 C1 2.5 N / mm OR 2500 N / m A1 (iii) from 50 mm extension, graph curves with no negative gradient B1 (iv) straight line through origin with smaller gradient than graph shown finishing at more than 50 mm B1 [Total: 7]

This question in 0625/31 Oct/Nov 2013

2 A train has a total mass of 7.5 × 105 kg. For Examiner’s (a) The train accelerates from rest at a constant rate along a straight, horizontal track. Use It reaches a speed of 24 m / s in 60 s. Calculate (i) the train’s acceleration, acceleration = … [2] (ii) the resultant force acting on the train. force = … [2] (b) The train now travels with a constant speed of 24 m / s along a straight, horizontal track. The total force opposing the motion due to friction and air resistance is 7.2 × 104 N. (i) By considering the work done by the train’s engine in 1.0 s, calculate its output power. power = … [2] (ii) The train begins to travel up a slope. For Examiner’s Explain why the power of the train’s engine must be increased to maintain the Use speed of 24 m / s. … … … … … [3] [Total: 9]

9 marks

Mark scheme: 2 (a) (i) v = u + at OR (a =) (v – u) / t OR 24 = a × 60 OR 24 / 60 C1 0.4(0) m / s2 A1 (ii) (F =) ma OR 7.5 × 105 × 0.40 C1 300 000 N OR 300 kN A1 (b) (i) in words or symbols (P =) W / t OR F x d / t OR Fv OR 7.2 × 104 × 24 / 1 OR OR 7.2 × 104 × 24 C1 1.7 × 106 W A1 (ii) gravitational/potential energy of train has to be increased OR force acts down the slope / backward force acts (on train) B1 (for the same distance moved) more work done has to be done OR energy has to be provided (by the engine) B1 in the same time (so needs more power) B1 [Total: 9]

This question in 0625/31 Oct/Nov 2013

3 (a) (i) Write down the names of three man-made devices in everyday use that depend, For for their action, upon the moments of forces. Examiner’s Use 1. … 2. … 3. … [2] (ii) Fig. 3.1 shows a uniform rod AB acted upon by three equal forces F. F F A B F Fig. 3.1 State two reasons why the rod is not in equilibrium. 1. … 2. … [2] (b) Fig. 3.2 shows a uniform rod PQ, supported at its centre and held in a horizontal position. For The length of PQ is 1.00 m. Examiner’s Use 1.00 m 0.30 m P Q 12 N S Fig. 3.2 A force of 12 N acts at a distance of 0.30 m from the support. A spring S, fixed at its lower end, is attached to the rod at Q. (i) Calculate the force exerted on PQ by the spring. force = … [2] (ii) Explain why it is not necessary to know the weight of PQ. … … [1] [Total: 7]

7 marks

Mark scheme: 3 (a) (i) 3 appropriate examples: e.g. spanner, scissors, tap etc. –1e.e.o.o. B2 (ii) there is a resultant force OR more force down than up B1 there is a resultant moment OR clockwise moment is not equal to anticlockwise moment B1 (b) (i) F × 0.5 = 12 × 0.3 C1 7.2 N A1 (ii) weight has no moment about centre of rod / has no perpendicular distance from centre of rod OR weight acts at centre of rod / pivot / centre of mass B1 [Total: 7] IGCSE – October/November 2013 0625 31

This question in 0625/31 Oct/Nov 2013

2 Fig. 2.1 shows a model fire engine used by a student to take measurements of force and For motion. Examiner’s Use model fire engine containing water tank jet of water FIRE light beams forcemeter 12 mm Fig. 2.1 The model projects a jet of water forwards. The forcemeter holds the model stationary. It indicates a force of 0.060 N acting on the model. The forcemeter is now disconnected and the model accelerates to the right at 0.030 m / s2. (a) The back of the model breaks a pair of light beams and the time to pass between them is measured electronically. The beams are 12 mm apart and the second beam is broken 0.080 s after the first. The student times with a stopwatch how long it takes from the release of the model until the beams are cut. Calculate the time he measures. time measured = … [4] (b) This experiment is carried out with the water tank in the model nearly full. For Examiner’s Calculate the mass of the model including the water in the tank. Use mass = … [2] (c) The student repeats the experiment with the same force but with the water tank nearly empty. State and explain how the acceleration will compare to that of the first experiment. … … … … … [2] [Total: 8]

8 marks

Mark scheme: 2 (a) evidence of division of 12 mm by 0.080 s C1 (v =) 0.15 m / s or 150 mm / s C1 uses t = his (∆)v/a in any form C1 (t = [0.15 – 0] / 0.03 = 0.15 / 0.03) = 5(.0) s accept 1sig. fig. allow e.c.f. from clearly identifiable wrong speed A1 [4] (b) use of F / a OR F = ma in any form, numbers or symbols, ignore g C1 (0.06/0.03=) 2(.0) kg accept 1 significant figure A1 [2] (c) greater M1 because mass is less, ignore comments about force A1 [2] [Total: 8]

This question in 0625/32 Oct/Nov 2013

3 (a) (i) State one similarity and one difference between vector and scalar quantities. For Examiner’s similarity … Use difference … [2] (ii) Give an example of each quantity. vector quantity … scalar quantity … [2] (b) Fig. 3.1 is an overhead view of two tractors pulling a tree trunk. tractors 30 000 N 20° tree trunk 20 000 N Fig. 3.1 The force exerted by each tractor is indicated in the diagram. In the space below, carefully draw a scale diagram to determine the resultant force on the tree trunk. State the scale you use. Write down the magnitude of the resultant force and the angle between the resultant force and one of the original forces. magnitude of resultant force = … direction of resultant force = … [4] [Total: 8]

8 marks

Mark scheme: 3 (a) (i) (both have) magnitude o.w.t.t.e. B1 (only) vector has direction B1 [2] (ii) valid example of vector quantity e.g. displacement, weight, force, velocity B1 valid example of scalar quantity e.g. distance, length, time, pressure, mass, energy accept height B1 [2] IGCSE – October/November 2013 0625 32 (b) each vector to scale and correct angle, larger vector clockwise by acute angle from smaller B1 parallelogram or correct two sides of triangle B1 resultant drawn correct, from his parallelogram or his sides of triangle M1 magnitude 4.5 – 5.4 ×104 N, accept 1 sig. fig. if exact AND direction 4 – 12° from 3 × 104 N force OR 8 – 16° from 2 × 104 N force accept values from diagram A1 [4] [Total: 8]

This question in 0625/32 Oct/Nov 2013

2 A spring S is suspended from a clamp stand in a school laboratory. For Examiner’s A student hangs various masses from the end of S and determines the extension x produced Use by each mass. (a) Calculate the weight of a 250 g mass. weight = … [2] (b) The student plots a graph of the force F applied to the spring against the extension x. Fig. 2.1 is the student’s graph. 6.0 5.0 F / N P 4.0 3.0 2.0 1.0 0 0 2 4 6 8 10 12 x / cm Fig. 2.1 At point P on the graph, the line begins to curve. (i) State the name given to point P. … … [1] (ii) Use the section of the graph where spring S obeys Hooke’s law (F = kx) to determine the spring constant k of the spring. k = … [2] (c) Fig. 2.2 shows a mass of 0.12 kg resting on the bottom of a box. For Examiner’s Use box mass spring Fig. 2.2 A spring that is identical to S connects the mass and one side of the box. Ignore friction between the mass and the box. (i) The box and the mass are at rest. State the resultant force acting on the mass. force = … [1] (ii) The box is firmly attached, in a horizontal position, to the body of a racing car. As the car accelerates the spring stretches by 2.0 cm. 1. Using Fig. 2.1, determine the tension in the spring. tension = … [1] 2. Calculate the acceleration of the mass produced by this tension. acceleration = … [2] [Total: 9]

9 marks

Mark scheme: 2 (a) (W =) mg or 0.25 × 10 or 250 × 10 or 2500 C1 2.5 N A1 [2] (b) (i) limit of proportionality or (the point where) proportionality between force and extension stops or Hooke’s Law no longer obeyed (condone elastic limit) B1 [1] (ii) gradient or numbers from graph divided e.g. 4.5 ÷ 10 C1 0.45 N / cm or 45 N / m A1 [2] (c) (i) 0 (N) or zero or no net force etc. (ignore absent unit; wrong unit loses mark) B1 [1] (ii) 1. 0.9 N (accept 0.8 N < value < 1.0 N) B1 [1] 2. (a =) F/m or 0.90/0.12 (e.c.f. from 2(c)(i)) C1 7.5 m / s2 (e.c.f. from 2(c)(i)) A1 [2] [Total: 9]

This question in 0625/33 Oct/Nov 2013

3 (a) On a day with no wind, a fountain in Switzerland propels 30 000 kg of water per minute to a height of 140 m. Calculate the power used in raising the water. power = … [4] (b) The efficiency of the pump which operates the fountain is 70%. Calculate the power supplied to the pump. power = … [3] (c) On another day, a horizontal wind is blowing. The water does not rise vertically. Explain why the water still rises to a height of 140 m. … … [1] [Total: 8]

8 marks

Mark scheme: 3 (a) Fd OR weight × d OR mgh OR 30 000 × 10 × 140 OR 4.2 × 107 seen anywhere C1 (P = ) E / t OR W / t OR mgh / t symbols or words C1 4.2 × 107 / 60 C1 7.0 ×105 W / 700 kW / 0.7 MW A1 (b) efficiency = output / input OR (Pin =) 100 × Pout / efficiency C1 (Pin =) 100 × 7 × 105 / 70 C1 1.0 × 106 W OR 1 000 000 W OR 1.0 MW A1 (c) (horizontal) wind has no effect on P.E gained / vertical force on water OR same upward / vertical force acts on water OR force from wind is horizontal B1 [Total: 8]

This question in 0625/31 May/June 2014

4 Fig. 4.1 shows a heavy ball B of weight W suspended from a fixed beam by two ropes P and Q. beam P Q 30 N 30 N 45° 45° B W Fig. 4.1 P and Q are both at an angle of 45° to the horizontal. The tensions in P and Q are each 30 N. (a) In the space below, draw a scale diagram to find the resultant of the tensions in P and Q. Use a scale of 1.0 cm to represent 5.0 N. Label the forces and show their directions with arrows. resultant = … [4] (b) State the direction of the resultant. … [1] (c) State the magnitude of W. magnitude of W = … [1] [Total: 6]

6 marks

Mark scheme: 4 (a) 2 lines at 90 ° to each other of same length labelled 30 N or 6 cm B1 both lines 6.0 ± 0.2 cm. B1 arrows on the two lines drawn, either head to tail B1 OR a complete square shown with diagonal and arrows on adjacent sides resultant in range 40–45 N B1 (b) (vertically) upwards B1 IGCSE – May/June 2014 0625 31 (c) same as value in (a), only if answer to (a) is a force OR 40–45 N B1 [Total: 6]

This question in 0625/31 May/June 2014

1 (a) State the two conditions necessary for a system of forces acting on a body to be in equilibrium. 1. … … 2. … … [2] (b) Fig. 1.1 shows a loaded wheelbarrow held in equilibrium by a gardener. The wheel of the wheelbarrow is in contact with the ground at point C. P Q C W Fig. 1.1 In Fig. 1.1, there are three vertical forces acting on the wheelbarrow. P is the upward force applied by the gardener. Q is the upward force of the ground on the wheel at point C. W is the weight of the wheelbarrow and its contents. Explain why the force P is less than the force W (i) by considering the forces P, Q and W, … … [2] (ii) by considering the moments of the forces P and W about point C. … … [2] (c) Fig. 1.2 shows a kitchen cupboard resting on a support and attached to a wall by a screw. wall screw cupboard F G 0.75 m P support 0.24 m 75 N Fig. 1.2 The weight of the cupboard and its contents is 75 N. G is the position of the centre of mass of the cupboard. The clockwise and anticlockwise moments about point P are equal. Calculate the force F exerted by the screw. F = … [3] [Total: 9]

9 marks

Mark scheme: 1 (a) no resultant/net force (acting) B1 no resultant/net moment (acting) OR clockwise moment = anticlockwise moment B1 (b) (i) W = P + Q in any form OR (total) upward force = (total) downward force B1 P = W – Q so P must be less than W OR P is not the only upward force B1 (ii) P × its distance (from C)=W × its distance (from C) OR P and W have equal moments (about C) OR clockwise moment = anticlockwise moment B1 P is farther from C/pivot (than W so P must be less than W) B1 (c) clockwise moment = 75 × 0.24 C1 anticlockwise moment = F × 0.75 C1 (moments equated gives F =) 24 N A1 [Total: 9] st

This question in 0625/32 Oct/Nov 2014

1 A free-fall parachutist jumps from a helium balloon, but does not open his parachute for some time. Fig. 1.1 shows the speed-time graph for his fall. Point B indicates when he opens his parachute. speed A B C D 0 0 time Fig. 1.1 (a) (i) State the value of the gradient of the graph immediately after time t = 0. gradient = … [1] (ii) Explain why the gradient has this value. … … [1] (b) State how Fig. 1.1 shows that the acceleration decreased between time t = 0 and the time to A. … … [1] (c) Explain, in terms of forces, what is happening in section AB of the graph in Fig. 1.1. … … … [2] (d) A second parachutist of the same size and mass jumps from the balloon with a larger parachute. He also opens his parachute at point B. On Fig. 1.1, sketch a possible speed-time graph for his fall after he opens his parachute. [3] [Total: 8]

8 marks

Mark scheme: 1 (a) (i) (gradient =) 10 (m / s2) B1 (ii) any linking of gradient to acceleration of freefall OR gravitational field strength B1 (b) gradient decreases B1 (c) speed / velocity stays constant OR terminal velocity / speed B1 no resultant force OR forces cancel / balance B1 (d) initially gradient steeper B1 graph lower in second half of BC B1 horizontal final section and lower than CD B1 [Total: 8]

This question in 0625/33 Oct/Nov 2014

2 The rocket shown in Fig. 2.1 is about to be launched. rocket Fig. 2.1 The total mass of the rocket and its full load of fuel is 2.8 × 106 kg. The constant force provided by the rocket’s motors is 3.2 × 107 N. (a) Calculate (i) the total weight of the rocket and the fuel, weight = … [1] (ii) the resultant force acting on the rocket, resultant force = … [2] (iii) the vertical acceleration of the rocket immediately after lift-off. acceleration = … [2] (b) Suggest why the acceleration of the rocket increases as it rises above the Earth’s surface. … … [1] [Total: 6]

6 marks

Mark scheme: 2 (a) (i) (W = mg = 2.8 × 106 × 10 =) 2.8 × 107 N B1 (ii) 3.2 × 107 – 2.8 × 107 C1 4.0 × 106 OR 0.4 × 107 N A1 (iii) F = ma in any form OR (a =) F÷m OR 4.0 × 106÷(2.8 × 106) C1 1.4 m / s2 A1 (b) Mass of rocket decreases (as fuel is used up) OR Value of g/gravitational force on rocket decreases as rocket rises B1 OR Air resistance decreases [Total: 6]

This question in 0625/32 Feb/March 2015

3 (a) Complete the following statement. An object is in equilibrium when both the … and the … on the object are zero. [2] (b) Fig. 3.1 shows a ladder AB. End A of the ladder rests against a vertical wall. End B rests on rough ground. A F ladder wall 3.2 m 1.2 m P B ground 240 N Fig. 3.1 Fig. 3.1 shows two of the forces acting on the ladder. The only force on the ladder at A is F, which acts at right-angles to the wall. The weight of the ladder is 240 N acting at the centre of mass of the ladder. (i) 1. Calculate the moment of the weight of the ladder about point B. moment = … [1] 2. Write an expression, in terms of F, for the moment of F about point B. moment = … [1] (ii) Use your answers from (i) to calculate F. F = … [2] (iii) Explain why there must be an upwards force acting on the ladder at B. … … [1] [Total: 7]

7 marks

Mark scheme: 3 (a) Note: answers in either order Resultant/net/total force B1 Resultant/net/ total turning effect/moment/torque/couple B1 (b) (i) 1. (240 × 1.2 =) 290 (N m) B1 2. F × 3.2 B1 (ii) F × 3.2 = 288 C1 90 N A1 (iii) To balance the weight B1 OR to make resultant (vertical) force zero OR to make resultant moment zero OR to keep the ladder in (vertical) equilibrium OR because there is a downward force OR because the ladder is pressing on the ground OR otherwise the ladder would fall / sink (into the ground) [Total: 7]

This question in 0625/32 Feb/March 2015

2 (a) (i) Mass is a scalar quantity. State another scalar quantity. … (ii) Force is a vector quantity. State another vector quantity. … [2] (b) A boat is floating on still water. The mass of the boat is 290 000 kg. A resultant force of 50 kN acts on the boat. Calculate the acceleration of the boat. acceleration = … [3] (c) Fig. 2.1, not to scale, shows the view from above of the boat, now on a fast-flowing river. The boat accelerates. Two forces are shown acting on the boat. The resultant of these forces is at right angles to the river banks. river bank direction of force from engine direction of river current direction of boat force on boat from river current river bank Fig. 2.1 (not to scale) Fig. 2.2 is an incomplete vector diagram of the forces acting on the boat. resultant force force from river current Fig. 2.2 The force from the river current is 80 kN. (i) Determine the scale that has been used in the vector diagram. scale is … (ii) On Fig. 2.2, complete the vector diagram to determine the magnitude and direction of the force from the engine. Measure the angle between the direction of the current and the force from the engine. magnitude of force from engine = … angle = … [4] [Total: 9]

9 marks

Mark scheme: 2 (a) (i) any scalar quantity other than mass B1 (ii) any vector quantity other than force B1 (b) F = ma in any form OR (a =) F / m C1 50 000 / 290 000 OR 50 / 290 C1 a = 0.17 m / s2 A1 (c) (i) 1 cm: 20 000 N / 20 kN B1 (ii) triangle completed B1 230 000 N OR 230 kN in range 220 000 N – 240 000 N / 220 kN – 240 kN B1 by calculation: 110o OR by measurement: 108o – 112o B1 [Total: 9]

This question in 0625/31 Oct/Nov 2015

4 (a) Fig. 4.1 shows a top view of a tourist vehicle in a game park and two elephants pushing against the vehicle. The two forces indicated are at right angles to each other. vehicle 4.0 kN 6.0 kN elephant elephant Fig. 4.1 In the space below, draw a scale vector diagram to determine the magnitude of the resultant force. Label the two forces applied and the resultant, and clearly state the scale you use. magnitude of resultant force = … [3] (b) Fig. 4.2 shows another elephant pushing horizontally against a vehicle with a force of 11 kN at a distance 1.8 m above the ground. Point M is the centre of mass of the vehicle. elephant vehicle 11 kN M 1.8 m A 1.25 m Fig. 4.2 (i) Calculate the moment about point A of the force exerted by the elephant. moment = … [2] (ii) The mass of the vehicle is 1900 kg, and it does not slide when pushed by the elephant. Determine whether the elephant tips the vehicle over. Show your working. calculation conclusion … [2] [Total: 7]

7 marks

Mark scheme: 4 (a) 2 vectors correct direction AND relative length by eye B1 correct triangle OR rectangle with resultant on correct diagonal B1 7.2 kN tolerance 7.0 – 7.4 kN B1 (b) (i) (moment =) force × distance C1 (moment = 11 000 × 1.8 =) 20 kNm A1 (ii) (moment of weight = 19 000 x 1.25 =) 24 (kNm) B1 correct statement based on two moments seen B1 [Total: 7]

This question in 0625/32 Oct/Nov 2015

1 A plastic ball is dropped from the balcony of a tall building and falls towards the ground in a straight line. Fig. 1.1 is the speed-time graph for the falling ball. 8.0 speed m / s 6.0 4.0 2.0 0 0 1.0 2.0 3.0 time / s Fig. 1.1 (a) State and explain, in terms of forces, what is happening to the speed of the ball between time t = 2.0 s and t = 3.0 s. … … … [2] (b) On Fig. 1.1, mark a point P on the line where the acceleration of the ball is not constant. [1] (c) Using Fig. 1.1, (i) calculate the acceleration of the ball between t = 0 s and t = 0.25 s, acceleration = … [2] (ii) estimate the distance that the ball falls in the first 3.0 s. distance = … [2] [Total: 7]

7 marks

Mark scheme: 1 (a) speed is constant / uniform / unchanging OR terminal velocity / speed B1 no net / resultant force OR air resistance cancels / equals weight B1 (b) P between 0.25 s and 1.90 s (inclusive) B1 (c) (i) (a =) ∆v / t OR 2.5 / 0.25 OR other point on correct section of line B1 9.6 to 10 m / s2 (inclusive) B1 (ii) area under graph OR attempt at counting squares OR between 16.2 and 17.5 m (inclusive) C1 between 16.5 and 17.1 m (inclusive) A1 [Total: 7] 3

This question in 0625/33 Oct/Nov 2015

3 Fig. 3.1 shows part of the extension-load graph for a spring. extension / cm 3.0 B A 00 12.0 load / N Fig. 3.1 The spring obeys Hooke’s law between points A and B. (a) (i) On Fig. 3.1, complete the graph between A and B. [1] (ii) State the name of point B. … [1] (b) The average value of the load between A and B is 6.0 N. Calculate the work done in extending the spring from A to B. work done = … [2] (c) The spring has an unstretched length of 4.0 cm. An object is hung on the spring and the spring length increases from 4.0 cm to 6.0 cm. (i) Calculate the mass of the object. mass = … [3] (ii) The object is immersed in a liquid but remains suspended from the spring. The liquid exerts an upward force on the object and the length of the spring decreases to 5.0 cm. Calculate the upward force exerted on the object by the liquid. upward force = … [2] [Total: 9]

9 marks

Mark scheme: 3 (a) (i) straight line between A and B B1 (ii) limit of proportionality B1 (b) (WD =) 12 F × d OR Fave × d OR 6.0 × 0.030 OR 18 (J) C1 0.18 J A1 (c) (i) (x =) 2.0 (cm) OR 6.0 – 4.0 OR F = kx OR 4.0 (N / cm) C1 12.0 × 2.0 / 3.0 OR 4.0 × 2.0 OR 8.0 (N) C1 0.80 kg OR 800 g A1 (ii) (e =) 1.0 (cm) OR (∆e = –)1.0 (cm) C1 4.0 N OR 4.0 N A1 [Total: 9]

This question in 0625/33 Oct/Nov 2015

3 Fig. 3.1 shows a see-saw. The see-saw is horizontal when not in use. P Fig. 3.1 A small child sits on one seat of the see-saw. This creates a turning effect about point P. (a) Which of these words means the turning effect of a force? Tick one box. equilibrium moment resultant [1] (b) State the scientific name for point P. … [1] (c) A much heavier boy sits on the other end of the see-saw, as shown in Fig. 3.2. P end A boy moves end B Fig. 3.2 The heavier boy moves slowly along the see-saw from end B until he reaches point P. Describe and explain what happens to the see-saw. … … … … … [4] [Total: 6]

6 marks

Mark scheme: 3 (a) middle box ticked – moment B1 (b) pivot / fulcrum B1 (c) any four from: B4 • (heavier) boy has greater force / weight / moment • when (heavier) boy lifts feet initially tips clockwise • as boy moves his (clockwise) moment (about P) becomes less • as distance (of boy’s weight) from the pivot decreases end B moves upward • see-saw level o.w.t.t.e (when) turning forces balanced / moments equal • then end A tips down as anticlockwise moment is greater [Total: 6]

This question in 0625/32 Feb/March 2016

2 A boy steps off a high board into a swimming pool. Fig. 2.1 shows the forces acting on the boy at one point in his fall. board 100 N 540 N swimming pool Fig. 2.1 (a) The 540 N force is caused by gravitational attraction. State the cause of the 100 N force. … [1] (b) Calculate the mass of the boy. mass of boy = … kg [2] (c) Calculate the resultant force on the boy. State its direction. resultant force = … N direction = … [2] [Total: 5]

5 marks

Mark scheme: 2(a) air resistance B1 2(b) W = m × g in any form B1 54(kg) B1 2(c) (540 – 100) = 440(N) B1 downwards B1 Total: 5

This question in 0625/31 May/June 2016

2 Fig. 2.1 shows the speed-time graph for a student cycling along a straight, flat road. 8 speed m / s 6 4 2 0 0 5 10 15 20 25 30 35 time / s Fig. 2.1 (a) Calculate the distance he travels in the first 10 s. distance = … m [3] (b) Fig. 2.2 shows three pairs of forces A, B and C. 60 N 20 N A backward force forward force 50 N 50 N B backward force forward force 20 N 70 N C backward force forward force Fig. 2.2 Identify which pair of forces, A, B or C, acts on the cyclist between 11 s and 16 s. Explain your choice. pair of forces … explanation … … … [3] (c) The cyclist pushes on one pedal with a force of 120 N. The area of his shoe in contact with the pedal is 16 cm2. Calculate the pressure on the pedal. Include the unit. pressure = … [4] [Total: 10]

10 marks

Mark scheme: 2(a) (distance) = area under (speed-time) graph in words, numbers or symbols OR distance = speed × time 4.4 × 10 44 (m) C1 C1 A1 2(b) C (cyclist is) accelerating (so) forward force must be greater than backward force OR there is a resultant (forward) force B1 B1 B1 2(c) P = F ÷ A 120 ÷ 16 7.5 N / cm2 C1 C1 A1 B1 Total: 10

This question in 0625/32 May/June 2016

3 Fig. 3.1 shows a girl and a boy on a see-saw. boy girl 2.4 m 1.6 m 300 N W log pivot Fig. 3.1 The see-saw pivots on the log. (a) Calculate the girl’s moment about the pivot. girl’s moment = … N m [2] (b) The see-saw is balanced horizontally. Calculate the weight W of the boy. weight of boy = … N [3] [Total: 5]

5 marks

Mark scheme: 3(a) force × distance (from pivot) OR 300 × 2.4 720 (N m) C1 A1 3(b) sum of clockwise moment = sum of anticlockwise moment 720 = W × 1.6 OR 720 ÷ 1.6 450 (N) C1 C1 A1 Total: 5

This question in 0625/32 May/June 2016

2 Fig. 2.1 shows the speed-time graph for a student cycling along a straight, flat road. 8 speed m / s 6 4 2 0 0 5 10 15 20 25 30 35 time / s Fig. 2.1 (a) Calculate the distance he travels in the first 10 s. distance = … m [3] (b) Fig. 2.2 shows three pairs of forces A, B and C. 60 N 20 N A backward force forward force 50 N 50 N B backward force forward force 20 N 70 N C backward force forward force Fig. 2.2 Identify which pair of forces, A, B or C, acts on the cyclist between 11 s and 16 s. Explain your choice. pair of forces … explanation … … … [3] (c) The cyclist pushes on one pedal with a force of 120 N. The area of his shoe in contact with the pedal is 16 cm2. Calculate the pressure on the pedal. Include the unit. pressure = … [4] [Total: 10]

10 marks

Mark scheme: 2(a) (distance) = area under (speed-time) graph in words, numbers or symbols OR distance = speed × time 4.4 × 10 44 (m) C1 C1 A1 2(b) C (cyclist is) accelerating (so) forward force must be greater than backward force OR there is a resultant (forward) force B1 B1 B1 2(c) P = F ÷ A 120 ÷ 16 7.5 N / cm2 C1 C1 A1 B1 Total: 10

This question in 0625/33 May/June 2016

3 Fig. 3.1 shows a girl and a boy on a see-saw. boy girl 2.4 m 1.6 m 300 N W log pivot Fig. 3.1 The see-saw pivots on the log. (a) Calculate the girl’s moment about the pivot. girl’s moment = … N m [2] (b) The see-saw is balanced horizontally. Calculate the weight W of the boy. weight of boy = … N [3] [Total: 5]

5 marks

Mark scheme: 3(a) force × distance (from pivot) OR 300 × 2.4 720 (N m) C1 A1 3(b) sum of clockwise moment = sum of anticlockwise moment 720 = W × 1.6 OR 720 ÷ 1.6 450 (N) C1 C1 A1 Total: 5

This question in 0625/33 May/June 2016

2 Fig. 2.1 shows a cyclist travelling along a level road. backwards forwards force force Fig. 2.1 (a) Describe the movement of the bicycle when the forces shown in Fig. 2.1 are of equal size. … [1] (b) Fig. 2.2 shows the forces on the bicycle later in the journey. 150 N 225 N Fig. 2.2 Determine the size and direction of the resultant force. size of resultant force = … N direction of resultant force = … [2] (c) At the end of the journey the cyclist applies the brakes to stop the bicycle. Name and describe the force that causes the bicycle to stop. force … description … … [2] [Total: 5]

5 marks

Mark scheme: 2(a)(i) constant speed/velocity B1 2(a)(ii) 75 N B1 forwards B1 2(b) friction B1 two surfaces rubbing together owtte B1 Total 5

This question in 0625/33 Oct/Nov 2016

3 Fig. 3.1 shows a beam on a pivot. The beam is balanced at its centre of mass. beam pivot bench Fig. 3.1 (a) Explain the meaning of centre of mass. … … [1] (b) Fig. 3.2 shows a load of 2.5 N on one side of the beam. The beam is balanced by a load of 1.5 N suspended by a thin string. load of 18 cm x 2.5 N string load of 1.5 N Fig. 3.2 (not to scale) (i) Calculate the distance x from the pivot to the string. distance from pivot = … cm [3] (ii) Calculate the mass of the 2.5 N load. mass = … kg [3] [Total: 7]

7 marks

Mark scheme: 3(a) point where all the weight seems to act owtte B1 3(b)(i) moments clockwise = moments anticlockwise C1 2.5 X 18 = 1.5 x ? OR 45 ÷ 1.5 C1 30 (cm) A1 3(b)(ii) w = m × g in any recognised form C1 2.5 ÷ 10 C1 0.25 (kg) A1 Total: 7

This question in 0625/32 Feb/March 2017

3 Fig. 3.1 shows a tyre hanging from the branch of a tree. branch 2.5 m P rope tyre Fig. 3.1 (a) The mass of the tyre is 15 kg. Calculate its weight. weight of tyre = … N [2] (b) The weight of the tyre exerts a moment on the branch, about point P where the branch joins the tree. (i) Explain what is meant by the term moment. … [1] (ii) A child sits on the tyre. The weight of the child and tyre together is 425 N. Calculate the moment of this force about point P. Use information given in Fig. 3.1. Include the unit. moment = … [4] (iii) A heavier child wants to sit on the tyre. Describe how the tyre position should be adjusted so that the moment is the same as in (b)(ii). … [1] [Total: 8]

8 marks

Mark scheme: 3(a) C1 150 (N) A1 3(b)(i) turning effect (of a force) B1 3(b)(ii) moment = force × distance C1 425 × 2.5 C1 1062.5 OR 1063 A1 N m B1 3(b)(iii) (move rope/tyre) closer to trunk owtte B1 Total: 8

This question in 0625/31 May/June 2017

5 A laboratory floor has a surface that prevents people from slipping when the floor is wet. (a) Name the force that prevents a person from slipping. … [1] (b) A stool has a round non-slip pad fitted to the bottom of each leg. (i) The stool has four legs. The area of each pad is 3 cm2. The weight of the stool is 75 N. A student sits on the stool. The weight of the student is 525 N. Calculate the pressure acting on the floor due to the student and the stool. pressure = … N / cm2 [5] (ii) The legs of the stool are made of hollow metal tubes. Fig. 5.1 shows the bottom of a stool leg with and without a pad. metal tube with pad without a pad Fig. 5.1 Explain why a stool leg without a pad does more damage to the floor. … … [2] [Total: 8]

8 marks

Mark scheme: 5(a) friction B1 5(b)(i) total area = 3 × 4 = 12 (cm2) C1 total weight = 525 + 75 N = 600(N) C1 P = F ÷ A in any form C1 600 ÷ 12 C1 50 (N / cm2) A1 5(b)(ii) less (surface) area (in contact with the ground) owtte B1 more pressure (results in more damage to the surface) B1 Total: 8

This question in 0625/31 May/June 2017

3 Fig. 3.1 shows a large sunshade. arm sunshade pivot support base Fig. 3.1 The arm holding the sunshade pivots about the end of a support. (a) The sunshade has a mass of 20.0 kg. Calculate the weight of the sunshade. weight = … N [3] (b) (i) Another sunshade is shown in Fig. 3.2. This sunshade weighs 180 N. The arm holding the sunshade extends 2.5 m from the pivot. 2.5 m pivot support 180 N base Fig. 3.2 Calculate the moment of the sunshade about the pivot. moment = … N m [3] (ii) How can the moment produced by the sunshade be reduced? Tick one box. by decreasing the height of the support by decreasing the length of the arm holding the sunshade by increasing the weight of the base by increasing the weight of the sunshade [1] [Total: 7]

7 marks

Mark scheme: 3(a) C1 20.0 × 10.0 C1 200 (N) A1 3(b)(i) moment = force × (perpendicular) distance (from pivot) in any form C1 180.0 × 2.5 C1 450 (N m) A1 3(b)(ii) 2nd box down ticked decrease the length of the arm holding the sun-shade B1 Total: 7

This question in 0625/32 May/June 2017

2 A student has a laptop computer. The computer is powered by a battery. (a) State the word used to describe the energy stored in the battery. … [1] (b) The student opens the laptop using a force of 3.0 N, as shown in Fig. 2.1. 3.0 N 25.0 cm pivot Fig. 2.1 (i) Calculate the moment of the 3.0 N force about the pivot. moment = … N cm [3] (ii) The student does work as he opens the laptop. Explain how the principle of conservation of energy applies to this example. … … [2] (c) The student is in a country with many hours of sunshine each day. He charges his laptop using a solar panel. Give two advantages of using a solar panel, compared with using a mains electrical supply. 1. … 2. … [2] (d) A mains battery charger has a power output of 80 W. The solar panel has a power output of 16 W. Describe one disadvantage of using the solar panel, compared with using the mains battery charger. … [1] [Total: 9]

9 marks

Mark scheme: 2(a) chemical B1 2(b)(i) Moment = force × (perpendicular) distance (from pivot) in any form C1 3.0 × 25.0 C1 75 (N cm) A1 2(b)(ii) any two from: idea that work done = energy gained total energy does not change the student loses chemical energy laptop gains (gravitational) PE (of lid) energy dissipated as thermal energy in the environment B2 2(c) any two from: laptop can be charged anywhere owtte cost of charging is zero (Sun is a) renewable energy (source)/not using fossil fuels B2 2(d) (Takes 5 times) longer to (re-)charge (battery) B1 Total: 9

This question in 0625/33 May/June 2017

2 Fig. 2.1 shows a river flowing through a village. There are two bridges across the river. bridge X direction of flow bridge Y Fig. 2.1 Two students plan to measure the speed of a stick as it floats on the river between bridge X and bridge Y. (a) The students plan to drop a stick into the middle of the river from bridge X. The stick moves with the water between bridge X and bridge Y. Describe how the students can determine the average speed of the stick. … … … … … … … … [4] (b) The stick moves with constant speed. One statement correctly describes the horizontal forces acting on the stick. Put a tick (✓) in the box next to the correct statement. Only a forward force acts. The forward force and the backward force are equal. The forward force is greater than the backward force. The backward force is greater than the forward force. [1] [Total: 5]

5 marks

Mark scheme: 2(a) Any four from: Measure the distance between the two bridges Start stopwatch when stick hits water / starts moving (with river) stop stopwatch when stick reaches bridge Y Use speed = distance ÷ time repeat procedure and find average B4 2(b) 2nd box ticked The forward force and the backward force are equal B1

This question in 0625/31 Oct/Nov 2017

3 Fig 3.1 shows a warning marker floating on the surface of a lake. warning marker surface of lake metal chain heavy object bottom of lake Fig. 3.1 The marker is attached by a metal chain to a heavy object on the bottom of the lake. (a) Fig. 3.2 shows the forces acting on the marker at one moment in time. 280 N 250 N Fig. 3.2 Calculate the resultant force on the marker. resultant force = … N direction = … [2] (b) Fig. 3.3 shows part of the metal chain. It is made from small metal loops. Fig. 3.3 A damaged loop is removed from the chain. Describe a method to determine the density of the metal from which the loops are made. … … … … … … … [5] [Total: 7]

7 marks

Mark scheme: 3(a) subtraction of forces to obtain resultant or 30 (N) B1 up(wards) B1 3(b) any five from: measure mass (on top pan balance) part fill measuring cylinder with water (and note volume) submerge link in measuring cylinder determine increase in volume increase in volume = volume of link use density = mass ÷ volume Only award full marks for a viable method B5

This question in 0625/31 Oct/Nov 2017

4 Fig. 4.1 shows a fairground ride. pylon pylon starting position path of cabin cabin Fig. 4.1 People sit inside a cabin suspended between two pylons. The cabin is lifted to the starting position shown in Fig. 4.1. (a) State the name of the type of energy gained by the cabin as it rises. … [1] (b) The cabin is released and swings down between the two pylons. The path of the cabin is shown on Fig. 4.1. The cabin has maximum kinetic energy at one point on its path. Draw this point on Fig. 4.1 and label this point X. [1] (c) A force opposes the motion of the cabin as it falls. State the name of this force. … [1] (d) After a few swings, a brake system stops the cabin (not shown on the diagram). Use ideas about energy transfer to suggest how the brake system stops the cabin. … … … … [3] [Total: 6]

6 marks

Mark scheme: 4(a) (gravitational) potential (energy) B1 4(b) arrow at the lowest point of swing B1 4(c) friction / air resistance / drag B1 4(d) any three from: cabin has kinetic energy two surfaces rub together / friction thermal energy generated / KE transferred to thermal dissipated to surroundings / air B3

This question in 0625/32 Oct/Nov 2017

3 Fig. 3.1 shows a large box with a heavy lid. Fig. 3.1 (a) The weight of the box is 2250 N. Calculate the mass of the box. mass = … kg [3] (b) A man wants to lift the lid of the box. He puts a strong metal bar between the box and the lid. He applies a force to the bar as shown in Fig. 3.2. metal bar 40 cm 400 N pivot Fig. 3.2 (i) Calculate the moment of his force about the pivot. State the unit. moment = … [4] (ii) The moment in (b)(i) is not sufficient to lift the lid. Describe how the man can increase the moment, using the same force. … … [1] [Total: 8]

8 marks

Mark scheme: 3(a) C1 2250 / 10 C1 225 (kg) A1 3(b)(i) moment = force × distance from pivot in any recognised form C1 400 × 0.4 OR 400 × 40 C1 160 OR 16 000 A1 Nm OR Ncm B1 3(b)(ii) apply force further from pivot owtte B1

This question in 0625/33 Oct/Nov 2017

3 (a) Fig. 3.1 shows the vertical forces on a rocket. thrust 74.2 N air resistance 2.4 N weight 43.0 N Fig. 3.1 Calculate the resultant force on the rocket. resultant force = … N direction = … [3] (b) Fig. 3.2 shows the speed and direction of motion of an object at a point in time. 150.0 m / s object Fig. 3.2 The resultant force on the object is zero for 10 seconds. Deduce the speed and direction of motion after 5 seconds. Indicate the speed and direction of the object by drawing a labelled arrow next to the object in Fig. 3.3. Fig. 3.3 [1] [Total: 4]

4 marks

Mark scheme: 3(a) 43.0 + 2.4 = 45.4 (N) 1 (74.2 – 45.4 =) 28.8 (N) 1 upwards 1 3(b) 1

This question in 0625/31 May/June 2018

3 A man uses a metal bar to remove an iron nail from a piece of wood, as shown in Fig. 3.1. 150 N nail 0.50 m wood pivot Fig. 3.1 (a) (i) The man applies a force of 150 N at a distance of 0.50 m from the pivot. Calculate the moment of this force about the pivot. Include a unit. moment = … [4] (ii) The force applied by the man produces a turning effect (moment) about the pivot. Describe another example of using the turning effect of a force. … … [1] (b) The man tries to use the metal bar to remove another nail from the piece of wood. He applies the same force of 150 N at a distance of 0.50 m from the pivot. The turning effect produced is not enough to remove this nail from the piece of wood. Describe how the man can increase the turning effect without increasing the force. … … [1] [Total: 6]

6 marks

Mark scheme: 3(a)(i) 1 150 × 0.5 1 75 1 N m 1 3(a)(ii) accept any example involving turning forces 1 3(b) increase distance (of force from pivot point ) 1

This question in 0625/32 May/June 2018

2 A student is using some 50 g masses. (a) Calculate the weight of one 50 g mass. weight of 50 g mass = … N [3] (b) The student uses the 50 g masses as loads to stretch a spring. Fig. 2.1 shows the apparatus the student uses to obtain readings for a load-extension graph. spring pin stand 50 g masses 50 g mass rule hanger Fig. 2.1 (NOT to scale) Describe how the student could use the apparatus and ensure that the readings are accurate. … … … … … … … … … … [4] [Total: 7]

7 marks

Mark scheme: 2(a) W = m × g 1 50 ÷ 1000 OR 0.05 seen 1 0.5 (N) 1 2(b) any 4 from: fix ruler vertically add weight/hanger to spring fix pin (horizontally) to (top/bottom) of weight hanger pin arranged so near ruler scale ensure load stationary eye level with pin to take reading (of length) determine extension for given load repeat for different loads 4

This question in 0625/33 May/June 2018

3 (a) The mass of a small steel ball is 120 g. The volume of the ball is 16.0 cm3. (i) Calculate the density of the steel ball. density = … g / cm3 [3] (ii) The ball falls to the ground from rest. At a time of 0.2 s after it started to fall, its acceleration is 10 m / s2. State the acceleration of the ball at a time of 0.1 s after it started to fall. … [1] (b) Fig. 3.1 shows the vertical forces that act on a large plastic ball as it is falling. 0.3 N large plastic ball 1.2 N Fig. 3.1 (not to scale) (i) State the name given to each of the forces shown in Fig. 3.1. 1. … 2. … [1] (ii) Calculate the size of the resultant force on the ball. resultant force = … N [1] [Total: 6]

6 marks

Mark scheme: 3(a)(i) (D =) m ÷ v in any form 1 120 ÷ 16.0 1 7.50 (g / cm3) 1 3(a)(ii) 10 (m / s2) 1 3(b)(i) (downward force) weight AND (upward force) air resistance/friction/drag 1 3(b)(ii) (1.2 – 0.3 =) 0.9 (N) 1

This question in 0625/33 May/June 2018

4 A lamp is attached to a wall, as shown in Fig. 4.1. 90.0 cm lamp pivot wall 25.0 N Fig. 4.1 Calculate the moment of the lamp about the pivot. Give the unit. moment = … [4] [Total: 4]

4 marks

Mark scheme: 4 1 25 × 90 or 25 × 0.9 1 2250 or 22.5 1 N cm or N m 1

This question in 0625/33 May/June 2018

2 Fig. 2.1 shows a raft floating on water. raft water Fig. 2.1 (a) A force of 20 000 N acts on the raft in the direction of the arrow shown in Fig. 2.1. (i) State the name given to the force shown in Fig. 2.1. … [1] (ii) Calculate the mass of the raft. mass = … kg [3] (b) A sail is added to the raft, as shown in Fig. 2.2. sail 800 N 1200 N Fig. 2.2 Fig. 2.2 shows the horizontal forces acting on the raft at one moment. Calculate the resultant horizontal force acting on the raft and state the direction of this force. force = … N direction = … [2] [Total: 6]

6 marks

Mark scheme: 2(a)(i) weight B1 2(a)(ii) W = m × g C1 m = 20 000 ÷ 10 C1 2000 (kg) A1 2(b) 400 (N) B1 forwards / to the right B1

This question in 0625/31 Oct/Nov 2018

3 A tower crane has a load W, as shown in Fig. 3.1. 8.0 m 5.0 m counterweight pivot 80 000 N tower load W Fig. 3.1 (a) The counterweight has a weight of 80 000 N. This acts at a distance of 5.0 m from the pivot, as shown in Fig. 3.1. Calculate the moment of the counterweight about the pivot. Give the unit. moment = … [3] (b) The tower crane in Fig. 3.1 balances horizontally when holding the load W. Calculate the weight of load W. weight = … N [3] [Total: 6]

6 marks

Mark scheme: 3(a) C1 400 000 A1 Nm B1 3(b) c.w. moment = a.c.w moment OR moment of load = moment of counterweight OR 5.0 × 80 000 = load × 8.0 C1 400 000 ÷ 8.0 = load C1 50 000 (N) A1

This question in 0625/31 Oct/Nov 2018

3 A load is attached to a spring, as shown in Fig. 3.1. Two arrows indicate the vertical forces acting on the load. The spring and the load are stationary. support spring 4.0 N load Fig. 3.1 (a) (i) State the name of the force acting vertically downwards. … [1] (ii) The vertical force that acts upwards is 4.0 N. State the value of the force acting vertically downwards. force = … N [1] (b) The load is pulled downwards and then released. The load moves up and down. Fig. 3.2 represents the vertical forces acting on the load at some time after it is released. 7.6 N 2.8 N Fig. 3.2 Calculate the resultant force on the load and state its direction. resultant force = … N direction = … [2] (c) (i) State the principle of conservation of energy. … … [1] (ii) Eventually the load stops moving up and down. Describe and explain why the load stops moving. Use your ideas about conservation of energy. … … … … [2] [Total: 7]

7 marks

Mark scheme: 3(a)(i) gravity OR weight B1 3(a)(ii) 4.0 (N) B1 Question Answer Marks 3(b) 4.8 (N) B1 Up(wards) B1 3(c)(i) energy cannot be created or destroyed (but can be transformed) B1 3(c)(ii) PE / KE / elastic energy of load / spring decreases / is transformed B1 Any one from: to thermal energy (which is) dissipated (to surroundings) B1

This question in 0625/32 Feb/March 2019

4 Fig. 4.1 shows a truck lifting a heavy load. load truck pivot Fig. 4.1 (a) (i) The truck is stationary. Identify the quantities that determine the work done as it lifts the load. Tick the box next to each correct quantity. distance force time [1] (ii) Draw a ring around the unit for work done from the list. joule newton pascal watt [1] (b) Identify the quantities that determine the power of the truck. Tick the box next to each correct quantity. energy transferred temperature time [1] (c) The truck has a pivot near the front wheel. Fig. 4.2 represents the pivot and the vertical forces acting on the truck. 1.5 m 1.0 m 30 000 N pivot load Fig. 4.2 The truck is in equilibrium. Calculate the load. load = … N [3] (d) Fig. 4.3 shows another truck lifting a pile of identical bricks. pile of bricks Fig. 4.3 (i) On Fig. 4.3, draw a cross to indicate the centre of mass of the pile of bricks. [1] (ii) The truck can tilt the pile of bricks backwards, as shown in Fig. 4.4. Fig. 4.4 Explain how tilting the pile of bricks backwards makes the truck more stable. … … … … [1] [Total: 8]

8 marks

Mark scheme: 4(a)(i) tick in top two boxes: distance AND force B1 4(a)(ii) first word circled: joule B1 4(b) tick in top AND bottom boxes: energy transferred AND time B1 4(c) clockwise moment = anticlockwise moment C1 1.5 × 30 000 = 1 × (load) C1 (load =) 45 000 (N) A1 4(d)(i) centre of mass in centre of load B1 4(d)(ii) centre of mass (moves) closer to pivot (point) B1

This question in 0625/32 Feb/March 2019

1 (a) A student has a metal object. (i) The student measures the mass of the object. State the name of the equipment used to measure the mass. … [1] (ii) The mass of the metal object is 1260 g. The volume of the metal is 150 cm3. Calculate the density of the metal. Include the unit. density = … [4] (iii) The mass of the metal object is given in grams. State the mass in kg. mass = … kg [1] (b) A vase is placed on a table. Forces X and Y act on the vase, as shown in Fig. 1.1. X vase Y Fig. 1.1 The mass of the vase is 0.25 kg. The vase is not moving. Calculate the value of force X and the value of force Y. X … Y … [4] [Total: 10]

10 marks

Mark scheme: 1(a)(i) balance B1 1(a)(ii) density = mass ÷ volume in any form C1 1260 ÷ 150 C1 8.4 A1 g / cm3 B1 1(a)(iii) 1.26 (kg) B1 1(b) W = mg in any form C1 0.25 × 10 C1 2.5 (N) A1 Both lines have 2.5 (N) B1

This question in 0625/31 May/June 2019

2 Fig. 2.1 shows a man pushing down on a lever to lift one end of a heavy log. lever heavy log Fig. 2.1 (a) State the term used to describe the turning force exerted by the man. … [1] (b) (i) Fig. 2.2 shows the forces acting as the man starts to lift the heavy log. 1.2 m 0.3 m lever F 400 N pivot Fig. 2.2 Calculate the force F, exerted by the lever on the heavy log. force F = … N [3] (ii) Describe how the man can use a smaller force to lift the heavy log. … … [1] [Total: 5]

5 marks

Mark scheme: 2(a) moment B1 2(b)(i) (sum of) clockwise moment(s) = (sum of) anticlockwise moment(s) C1 1.2 × 400 = 0.3 × F C1 1600 (N) A1 2(b)(ii) use a longer lever OR pivot closer to log / force F B1

This question in 0625/31 May/June 2019

3 Fig. 3.1 shows a wheelbarrow and Fig. 3.2 shows the dimensions of its wheel. load 35 cm = diameter of wheel 1.50 m pivot 25 mm = diameter of axle Fig. 3.1 Fig. 3.2 (a) Complete the table to show the diameter of the wheel and axle in metres. measurement measurement in metres diameter of wheel 35 cm diameter of axle 25 mm [2] (b) The mass of the wheelbarrow is 20 kg. The mass of the load in the wheelbarrow is 30 kg. Calculate the total weight of the wheelbarrow and its load. weight of wheelbarrow and load = … N [3] (c) A man lifts the handle of the wheelbarrow. He applies a force of 140 N, as shown in Fig. 3.3. wheelbarrow 140 N handle 1.30 m pivot Fig. 3.3 Calculate the moment of the force about the pivot. Include the unit. moment = … [4] [Total: 9]

9 marks

Mark scheme: 3(a) 0.35 (m) B1 0.025 (m) B1 3(b) (weight =) mass × gravity in any form C1 50 × 10 OR (20 × 10) + (30 × 10) C1 500 (N) A1 3(c) moment = force × distance from pivot C1 140 × 1.3 C1 180 A1 Nm B1

This question in 0625/33 May/June 2019

3 Fig. 3.1 shows a spring with no load attached. Fig. 3.2 shows the same spring with a load attached. stand spring load Fig. 3.1 Fig. 3.2 (a) Describe how a student can determine the extension of the spring. You may draw on Fig. 3.1 and Fig. 3.2 as part of your answer. … … … … … [3] (b) The student plots a graph of load against extension, as shown in Fig. 3.3. 10.0 load / N 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 0 4 8 12 16 20 24 28 32 36 40 extension / cm Fig. 3.3 (i) Determine the extension produced by a load of 7.5 N. extension = … cm [1] (ii) Determine the load that would produce an extension of 10.0 cm. load = … N [1] (c) Calculate the mass that has a weight of 6.0 N. mass = … kg [3] [Total: 8]

8 marks

Mark scheme: 3(a) measure without any load / weights AND measure with load / weights B1 measure length OR ruler stated or seen B1 (extension =) difference in two values B1 3(b)(i) 30 (cm) B1 3(b)(ii) 2.5 (N) B1 3(c) W = m × g OR W = m × 10 OR (m =) W ÷ g in any form C1 6.0 ÷ 10 C1 0.6(0) (kg) A1

This question in 0625/31 Oct/Nov 2019

4 Fig. 4.1 shows a tractor fitted with a device for breaking up soil in a field. tractor device heavy weight pivot point soil Fig. 4.1 (a) (i) The tractor has a heavy weight at the front. Explain why the heavy weight is needed. … … [1] (ii) Fig. 4.2 represents the weight of the device and its distance from the pivot. pivot 2.1 m 6000 N Fig. 4.2 Calculate the moment of the weight of the device about the pivot. State the unit. moment = … [4] (b) Fig. 4.3 shows a tractor fitted with narrow tyres and the same tractor fitted with wide tyres. narrow tyre wide tyre tractor fitted with same tractor fitted with narrow tyres wide tyres Fig. 4.3 (view from the front) Explain why wide tyres are more suitable for the tractor on soft soil. … … … … [3] [Total: 8]

8 marks

Mark scheme: 4(a)(i) stop the tractor tipping up/keep tractor level owtte B1 4(a)(ii) moment = force × (perp.) distance from pivot in any form C1 6000 × 2.1 C1 12 600 A1 Nm B1 4(b) Any three from: (wide tyres have) greater area (in contact with ground) pressure = force ÷ area in any form the bigger the area the smaller the pressure so tractor less likely to sink/become stuck (in soft ground) B3

This question in 0625/31 Oct/Nov 2019

3 (a) Fig. 3.1 shows the horizontal forces acting on a swimmer. 120 N 110 N Fig. 3.1 (i) Calculate the size and direction of the resultant horizontal force on the swimmer. size of resultant horizontal force = … N direction of resultant horizontal force = … [1] (ii) State the name of the 110 N force on the swimmer. … [1] (iii) Fig. 3.2 shows the horizontal forces acting on the swimmer as he moves forwards a short time later. 120 N 120 N Fig. 3.2 Describe and explain the motion of the swimmer. … … [2] (b) Another swimmer weighs 700 N. He stands on a diving board, as shown in Fig. 3.3. P 3.5 m diving board 700 N Fig. 3.3 Calculate the moment of the swimmer’s weight about point P. moment = … N m [3] [Total: 7]

7 marks

Mark scheme: 3(a)(i) 10 (N) AND forwards/to the right B1 3(a)(ii) friction (between swimmer and water) B1 3(a)(iii) (now) moving at steady/constant speed B1 forces (now)balanced / in equilibrium OR forward force = backward force OR no resultant force B1 3(b) moment = force × (perp.) distance (from pivot) C1 700 × 3.5 C1 2450 (N m) A1

This question in 0625/32 Oct/Nov 2019

4 (a) Fig. 4.1 shows a metal triangle suspended from a thread. thread metal triangle Fig. 4.1 Complete the sentence. Choose the correct word or phrase from the box. above below to the left of to the right of The metal triangle will come to rest with its centre of mass directly … the point of suspension. [1] (b) A student finds the centre of mass of a shape made of thin card. Fig. 4.2 shows the equipment. clamp stand and clamp plumbline nail or pin shape made of thin card Fig. 4.2 (NOT to scale) Describe how the student finds the centre of mass of the card. Choose from these sentences. A A line is drawn on the card showing the position of the string. B A pin held in a clamp is put through the hole in the card. C The centre of mass is where the lines cross on the card. D The process is repeated using holes near the other two edges. Complete the flow chart. Write the letter for the correct sentence in each box. A small hole is made near one edge of the card The plumbline is attached to the pin [3] [Total: 4]

4 marks

Mark scheme: 4(a) below B1 4(b) B A D C B3

This question in 0625/32 Oct/Nov 2019

3 A model aircraft is flying through air. Fig. 3.1 shows the forces acting on the model aircraft. The weight of the model aircraft is 15.0 N. 15.0 N direction of motion 11.0 N 19.0 N 15.0 N Fig. 3.1 (a) (i) Determine the size and direction of the resultant horizontal force acting on the model aircraft. size of resultant horizontal force = … N direction of resultant horizontal force = … [1] (ii) Describe the change in the motion of the model aircraft. … … [2] (b) The horizontal forces acting on the model aircraft become balanced. Suggest how the horizontal forces acting on the model aircraft have changed. … … [1] [Total: 4]

4 marks

Mark scheme: 3(a)(i) 8.0 (N) AND forwards / to the left B1 3(a)(ii) accelerating forwards / to the left B1 B1 3(b) greater drag force / air resistance OR lower thrust owtte B1

This question in 0625/33 Oct/Nov 2019

2 A 50 cm rule is balanced at its mid-point. A force of 8.0 N acts at a distance of 10 cm from one end of the rule. Fig. 2.1 shows the arrangement. 10 cm 25 cm 50 cm rule 8.0 N pivot Fig. 2.1 (a) Calculate the moment of the 8.0 N force about the pivot. Give the unit. moment = … unit = … [5] (b) Another force acts at a point 10 cm from the pivot. It makes the rule balance. On Fig. 2.1, draw an arrow to show the position and direction of this force. [2] [Total: 7]

7 marks

Mark scheme: 2(a) moment = force × distance from pivot in any form C1 (distance of force from pivot = (25 – 10) =) 15 (cm) C1 8 × 15 C1 120 A1 N cm B1 2(b) arrow giving clockwise moment B1 arrow drawn 10 cm from pivot by eye B1

This question in 0625/32 Feb/March 2020

2 (a) A student stretches a spring by adding different loads to it. She measures the length of the spring for each load. She plots a graph of the results. Fig. 2.1 shows the graph of her results. 16.0 length / cm 12.0 8.0 4.0 0 0 1.0 2.0 3.0 4.0 5.0 6.0 load / N Fig. 2.1 Use the graph to determine: (i) the length of the spring without a load length = … cm [1] (ii) the length of the spring with a load of 4.0 N length = … cm [1] (iii) the extension due to a 4.0 N load. extension = … cm [1] (b) Complete the sentence about effects of forces. Choose words from the box. colour friction pressure shape size speed Stretching a spring with a load is an example of how a force can change the … and the … of an object. [2] [Total: 5]

5 marks

Mark scheme: 2(a)(i) 6.0 (cm) B1 2(a)(ii) 13.0 (cm) B1 2(a)(iii) (ii) – (i) B1 2(b) shape B1 size B1

This question in 0625/31 May/June 2020

5 A metre rule is balanced on a pivot by three vertical forces, as shown in Fig. 5.1. 100 cm 40 cm 10 cm 5.0 N pivot weight F = 2.0 N of rule Fig. 5.1 (not to scale) (a) Show that the moment of the 5.0 N force about the pivot is 200 N cm. [2] (b) Calculate the size of force F. F = … N [4] [Total: 6]

6 marks

Mark scheme: 5(a) (moment =) force × distance (from pivot) B1 (moment =) 5.0 × 40 B1 5(b) (sum of) clockwise moments = (sum of) anticlockwise moments C1 200 = (2.0 × 10) + (F × 60) C1 F = (200 – 20) ÷ 60 OR 180 ÷ 60 C1 (F =) 3.0 (N) A1

This question in 0625/31 May/June 2020

3 Fig. 3.1 shows three horizontal forces acting on a car as it moves along a straight road. The horizontal forces act along the same straight line. 250 N 900 N 300 N Fig. 3.1 (a) (i) Calculate the size of the resultant horizontal force on the car and state its direction. size of resultant force = … N direction of resultant force … [3] (ii) The driver presses the brake pedal and the car slows down. As the car slows down, the kinetic energy of the car decreases by 100 kJ. Describe and explain what happens to this 100 kJ of energy. … … [2] (b) Fig. 3.2 shows the force applied to the brake pedal by the driver’s foot. pivot 20 cm 35 N brake pedal Fig. 3.2 Calculate the moment of the force about the pivot. Include the unit. moment = … unit … [4] [Total: 9]

9 marks

Mark scheme: 3(a)(i) 900 – (300 + 250) C1 350 (N) A1 (direction of resultant force =) forwards B1 3(a)(ii) any two from: • friction (in the brakes) • (transfers 100 kJ OR kinetic energy) into thermal energy (store) OR internal energy (store) • of brakes / car / surroundings OR is dissipated OR (transferred) into surroundings / environment B2 3(b) (moment =) force × (perpendicular) distance (from pivot) C1 (moment =) 35 × 20 C1 (moment =) 700 A1 Ncm B1

This question in 0625/31 Oct/Nov 2020

1 Fig. 1.1 shows a box attached to a parachute. The box and the parachute are falling through the air. 16.0 N parachute 2.5 N box 20.0 N Fig. 1.1 (a) Fig. 1.1 shows three vertical forces acting on the box and the parachute. (i) Calculate the resultant vertical force and state its direction. resultant vertical force = … N direction … [3] (ii) Suggest and explain what happens to the size of the upward vertical force on the parachute if the area of the parachute used is increased. suggestion … explanation … … [2] (b) Fig. 1.2 shows the speed–time graph for the box before the parachute is opened. 45 40 35 speed m / s 30 25 20 15 10 5 0 0 10 20 30 40 time / s Fig. 1.2 (i) Determine the time when the speed of the box is 30 m / s. time = … s [1] (ii) Deduce the size of the resultant vertical force on the box when the time is 35 s. Explain your answer. size of resultant vertical force … explanation … … [2] (iii) Calculate the distance the box moves between time = 30 s and time = 40 s. distance = … m [3] [Total: 11]

11 marks

Mark scheme: 1(a)(i) 20.0 – (2.5 + 16.0) C1 1.5 (N) A1 (vertically) down B1 1(a)(ii) (upwards force) increases B1 increases air resistance B1 1(b)(i) 6.5 (s) B1 (b)(ii) (resultant force is) zero B1 (because the) speed (of parachute) is constant / steady / uniform B1 1(b)(iii) (dist. travelled =) area under line (of speed-time graph) C1 45 x 10 C1 450 (m) A1

This question in 0625/32 Oct/Nov 2020

2 A car driver needs to remove one of the wheels on his car. He puts a spanner on a wheel nut. wheel 50 cm 200 N wheel nut spanner tyre Fig. 2.1 (a) The driver applies a force of 200 N, as shown in Fig. 2.1. Calculate the moment of the 200 N force about the centre of the wheel nut. moment of force = … N cm [3] (b) The moment in (a) does not release the wheel nut. The driver cannot increase the force but can increase its moment. State and explain how the driver can increase the moment of the force. statement … explanation … … [2] (c) The driver releases a second wheel nut in a shorter time than the first wheel nut. The driver uses the same amount of energy in releasing both wheel nuts. less than the same as greater than Complete the sentences using the phrases in the box. Each phrase may be used once, more than once or not at all. The work done in releasing the second wheel nut is … the work done in releasing the first wheel nut. The power produced in releasing the second wheel nut is … the power produced in releasing the first wheel nut. [2] [Total: 7]

7 marks

Mark scheme: 2(a) (Moment) = F × d C1 200 × 50 C1 10 000 (Ncm) A1 2(b) use a longer spanner / move force to end of spanner owtte B1 (to) increase the distance (from force to wheel nut or pivot) OR distance (from force to wheel nut or pivot) is greater than 50 cm B1 2(c) (work done is the) same (as) B1 (power produced is) greater (than) B1

This question in 0625/32 Oct/Nov 2020

1 (a) Fig. 1.1 shows a lorry moving on a straight road. The arrows represent the horizontal forces acting on the lorry. These forces act along the same straight line. 500 N 1000 N 1500 N Fig. 1.1 (i) Calculate the size of the resultant horizontal force on the lorry. size of resultant force = … N [2] (ii) Describe the effect of a horizontal resultant force of zero on the speed of the lorry. Put a tick (3) in one box. speed increases to a higher constant speed speed stays the same speed decreases to a lower constant speed speed decreases to zero [1] (b) The speed of the motorcycle in Fig. 1.2 is 20 m / s. Fig. 1.2 The rider reacts to a sudden change in the traffic ahead. He stops as quickly as possible by applying the brakes. The total stopping distance is made up of the distance travelled while the rider is reacting and the distance travelled when the brakes are applied. Fig. 1.3 shows information about stopping when the speed of the motorcycle is 20 m / s. 15 m 38 m distance travelled distance travelled while braking while reacting Fig. 1.3 (i) Calculate the total stopping distance when the speed of the motorcycle is 20 m / s. total stopping distance = … m [1] (ii) Suggest one factor that could increase the total stopping distance. … [1] [Total: 5]

5 marks

Mark scheme: 1(a)(i) 1000 + 500 OR 1500 OR 1500 – 1500 OR 1500 – their ‘1500’ OR 1500 – 1000 OR 1500 – 500 C1 0 / zero A1 1(a)(ii) 2nd box (speed stays the same) B1 1(b)(i) (15 + 38) = 53 (m) B1 1(b)(ii) reduced friction / wet / icy (conditions) / worn tyres tiredness / drugs / alcohol / higher speed / going down hill B1

This question in 0625/33 Oct/Nov 2020

2 (a) Fig. 2.1 shows two children sitting on a see-saw. child A child B X 0.5 m beam 125 N pivot 250 N Fig. 2.1 (not to scale) (i) The weight of child A is 125 N. Calculate the mass of child A. Include the unit in your answer. mass of child A = … unit … [3] (ii) Fig. 2.1 shows child A and child B sitting in positions which balance the see-saw horizontally. Using the information in Fig. 2.1, determine the distance X. distance X = … m [3] (b) The person in Fig. 2.2 is pushing a child on a swing. Fig. 2.2 State the name of the force that acts against the motion of the swing. … [1] [Total: 7]

7 marks

Mark scheme: 2(a)(i) W = mg OR W = 10 × m C1 12.5 A1 kg B1 2(a)(ii) 250 × 0.5 OR X × 125 C1 clockwise moment = anticlockwise moment C1 (X=) 1.0 (m) A1 2(b) air resistance / drag B1

This question in 0625/33 Oct/Nov 2020

1 Fig. 1.1 shows a box dropped from an aeroplane. The box contains supplies. A parachute is attached to the box. The parachute is opened when the time is 6.0 s. parachute box containing supplies Fig. 1.1 The graph in Fig. 1.2 shows the vertical speed of the box as it falls. 50 40 30 speed m / s 20 10 0 0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 time / s Fig. 1.2 (a) State and explain what happens to the kinetic energy of the box during the first 6.0 s of its descent. … … … [2] (b) State and explain what happens to the gravitational potential energy of the box during the first 6.0 s. … … … [2] (c) (i) Use the graph in Fig. 1.2 to determine the speed of the object when the object is moving with a constant speed. speed of the object at constant speed = … m / s [2] (ii) State the size of the resultant vertical force on the box when it is falling at a constant speed. … [1] (d) Use the graph in Fig. 1.2 to determine the distance travelled by the box during the first 6.0 s. distance travelled in first 6.0 s = … m [3] (e) Without calculation, describe how Fig. 1.2 shows that the deceleration of the box is greater than the acceleration of the box. … … [1] [Total: 11]

11 marks

Mark scheme: 1(a) (kinetic energy / it) increases B1 (because) speed / velocity (of box) increases OR faster B1 1(b) (gravitational potential energy) decreases M1 (because) height (of box) decreases A1 1(c)(i) any indication on graph / in text that horizontal section represents steady speed C1 10 (m / s) A1 1(c)(ii) (resultant vertical force =) zero OR 0 (N) B1 1(d) distance = area under graph OR ½ × b × h C1 (distance =) ½ × 6.0 × 45 C1 135 (m) A1 1(e) deceleration (line) is steeper OR higher gradient than acceleration (line) B1

This question in 0625/32 Feb/March 2021

3 Fig. 3.1 shows the forces acting on a uniform balanced beam. The beam is pivoted at its centre. P 2.0 cm pivot 6.0 cm 2.0 cm 5.2 N 8.1 N Fig. 3.1 (a) Calculate the moment of the 5.2 N force about the pivot and show that its value is close to 30 Ncm. [3] (b) The beam is balanced. Calculate force P. force P = … N [4] [Total: 7]

7 marks

Mark scheme: 3(a) (moment of force =) force × (perpendicular) distance of force from pivot B1 5.2 × 6.0 B1 31.2 B1 3(b) (sum of) clockwise moment(s) = (sum of) anticlockwise moment(s) C1 P × 2.0 + 8.1 × 2.0 = 5.2 × 6.0 OR 31.2 OR answer from (a) C1 P = (31.2 – 16.2) ÷ 2.0 OR 15 ÷ 2.0 C1 7.5 (N) A1

This question in 0625/32 Feb/March 2021

3 A plank balances horizontally on a log of wood, which acts as a pivot. (a) A girl sits on one end of the plank, and her brother pushes down on the other end to make the plank balance horizontally. Fig. 3.1 shows the arrangement. pivot 1.2 m 1.6 m weight = 404 N F Fig. 3.1 (not to scale) Calculate the moment of the girl’s weight about the pivot and show that it is close to 480 N m. [3] (b) The plank balances horizontally when the boy pushes down with a force F at a distance of 1.6 m from the pivot. Calculate the size of force F. force F = … N [3] [Total: 6]

6 marks

Mark scheme: 3(a) moment = force × distance (of direction of force from pivot) C1 404 × 1.2 C1 484.8 (N m) (which is about 480 N m) A1 3(b) c.w. moment = a.c.w moment OR moment of weight = moment of force/F C1 404 × 1.2 = F × 1.6 OR (F =) 484.8 ÷ 1.6 C1 (F = ) 300 (N) A1

This question in 0625/31 May/June 2021

5 (a) A man starts pulling his suitcase across the floor. suitcase 12 N 20 N Fig. 5.1 (not to scale) (i) Fig. 5.1 shows the horizontal forces acting on the suitcase. Calculate the resultant horizontal force on the suitcase. size of force = … N direction … [2] (ii) After a short time, the suitcase is moving at a constant speed. Suggest values for the sizes of the two horizontal forces on the suitcase when it is moving at a constant speed. pulling force = … (N) friction force = … (N) [1] (b) The total downward force of the suitcase on the ground is 150 N. The suitcase has two wheels. Each wheel has an area of 0.60 cm2 touching the ground. Calculate the pressure of the suitcase on the ground. pressure on the ground = … N / cm2 [4] [Total: 7]

7 marks

Mark scheme: 5(a)(i) 8 (N) B1 forwards B1 5(a)(ii) same non-zero values for pulling and friction force B1 5(b) (area = 2 × 0.60) = 1.2 (cm2) B1 (P =) F ÷ A C1 150 ÷ 1.2 OR 150 ÷ 0.60 C1 125 (N / cm2) A1

This question in 0625/31 May/June 2021

3 Fig. 3.1 shows a barrier used at a car park. The beam can be raised and lowered by a man rotating it about its pivot. heavy weight pivot beam 1.8 m 150 N W Fig. 3.1 (not to scale) (a) The weight of the beam is 150 N. This acts at a distance of 1.8 m from the pivot as shown in Fig. 3.1. Calculate the moment of the weight of the beam about the pivot. Include the correct unit in your answer. moment of weight of beam = … unit … [4] (b) When the weight W of the heavy weight acts at a distance of 0.6 m from the pivot, the barrier is horizontal and balanced as shown in Fig. 3.1. The man raises the barrier and the heavy weight slips to a distance of 0.8 m from the pivot. This causes a problem for the man trying to lower the barrier. Describe and explain the problem this causes for the man lowering the barrier. … … … … … [3] [Total: 7]

7 marks

Mark scheme: 3(a) (moment of weight =) weight × (perpendicular) distance (of weight from pivot) (moment of weight =) 150 × 1.8 C1 270 A1 N m B1 3(b) barrier no longer balanced OR cannot be lowered (easily) B1 (more) force needed to lower barrier B1 (because) moment of heavy weight (has) increased B1

This question in 0625/32 May/June 2021

3 (a) A girl and her brother sit on opposite sides of a see-saw as shown in Fig. 3.1. girl brother 1.9 m 1.2 m 240 N pivot W Fig. 3.1 (i) Calculate the girl’s moment about the pivot and show that it is close to 460 N m. [3] (ii) The see-saw is balanced horizontally. Calculate the weight W of the brother. W = … N [3] (b) The weight of the girl in Fig. 3.1 is 240 N. Calculate the mass of the girl. Include the unit in your answer. mass of girl = … unit … [4] [Total: 10]

10 marks

Mark scheme: 3(a)(i) force × (perp) distance C1 240 × 1.9 C1 456 (Nm) (which is close to 460 Nm) A1 3(a)(ii) c.w. moment = a.c.w moment OR girl’s moment = brother’s moment C1 240 × 1.9 = W × 1.2 OR 456 ÷ 1.2 OR 460 ÷ 1.2 C1 380 (N) A1 3(b) W = mg in any form OR (m =) W ÷ g C1 240 ÷ 10 C1 24 A1 kg B1

This question in 0625/33 May/June 2021

3 (a) A student determines the centre of mass of a piece of card. Fig. 3.1 shows the equipment the student uses. pin cork A stand card ruler B C cotton thread weight Fig. 3.1 Describe how the student determines the centre of mass of the card using the equipment in Fig. 3.1. … … … … … … [3] (b) Another card is pivoted at point P. The weight of the card is 1.4 N and acts through a point 20 cm from P. Fig. 3.2 shows the arrangement. card P 20 cm pivot 1.4 N Fig. 3.2 Calculate the moment of the weight of the card about point P. moment of weight = … N cm [3] [Total: 6]

6 marks

Mark scheme: 3(a) any three from: line drawn alongside cotton thread / string hang triangle from a different corner (B or C) repeat marking of string (position on the card) centre of mass is where lines intersect B3 3(b) (moment of weight =) weight × distance (of direction of force from pivot) C1 (moment =) 1.4 × 20 C1 28 (N cm) A1

This question in 0625/31 Oct/Nov 2021

3 Fig. 3.1 shows the vertical forces acting on a toy rocket as it leaves the ground. upward force = 70 N smooth surface weight of rocket = 15 N Fig. 3.1 (a) Calculate the size of the resultant vertical force on the rocket. resultant force = … N [2] (b) Explain why the top of the rocket is pointed and has a smooth surface. … … [2] [Total: 4]

4 marks

Mark scheme: 3(a) 70 – 15 C1 55 (N) A1 3(b) streamline / friction / drag / air resistance M1 reduce (owtte) friction / drag / air resistance A1

This question in 0625/32 Oct/Nov 2021

3 Fig. 3.1 shows the horizontal forces acting on a skateboarder. backward force = 60 N forward force = 100 N skateboarder skateboard Fig. 3.1 (a) Calculate the resultant force acting on the skateboarder. resultant force = … N direction = … [2] (b) Describe the effect of the resultant force in (a) on the motion of the skateboarder. … [1] (c) The skateboarder is moving along a horizontal path. The backward force is 100 N. The forward force is 100 N. Describe the motion of the skateboarder. … [1] [Total: 4]

4 marks

Mark scheme: 3(a) 40 (N) B1 forward / to the right B1 3(b) accelerates / speed increases B1 3(c) constant / uniform speed / zero acceleration B1

This question in 0625/33 Oct/Nov 2021

2 (a) A student is doing some physical exercise. Fig. 2.1 shows the student holding a 50 N weight. 0.90 m pivot 50 N Fig. 2.1 The pivot in the shoulder is 0.90 m from the centre of mass of the weight. Calculate the moment of the weight about this pivot. moment of the weight = … N m [3] (b) The student does some running exercises. Fig. 2.2 shows the speed–time graph for one exercise. B CC 8.0 speed m / s 6.0 4.0 2.0 AA D E 0 D 0 5.0 10.0 15.0 20.0 time / s Fig. 2.2 (i) Describe the motion of the athlete in sections AB and DE. section AB … section DE … [2] (ii) Calculate the distance moved by the athlete from time = 0 to time = 5.0 s. distance = … m [3] [Total: 8]

8 marks

Mark scheme: 2(a) 45 (Nm) A3 (moment of force =) force × (perpendicular) distance (of force from pivot) (C1) 50 × 0.9 (C2) 2(b)(i) (section AB) increasing speed OR acceleration B1 (section DE) stationary OR stopped OR at rest B1 2(b)(ii) 20 (m) A3 (distance =) ½ × 8(.0) × 5(.0) (C2) distance travelled = area under graph OR (d = ) speed × time (C1)

This question in 0625/32 Feb/March 2022

5 A woman starts to push a trolley across the floor. Fig. 5.1 shows the horizontal forces acting on the trolley. trolley 120 N 90 N Fig. 5.1 (a) Determine the resultant horizontal force on the trolley. resultant force = … N direction of resultant force = … [3] (b) The total weight of the trolley and boxes is 900 N. The area of each wheel in contact with the ground is 8.0 cm2. The trolley has four wheels. Calculate the pressure on the ground due to the total weight of the trolley and boxes. Include the correct unit in your answer. pressure on the ground = … unit … [5] [Total: 8]

8 marks

Mark scheme: 5(a) (resultant force =) 30 (N) A2 (resultant force =) 120 – 90 (A1) to the left OR forwards B1 5(b) (pressure =) 28 A4 (pressure =) 900 ÷ 32 (C3) (total area =) (4 × 8 =) 32 (cm2) (B1) (pressure =) force ÷ area (C1) N / cm2 B1

This question in 0625/32 Feb/March 2022

4 Fig. 4.1 shows an electric motor and pulley wheel being used to raise a load M. The electric motor uses a belt to turn the pulley wheel. pulley pivot wheel belt electric motor load M Fig. 4.1 (a) When the electric motor lifts the load, it transfers energy. Fig. 4.2 shows the energy transfers. Write on Fig. 4.2 to complete the label in each box. The first label is done for you. useful energy electrical transfers … … …………… + … …………… energy energy energy wasted energy … …………… energy Fig. 4.2 [3] (b) Fig. 4.3 shows the force on the pulley from the load M. pulley pivot wheel 20 cm 2.5 N Fig. 4.3 The weight of load M is 2.5 N and the weight acts at a distance of 20 cm from the pivot of the pulley wheel. Calculate the moment of the weight of load M about the pivot. moment = … N cm [3] [Total: 6]

6 marks

Mark scheme: 4(a) (useful energy transfers:) kinetic (energy) B1 in either order gravitational potential (energy) B1 (wasted energy transfer:) thermal (energy) B1 4(b) 50 (N cm) A3 2.5  20 (C2) (moment of force =) force  (perpendicular) distance (of force from pivot) (C1)

This question in 0625/31 May/June 2022

3 (a) Fig. 3.1 shows an aeroplane flying. There are horizontal forces acting on the aeroplane, as shown in Fig. 3.1. 12 000 N 8000 N Fig. 3.1 (not to scale) (i) Calculate the resultant horizontal force on the aeroplane. resultant force = … N direction of resultant force … [3] (ii) State the name of the effect producing the 8000 N force on the aeroplane. … [1] (iii) At a later time in the flight, the resultant horizontal force on the aeroplane is zero. Describe the horizontal motion of the aeroplane. … [1] (b) Fig. 3.2 shows the handle used to open and close a cupboard door on the aeroplane. 60 N pivot 20 cm Fig. 3.2 (not to scale) A force of 60 N acts at a distance of 20 cm from the pivot of the handle. Calculate the moment of the 60 N force about the pivot. moment = … N cm [3] [Total: 8]

8 marks

Mark scheme: 3(a)(i) 4000 (N) A2 (resultant force =) force to R – force to L OR 12 000 – 8000 (C1) (to the) left or forwards B1 3(a)(ii) air resistance B1 3(a)(iii) constant/steady speed B1 3(b) 1200 (N cm) A3 (moment of force =) 60  20 (C2) (moment of force =) force  (perpendicular) distance of force from pivot (C1)

This question in 0625/32 May/June 2022

2 Fig. 2.1 shows the horizontal forces acting on a car. 900 N 1200 N Fig. 2.1 (not to scale) (a) Calculate the resultant horizontal force on the car. size of force = … N direction … [3] (b) A student uses a digital stop-watch to measure the time for the car to travel 100 m. Fig. 2.2 shows the time reading on the stop-watch. 1 s min s 100 0 : 07 20 Fig. 2.2 (i) Using the information in Fig. 2.2, state the time taken to travel 100 m. time to travel 100 m = … s [1] (ii) The car takes 12.8 s to travel the next 200 m. Calculate the average speed of the car for this 200 m. average speed = … m / s [3] (c) Fig. 2.3 shows the speed–time graph for another car. 20.0 speed 18.0 m / s 16.0 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 0.0 2.0 4.0 6.0 time / s Fig. 2.3 Calculate the distance travelled by this car between time = 2.0 s and time = 6.0 s. distance travelled = … m [3] [Total: 10]

10 marks

Mark scheme: 2(a) 300 (N) A2 (resultant force =) force to right – force to left OR 1200 – 900 C1 to the right OR in forward direction B1 2(b)(i) 7.20 (s) B1 2(b)(ii) 16 (m / s) A3 200 / 12.8 C2 (average speed =) (total) distance / (total) time in any form C1 2(c) 48 (m) A3 1 1 C2 (6 + 18)  4.0 OR 6  4 +  12  4 2 2 distance = area under graph C1 1 OR area = (sum of parallel sides)  base 2

This question in 0625/31 Oct/Nov 2022

3 A sailor uses a winch to raise a sail on a boat. Fig. 3.1 shows the sailor turning the winch. sail winchwinch Fig. 3.1 (a) The sailor applies a force of 200 N at a distance of 30 cm from the pivot in the winch, as shown in Fig. 3.2. 200 N winch pivot 30 cm Fig. 3.2 Calculate the moment of this force about the pivot. moment of force = … N cm [3] (b) (i) Describe two useful energy transfers when the sailor uses the winch to raise the sail. 1 … 2 … [2] (ii) Describe one non-useful energy transfer when the sailor uses the winch to raise the sail. … [1] [Total: 6]

6 marks

Mark scheme: 3(a) 6000 (N cm) A3 (moment of force =) 200  30 C2 (moment of force =) force  (perpendicular) distance (of force from pivot) C1 3(b)(i) any two from: B2 • chemical energy to (gravitational) potential energy (of sail) • chemical energy to kinetic energy • kinetic energy (of winch) to kinetic energy (of rope / sail) • kinetic energy (of rope / sail) to (gravitational) potential energy (of sail). 3(b)(ii) chemical energy OR kinetic energy to thermal OR sound (energy) B1

This question in 0625/31 Oct/Nov 2022

1 A skydiver jumps from an aeroplane. She falls freely with her parachute closed; then she opens her parachute. Fig. 1.1 shows the skydiver falling freely with her parachute closed. Fig. 1.2 shows the skydiver falling with the parachute open. Fig. 1.1 Fig. 1.2 Fig. 1.3 shows the speed–time graph for the skydiver’s vertical motion, from leaving the aeroplane to landing on the ground. 60 vertical speed m / s BB CC 50 40 30 20 10 DD EE A 0 0 10 20 30 40 50 60 70 80 time / s Fig. 1.3 (a) Using the information from Fig. 1.3: (i) Describe the vertical motion of the skydiver between time = 0 and time = 20 s. … [1] (ii) Determine the maximum vertical speed of the skydiver. maximum speed = … m / s [1] (iii) Determine the point, A, B, C, D or E, at which the skydiver opens her parachute. … [1] (iv) Determine the distance the skydiver falls between time = 50 s and time = 80 s. distance = … m [3] (b) The weight of the skydiver is 750 N. The weight of the skydiver acts downwards, as shown in Fig. 1.4. While the skydiver is falling, another force acts upwards. The upward force varies as the skydiver falls. … weight = 750 N Fig. 1.4 (not to scale) (i) On Fig. 1.4, write the name of the upward force on the dotted line above the upward force. [1] (ii) Suggest a value for the upward force on the skydiver at time = 10 s. … N [1] (iii) Determine the value of the upward force on the skydiver at time = 30 s. … N [1] (c) The weight of the skydiver is 750 N. Calculate the mass of the skydiver. mass = … kg [3] [Total: 12]

12 marks

Mark scheme: Question Answer Marks 1(a)(i) accelerating / increasing speed B1 1(a)(ii) 50 (m / s) B1 1(a)(iii) C B1 1(a)(iv) 150 (m) A3 5  30 (C2) (distance =) area under graph (C1) 1(b)(i) friction / air resistance / drag B1 1(b)(ii) number greater than 0 AND smaller than 750 (N) B1 1(b)(iii) 750 (N) B1 1(c) 75 (kg) A3 750 ÷ 10 (C2) W = mg OR (m =) W ÷ g OR W ÷ 10 in any form (C1)

This question in 0625/32 Oct/Nov 2022

2 (a) A student has a spring of length 14.0 cm. She stretches the spring by adding different loads to the spring. She measures the length of the spring for each load. She plots a graph of the results. Fig. 2.1 shows the graph of her results. 24.0 22.0 20.0 18.0 16.0 14.0 length of spring / cm 12.0 10.0 8.0 6.0 4.0 2.0 0 0 2.0 4.0 6.0 8.0 10.0 12.0 load / N Fig. 2.1 (i) Use the graph to determine the length of the spring when the student adds a load of 8.0 N to the spring. length of spring = … cm [1] (ii) Use the graph to determine the load added to the spring when the extension of the spring is 7.0 cm. load for an extension of 7.0 cm = … N [2] (b) Complete the sentence about effects of forces. Choose a word from the box. charge mass power shape velocity A load stretching a spring is an example of a force changing the size and the … of an object. [1] (c) A clamp stand used in the experiment has a weight of 8.6 N. Calculate the mass of the clamp stand. mass of clamp stand = … kg [3] [Total: 7]

7 marks

Mark scheme: 2(a)(i) (length of spring with 8.0 N load =) 20 (cm) B1 2(a)(ii) (load for length of 21 cm =) 9.3 (N) A2 (extension of 7 cm = length of) 21 cm (C1) 2(b) shape B1 2(c) (m = ) 0.88 (kg) A3 (m = ) 8.6 ÷ 9.8 (C2) W = mg OR (m =) W ÷ g (C1)

This question in 0625/32 Feb/March 2023

2 Fig. 2.1 shows a concrete beam resting on the ground. concrete beam 12 cm 160 cm ground Fig. 2.1 (not to scale) (a) The weight of the concrete beam is 1540 N. Calculate the pressure on the ground due to the concrete beam. pressure = … N / cm2 [4] (b) A builder starts to raise one end of the beam. He uses a force of 1030 N at a perpendicular distance of 120 cm from the pivot. Fig. 2.2 shows the arrangement. 1030 N one end of pivot the beam 120 cm ground Fig. 2.2 (not to scale) Calculate the moment of the 1030 N force about the pivot. moment = … N cm [3] (c) Describe how the builder can use a smaller force to lift the beam. … [1] (d) The builder positions the beam as shown in Fig. 2.3. 160 cm ground Fig. 2.3 (not to scale) State why the beam shown in Fig. 2.3 is less stable than the beam shown in Fig. 2.1. … [1] [Total: 9]

9 marks

Mark scheme: 2(a) (pressure =) 0.8(0) (N / cm2) A4 (pressure =) 1540  1920 OR 1540  (160  12) (C3) (pressure =) force  area (C1) (area in contact with ground =) 12  160 = 1920 (cm2) (C1) 2(b) (moment =) 120 000 (N cm) OR 1.2  105 (N cm) A3 (moment =) 1030  120 (C2) (moment =) force  (perpendicular) distance from pivot (C1) 2(c) move (lifting) force further from pivot owtte B1 2(d) centre of gravity / mass is high(er) OR idea that area of base is small(er) B1

This question in 0625/31 May/June 2023

2 Fig. 2.1 shows the speed–time graph for a cyclist. 14 W X 12 speed 10 m / s S T 8 6 4 2 Y Z 0 0 10 20 30 40 50 time / s Fig. 2.1 (a) In Fig. 2.1, the sections ST, TW, WX, XY and YZ indicate stages of the cyclist’s journey. State one section which shows the cyclist moving with: (i) constant speed … [1] (ii) constant deceleration … [1] (iii) constant non-zero acceleration. … [1] (b) Calculate the distance travelled by the cyclist in section ST. distance travelled = … m [3] (c) Fig. 2.2 shows the horizontal forces on a cyclist. 160 N 220 N Fig. 2.2 (i) Calculate the size of the resultant force on the cyclist. resultant force = … N [1] (ii) State the effect, if any, of the resultant force on the motion of the cyclist. … [1] [Total: 8]

8 marks

Mark scheme: 2(a)(i) ST OR WX B1 2(a)(ii) XY B1 2(a)(iii) TW OR XY B1 2(b) (distance travelled =) 100 (m) A3 (distance travelled =) 8  13 (C2) (distance travelled =) area under graph OR b  h (C1) 2(c)(i) 60 (N) B1 2(c)(ii) accelerates OR increases speed B1

This question in 0625/32 May/June 2023

3 A student balances a beam on a pivot. They then balance block A and block B on the beam, as shown in Fig. 3.1. 5.5 cm d block B block A beam pivot 0.14 N 0.19 N Fig. 3.1 (not to scale) (a) (i) The weight of block A is 0.14 N. Show that the moment of block A about the pivot is approximately 0.8 N cm. [3] (ii) The weight of block B is 0.19 N. Calculate the distance d between the pivot and the centre of block B. distance d = … cm [3] (b) The weight of block B is 0.19 N. Calculate the mass of block B. mass of block B = … kg [3] [Total: 9]

9 marks

Mark scheme: 3(a)(i) C1 0.14  5.5 C1 0.77 (N cm) A1 3(a)(ii) (sum of) ACM = (sum of) CM C1 0.19  X = 0.77 OR 0.19  X = 0.8 OR 0.19  X = 0.14  5.5 C1 4.1 (cm) OR 4.2 (cm) A1 3(b) (m =) W ÷ g OR W ÷ 9.8 in any form C1 0.19 ÷ 9.8 C1 0.019 (kg) A1

This question in 0625/33 May/June 2023

2 The mass of a solid metal cylinder is 400 g and its volume is 52 cm3. (a) Calculate the density of the metal. Include the unit. density = … [4] (b) The cylinder is falling at constant speed through the air. Fig. 2.1 shows the vertical forces acting on the cylinder. … … N cylinder 3.9 N weight Fig. 2.1 (not to scale) On Fig. 2.1, write the name and the size of the upward force on the cylinder. [2] (c) The student balances a beam on a pivot. On the beam, he positions the cylinder and a block so that the beam remains balanced. The arrangement is shown in Fig. 2.2. cylinder block pivot 42 cm 25 cm weight of block 3.9 N Fig. 2.2 (not to scale) Calculate the weight of the block. weight of block = … N [4] [Total: 10]

10 marks

Mark scheme: 2(a) 7.7 A3 400 ÷ 52 (C2) (density =) mass ÷ volume OR m / V (C1) g / cm3 B1 2(b) friction OR drag OR (air) resistance B1 3.9 (N) B1 2(c) (weight = 97.5 ÷ 42) = 2.3 (N) A4 W  42 = 3.9  25 {OR 97.5} OR (W =) 3.9  25 / 42 (C3) (moment of cylinder =) 3.9  25 OR 97.5 (C1) clockwise moment = anticlockwise moment OR moment of cylinder = moment of block (C1)

This question in 0625/31 Oct/Nov 2023

3 A platform rests on a pivot as shown in Fig. 3.1. A diver sits at a distance of 1.8 m from the pivot. The weight of the diver is 1100 N. 1.8 m diver pivot spring platform 1100 N water Fig. 3.1 (not to scale) (a) Using the information in Fig. 3.1, calculate the moment of the diver about the pivot. moment of diver = … N m [3] (b) (i) Fig. 3.2 represents the platform without the diver. 1.2 m 0.40 m platform 62 N spring pivot W Fig. 3.2 (not to scale) The moment of the weight W of the platform is balanced by the moment of the spring. The spring exerts a downward force of 62 N. Using the information in Fig. 3.2, calculate the weight W of the platform. W = … N [3] (ii) The graph of load against extension for a spring is shown in Fig. 3.3. 500 load / N 400 300 200 100 0 0 1 2 extension / cm Fig. 3.3 The unstretched length of the spring is 16 cm. Determine the length of the spring when the load on the spring is 240 N. length of spring = … cm [2] [Total: 8]

8 marks

Mark scheme: 3(a) 2000 (N m) A3 1100  1.8 (C2) (moment =) force  (perpendicular) distance (C1) 3(b)(i) 190 (N) A3 (W =) {62  1.2} ÷ 0.4 OR 74.4 ÷ 0.4 (C2) (moment of spring =) 62  1.2 OR 74.4 (C1) 3(b)(ii) (length of spring =) 17 (cm) A2 (extension =) 1.0 (cm) (C1)

This question in 0625/32 Oct/Nov 2023

2 (a) A student determines the volume of a piece of metal. The student pours 30 cm3 of water into a measuring cylinder. The piece of metal is submerged in the water and the new volume reading on the measuring cylinder is 42 cm3. Calculate the volume of the piece of metal. volume = … cm3 [1] (b) The mass of another piece of metal is 320 g. The volume of the piece of metal is 40 cm3. Calculate the density of the metal. Give the correct unit. density = … unit … [4] (c) The student drops the piece of metal into a tank of water. Two vertical forces act on the piece of metal as it falls through the water in the tank. On Fig. 2.1, each arrow represents a vertical force. … piece of metal … Fig. 2.1 (i) Complete the diagram in Fig. 2.1 by labelling the two forces. [2] (ii) The upward force is the same size as the downward force. Describe the motion of the piece of metal as it falls through the water. … [1] [Total: 8]

8 marks

Mark scheme: 2(a) (42 – 30 = ) 12 (cm3) B1 2(b) (=) 8(.0) A3 (=) 320 ÷ 40 (C2) (density =) mass ÷ volume OR (=) m / V in any form (C1) g / cm3 B1 2(c)(i) friction / drag (upward arrow) B1 weight (downward arrow) B1 2(c)(ii) (falling with) {constant / steady / uniform} speed B1

This question in 0625/32 Feb/March 2024

3 Fig. 3.1 shows a computer on the surface of a desk. computer surface of desk Fig. 3.1 (a) The weight of the computer is 48 N. The area of the computer in contact with the surface of the desk is 20 cm2. Calculate the pressure due to the computer on the surface of the desk. pressure = … N / cm2 [3] (b) A student uses a force of 12 N to tilt the computer as shown in Fig. 3.2. 12 N 32 cm pivot Fig. 3.2 Calculate the moment of the 12 N force about the pivot. moment = … N cm [3] [Total: 6]

6 marks

Mark scheme: 3(a) (P =) 2.4 (N / cm2) A3 (P =) 48 ÷ 20 (C2) (P =) F ÷ A (C1) 3(b) (moment = ) 380 (N cm) A3 (moment = ) 12  32 (C2) moment = force  (perp.) distance from pivot (C1)

This question in 0625/32 Feb/March 2024

3 Fig. 3.1 shows the horizontal forces acting on a boat. 80 N (backwards) 200 N (forwards) boat Fig. 3.1 (a) (i) Calculate the resultant horizontal force on the boat in Fig. 3.1. size of resultant force = … N direction of resultant force … [2] (ii) Suggest what causes the 80 N force on the boat in Fig. 3.1. … … [1] (iii) Another boat is travelling and the horizontal forces on this boat are balanced. Describe the horizontal motion of this boat. … [1] (b) Fig. 3.2 shows the wheel used to steer a boat. 60 N 50 cm50 cm pivot Fig. 3.2 A force of 60 N acts at a perpendicular distance of 50 cm from the wheel’s pivot. Calculate the moment of the 60 N force about the pivot. Include the unit. moment = … unit … [4] [Total: 8]

8 marks

Mark scheme: 3(a)(i) (200 – 80 =) 120 (N) B1 forwards OR to the right OR in same direction as 200 (N force) B1 3(a)(ii) friction OR air / water / wind resistance OR drag (from water) B1 3(a)(iii) constant / steady / uniform velocity B1 3(b) 3000 OR 30 A3 60  50 OR 60  0.5(0) (C2) moment = force  distance from pivot (C1) N cm OR N m B1

This question in 0625/31 May/June 2024

3 Fig. 3.1 shows two solid shapes, a cylinder and a cone, which are made from the same material. cylinder cone Fig. 3.1 (a) State and explain which shape is the more stable. the more stable shape is … explanation … … [1] (b) The mass of the cylinder is 0.25 kg. Calculate the weight of the cylinder. weight = … N [2] (c) A horizontal force of 3.0 N tilts the cone. The cone balances on one edge, as shown in Fig. 3.2. 3.0 N 22 cm pivot Fig. 3.2 (i) Calculate the moment of the 3.0 N force about the pivot in Fig. 3.2. moment = … N cm [3] (ii) Determine the moment of the weight of the cone about the pivot. Use ideas about the principle of moments. moment of weight about pivot = … N cm [1] [Total: 7]

7 marks

Mark scheme: 3(a) cone M0 (because it has) lower centre of mass/gravity A1 3(b) (weight =) 2.5 (N) A2 (weight =) mass  g OR 0.25  9.8 (C1) 3(c)(i) (moment =) 66 (N cm) A3 (moment =) 3(.0)  22 (C2) moment = force  (perpendicular) distance (from pivot) (C1) 3(c)(ii) (moment of weight =) answer to (c)(i) OR 66 (N cm) B1

This question in 0625/32 May/June 2024

1 A girl is cycling along a straight horizontal road. Fig. 1.1 shows the directions of the forces acting on the cyclist as she cycles in the direction of force C. force B force A force C force D Fig. 1.1 (a) State which force shows the direction of: (i) the force due to gravity … [1] (ii) the force due to air resistance. … [1] (iii) Force A changes and becomes larger than force C. State any effect this change has on the motion of the cyclist. … [1] (b) Another cyclist travels a distance of 250 m in a time of 21 s. (i) Calculate the average speed of the cyclist. average speed = … m/s [3] (ii) The cyclist exerts a force of 36 N to move the cycle forwards. Calculate the work done by this force when the cyclist travels 250 m. Include the unit. work done = … unit … [4] [Total: 10]

10 marks

Mark scheme: 1(a)(i) D B1 1(a)(ii) A B1 1(a)(iii) decelerating / slowing down / less speed owtte B1 1(b)(i) 12 (m / s) A3 250  21 (C2) (average speed =) (total) distance (travelled)  (total) time(taken) in any form (C1) 1(b)(ii) 9000 A3 36  250 (C2) (work =) force  distance in any form (C1) J B1

This question in 0625/33 May/June 2024

2 Fig. 2.1 shows a road sign on the ground. STOP base ground front view side view Fig. 2.1 (a) A strong wind blows and the sign begins to fall over. A man catches the sign before it falls completely. Fig. 2.2 shows the force applied to the sign by the man. 5.6 N 82 cm pivot Fig. 2.2 Calculate the moment of the 5.6 N force about the pivot. Use the information in Fig. 2.2. moment = … N cm [3] (b) The sign needs to be easy to move and stable. The base cannot be fixed to the ground. Suggest how to change the base so that the sign is more stable. Explain your answer. suggestion … explanation … [2] [Total: 5]

5 marks

Mark scheme: 2(a) 460 (N cm) A3 5.6  82 (C2) (moment =) force  (perpendicular) distance in any form (C1) 2(b) heavier base OR increases area of base B1 lowers centre of mass / gravity B1

This question in 0625/33 May/June 2024

2 A person pushes a pushchair. A young child rides in the pushchair. Fig. 2.1 shows horizontal forces acting on the front wheel of the pushchair. 30 N 10 N Fig. 2.1 (not to scale) (a) Calculate the resultant of the horizontal forces shown in Fig. 2.1. resultant force = … N direction = … [2] (b) (i) Another person pushes a shopping trolley with a force of 40 N. The shopping trolley moves at a constant speed along a horizontal path. Calculate the work done by the 40 N force to move the shopping trolley a distance of 50 m. work done = … J [3] (ii) The work done on the shopping trolley as it starts moving is transferred into other energy stores. State two such energy stores. 1 … 2 … [2] (c) In (a), the weight of the pushchair and child is 240 N. The total area of contact with the ground is 38 cm2. Calculate the pressure on the ground due to the pushchair and child. pressure on ground = … N / cm2 [3] [Total: 10]

10 marks

Mark scheme: 2(a) (30 – 10 =) 20 (N) B1 forwards OR in direction of 30 N force B1 2(b)(i) (work done =) 2000 (J) A3 (work done =) 40  50 (C2) (work done =) force  distance (moved in direction of force) OR (W) = F × d (C1) 2(b)(ii) internal OR thermal energy (of surroundings / tyres) B1 kinetic energy B1 2(c) (pressure =) 6.3 (N / cm2) A3 (pressure =) 240  38 (C2) (pressure =) force  area OR (p) = F  A (C1)

This question in 0625/31 Oct/Nov 2024

3 A car has a fault. A mechanic uses a machine to pull the car onto a recovery vehicle as shown in Fig. 3.1. handle mechanic machine rope recovery vehicle ramp Fig. 3.1 (a) Fig. 3.2 shows how the mechanic applies a force to the handle of the machine. handle 26 N machine 0.94 m pivot Fig. 3.2 (i) Calculate the moment of the 26 N force about the pivot. Use the information in Fig. 3.2. moment = … N m [3] (ii) Describe one way the mechanic can increase the moment of the 26 N force about the pivot. … [1] (b) The car is lifted vertically 0.78 m onto the recovery vehicle, as shown in Fig. 3.3. 0.78 m ground Fig. 3.3 The weight of the car is 14 000 N. Calculate the minimum work done on the car in lifting it onto the recovery vehicle from the ground. Include the unit. work done = … unit … [4] [Total: 8]

8 marks

Mark scheme: 3(a)(i) 24 (N m) A3 26  0.94 (C2) (moment =) force  (perpendicular) distance (from pivot) (C1) 3(a)(ii) increase distance between pivot and force owtte B1 3(b) 11 000 A3 14 000  0.78 (C2) (work =) force  distance OR (W =) F  d (C1) J B1

This question in 0625/32 Oct/Nov 2024

1 (a) Fig. 1.1 shows the speed–time graph for a cyclist at the beginning of a race. 5.0 speed m / s 4.0 3.0 2.0 1.0 0 0 2.0 4.0 6.0 8.0 10.0 time / s Fig. 1.1 (i) Describe the motion of the cyclist from time = 6.0 s to time = 10.0 s. … [1] (ii) Determine the distance moved by the cyclist from time = 6.0 s to time = 10.0 s. distance = … m [3] (b) Another cyclist travels 1000 m in 1 minute and 29 seconds. (i) Determine how many seconds there are in 1 minute and 29 seconds. number of seconds = … s [1] (ii) Calculate the average speed of this cyclist. average speed = … m / s [3] (c) The horizontal forces acting on a cyclist vary. Fig. 1.2 shows the horizontal forces at one moment during a race. backward force = 50 N forward force = 90 N Fig. 1.2 (not to scale) (i) Calculate the resultant horizontal force acting on the cyclist. resultant horizontal force = … N direction = … [2] (ii) Describe the effect of the resultant horizontal force in (c)(i) on the motion of the cyclist. … [1] [Total: 11]

11 marks

Mark scheme: Question Answer Marks 1(a)(i) constant speed or moving with zero acceleration B1 1(a)(ii) 18 (m) A3 4.5  4(.0) (C2) (distance =) area under graph OR (distance=) speed  time OR (d =) s × t (C1) 1(b)(i) 89 (s) B1 1(b)(ii) 11 (m / s) A3 1000  89 (C2) (speed =) distance  time OR (s =) d  t (C1) 1(c)(i) 40 (N) B1 forwards / to the right B1 1(c)(ii) accelerate / speed increase owtte B1

This question in 0625/33 Oct/Nov 2024

2 A teacher uses a spring in a demonstration. The spring is shown in Fig. 2.1. spring wire Fig. 2.1 (a) The spring is made from wire. Describe how to determine the diameter of the wire accurately. You may include a diagram as part of your answer. … … … … [3] (b) A student adds loads to the spring. She measures the extension for each load. Fig. 2.2 shows the graph of extension against load for the spring. extension / cm 2.0 1.0 0 0 1.0 2.0 3.0 4.0 5.0 load / N Fig. 2.2 (i) Using Fig. 2.2, determine the extension of the spring with a load of 4.0 N. extension = … cm [2] (ii) The length of the spring without a load is 8.0 cm. Calculate the length of the spring with a load of 4.0 N. length = … cm [1] [Total: 6]

6 marks

Mark scheme: 2(a) measure the diameter of n loops / turns with rule(r) B1 n = 5 or more loops / loops B1 (diameter of wire =) their measurement  n if n  1 B1 2(b)(i) 1.6 (cm) A2 any indication of correct attempt on graph (C1) 2(b)(ii) 9.6 (cm) B1

This question in 0625/33 Oct/Nov 2024

11 (a) Fig. 11.1 shows the Moon orbiting the Earth. the Moon the Earth Fig. 11.1 (i) State the name of the force that keeps the Moon in orbit around the Earth. … [1] (ii) State the time taken by the Moon to complete one orbit of the Earth. Include the unit. time for one orbit = … unit … [1] (iii) A device on the Moon sends a radio signal to the Earth. The distance from the Moon to the Earth is 3.8 × 108 m. The speed of a radio wave is 3.0 × 108 m / s. Calculate the time the radio signal takes to travel from the Moon to the Earth. time = … s [3] (b) An astronomer observes light from a distant galaxy. She writes this conclusion: ‘The observed wavelength of light from the distant galaxy is longer than the wavelength of the light emitted by the galaxy.’ Explain how this conclusion supports the Big Bang Theory of the Universe. … … … … … … [3] [Total: 8]

8 marks

Mark scheme: 11(a)(i) gravity B1 11(a)(ii) 1 month OR 4 weeks OR 27 to 31 days B1 11(a)(iii) 1.3 (s) A3 3.8  108  3(.0)  108 (C2) (time =) distance  speed OR (t =) d  s (C1) 11(b) any three from: B3 • redshift (of light) • (galaxy is) moving away / receding • Universe is expanding • Big Bang theory predicts expanding Universe

This question in 0625/33 Oct/Nov 2024

1 Fig. 1.1 shows the speed-time graph for a ball falling through the air. 4.0 speed 3.0 m / s 2.0 1.0 0.0 0.0 1.0 2.0 3.0 4.0 5.0 time / s Fig. 1.1 (a) Determine the speed of the ball at time = 1.0 s. speed = … m / s [1] (b) Describe the speed of the ball between time = 2.0 s and time = 5.0 s. … [1] (c) Calculate the distance moved by the ball between time = 2.0 s and time = 5.0 s. distance = … m [3] (d) Fig. 1.2 shows the two vertical forces acting on the ball at time = 3.0 s. X ball Y Fig. 1.2 Give the term used to describe each force. upward force X is ………………………………… downward force Y is ………………………………… [2] [Total: 7]

7 marks

Mark scheme: Question Answer Marks 1(a) 3.55 (m/s) B1 1(b) constant/steady (speed) OR B1 zero acceleration 1(c) (distance travelled =) 12 (m) A3 (distance travelled =) 4(.0)  3(.0) (C2) (distance travelled =) area under graph OR b  h (C1) 1(d) (upward force  is) air resistance OR friction OR drag B1 (downward force Y is) weight OR (force of) gravity B1

This question in 0625/32 Feb/March 2025

3 Fig. 3.1 shows a cross-section through the centre of a solid cone and a cross-section through the centre of a solid cuboid. The cone and the cuboid are placed on the ground as shown in Fig. 3.1. The solid cone is more stable than the solid cuboid. The symbol × indicates the position of the centre of gravity of each object. cone cuboid ground Fig. 3.1 (a) Explain why the solid cone in Fig. 3.1 is more stable than the solid cuboid. … … [1] (b) The weight of the solid cuboid is 48 N. The area of the solid cuboid in contact with the ground is 32 cm2. Calculate the pressure exerted by the solid cuboid on the ground. pressure on ground = … N / cm2 [3] (c) A horizontal force of 15 N tilts the solid cone, as shown in Fig. 3.2. 15 N 18 cm pivot Fig. 3.2 Calculate the moment of the 15 N force about the pivot. moment = … N cm [3] [Total: 7]

7 marks

Mark scheme: 3(a) (has) lower centre of gravity owtte B1 3(b) (P =) 1.5 (N / cm2) A3 (P =) 48 ÷ 32 (C2) (P =) F ÷ A (C1) 3(c) (moment =) 270 (Ncm) A3 (moment =) 15  18 (C2) moment = force  (perp.) distance (from pivot) (C1)

This question in 0625/32 Feb/March 2025

12 Fig. 12.1 represents the Earth’s orbit of the Sun. Sun Earth Fig. 12.1 (Not to scale) (a) State the term used to describe the force that keeps the Earth in orbit around the Sun. … [1] (b) Explain why the Earth has an annual cycle of seasons. … … … [1] (c) The distance of the Earth from the Sun is 1.48 × 1011 m. The speed of light is 3.0 × 108 m / s. Calculate the time it takes light from the Sun to reach the Earth. time taken to reach Earth = … s [3] [Total: 5]

5 marks

Mark scheme: 12(a) gravitational (attraction) B1 12(b) Earth’s axis is tilted (relative to its orbital plane) B1 OR idea that in part of orbit, N or S pole points towards/away from Sun 12(c) 490 (s) A3 1.48  1011 ÷ 3(.0)  108 (C2) speed = distance ÷ time OR (t =) d ÷ s (C1)

This question in 0625/32 Feb/March 2025

3 A wooden beam is used in a see-saw. (a) The mass of the wooden beam is 40 kg. Calculate the weight of the wooden beam. weight of beam = … N [2] (b) Two children balance the wooden beam horizontally on a log. The wooden beam pivots on the log to make the see-saw. Fig. 3.1 shows the children on the see-saw. child B child A 1.6 m 1.2 m wooden beam W 360 N log pivot Fig. 3.1 (not to scale) The see-saw balances horizontally, as shown in Fig. 3.1. The weight of child B is 360 N. Calculate the weight W of child A. State and use the principle of moments in your answer. weight W of child A = … N [4] [Total: 6]

6 marks

Mark scheme: 3(a) 390 (N) A2 (weight =) mass  g OR (weight =) 40  9.8 C1 3(b) 270 (N) A4 W  1.6 = 360  1.2 OR (W =) 432 ÷ 1.6 OR (W =) {360  1.2} ÷ 1.6 C3 (clockwise moment OR moment of child B =) 360  1.2 OR 430 OR 432 seen C1 (total) clockwise moment = (total) anticlockwise moment C1

This question in 0625/31 May/June 2025

1 (a) Fig. 1.1 shows the speed–time graphs for two racing cars, X and Y, at the beginning of a race. 25 speed m / s 20 racingracing carcar XX 15 racingracing carcar YY 10 5 0 0 2 4 6 8 10 12 time / s Fig. 1.1 (i) Using the information on Fig. 1.1, state and explain which racing car, X or Y, has the greater acceleration between time = 2 s and time = 4 s. racing car … explanation … … [1] (ii) Determine the speed of racing car Y at time = 10 s. speed = … m / s [1] (iii) Determine the distance moved by racing car X from time = 0 to time = 4.0 s. distance = … m [3] (b) Fig. 1.2 shows the directions of four forces, A, B, C and D, acting on a racing car. A D B C Fig. 1.2 (i) Force B is described as ‘the driving force’. Describe: force C … force D … [2] (ii) The racing car is decelerating along a straight horizontal track. The value of force D is 2800 N. Suggest a value for force B. force B = … N [1] [Total: 8]

8 marks

Mark scheme: Question Answer Marks 1(a)(i) racing car X M0 steeper slope owtte A1 1(a)(ii) 24 (m / s) B1 1(a)(iii) 44 (m) A3 ½  22  4 C2 (distance =) area under graph OR ½  b  h OR speed  time C1 1(b)(i) force C – weight OR (force of) gravity OR gravitational pull B1 force D – (air) resistance/drag/friction B1 1(b)(ii) a value less than 2800 (N) B1

This question in 0625/32 May/June 2025

3 (a) A student determines the weight W of a metal block by using a 1.5 N load and a uniform metre ruler. She places the centre of the uniform metre ruler on a pivot. metal 0.44 m 0.21 m block load 1.5 N pivot W uniform metre ruler Fig. 3.1 (not to scale) She moves the metal block and the 1.5 N load until the uniform metre ruler balances horizontally as shown in Fig. 3.1. Calculate the weight W of the metal block. Use the principle of moments in your answer. weight W = … N [4] (b) A different metal block is lying on the ground, as shown in Fig. 3.2. 0.54 m 0.18 m ground Fig. 3.2 (not to scale) The weight of the metal block is 890 N. Calculate the pressure on the ground caused by the block in the position shown in Fig. 3.2. pressure = … N / m2 [3] [Total: 7]

7 marks

Mark scheme: 3(a) 3.1 (N) A4 1.5  0.44 = W  0.21 OR (W =) (1.5  0.44) ÷ 0.21 C3 OR (W =) 0.66 ÷ 0.21 (anticlockwise moment =) 1.5  0.44 OR 0.66 seen C1 (sum of) anticlockwise moment = (sum of) clockwise moment C1 3(b) 9200 (N / m2) A3 890 ÷ (0.54  0.18) OR 890 ÷ 0.0972 C2 (pressure =) force ÷ area C1

This question in 0625/32 May/June 2025

1 (a) Fig. 1.1 shows the speed–time graphs for two racing cars, X and Y, at the beginning of a race. 25 speed m / s 20 racingracing carcar XX 15 racingracing carcar YY 10 5 0 0 2 4 6 8 10 12 time / s Fig. 1.1 (i) Using the information on Fig. 1.1, state and explain which racing car, X or Y, has the greater acceleration between time = 2 s and time = 4 s. racing car … explanation … … [1] (ii) Determine the speed of racing car Y at time = 10 s. speed = … m / s [1] (iii) Determine the distance moved by racing car X from time = 0 to time = 4.0 s. distance = … m [3] (b) Fig. 1.2 shows the directions of four forces, A, B, C and D, acting on a racing car. A D B C Fig. 1.2 (i) Force B is described as ‘the driving force’. Describe: force C … force D … [2] (ii) The racing car is decelerating along a straight horizontal track. The value of force D is 2800 N. Suggest a value for force B. force B = … N [1] [Total: 8]

8 marks

Mark scheme: Question Answer Marks 1(a)(i) racing car X M0 steeper slope owtte A1 1(a)(ii) 24 (m / s) B1 1(a)(iii) 44 (m) A3 ½  22  4 C2 (distance =) area under graph OR ½  b  h OR speed  time C1 1(b)(i) force C – weight OR (force of) gravity OR gravitational pull B1 force D – (air) resistance/drag/friction B1 1(b)(ii) a value less than 2800 (N) B1

This question in 0625/33 May/June 2025

3 (a) A student determines the weight W of a metal block by using a 1.5 N load and a uniform metre ruler. She places the centre of the uniform metre ruler on a pivot. metal 0.44 m 0.21 m block load 1.5 N pivot W uniform metre ruler Fig. 3.1 (not to scale) She moves the metal block and the 1.5 N load until the uniform metre ruler balances horizontally as shown in Fig. 3.1. Calculate the weight W of the metal block. Use the principle of moments in your answer. weight W = … N [4] (b) A different metal block is lying on the ground, as shown in Fig. 3.2. 0.54 m 0.18 m ground Fig. 3.2 (not to scale) The weight of the metal block is 890 N. Calculate the pressure on the ground caused by the block in the position shown in Fig. 3.2. pressure = … N / m2 [3] [Total: 7]

7 marks

Mark scheme: 3(a) 3.1 (N) A4 1.5  0.44 = W  0.21 OR (W =) (1.5  0.44) ÷ 0.21 C3 OR (W =) 0.66 ÷ 0.21 (anticlockwise moment =) 1.5  0.44 OR 0.66 seen C1 (sum of) anticlockwise moment = (sum of) clockwise moment C1 3(b) 9200 (N / m2) A3 890 ÷ (0.54  0.18) OR 890 ÷ 0.0972 C2 (pressure =) force ÷ area C1

This question in 0625/33 May/June 2025

3 A student stretches a spring by suspending it and attaching metal discs to it, as shown in Fig. 3.1. clampstand ruler spring metal discs Fig. 3.1 (a) The mass of a metal disc is 0.25 kg. Calculate the weight of the metal disc. weight of metal disc = … N [2] (b) Fig. 3.2 shows the results from the student’s experiment. 60 50 length of 40 spring / cm 30 20 0 2.0 4.0 6.0 8.0 10 load on spring / N Fig. 3.2 (i) Determine the length of the spring when the load attached to the spring is 7.0 N. Show your working on Fig. 3.2. length of spring = … cm [2] (ii) Determine the length of the spring when the load attached to the spring is zero. Show your working on Fig. 3.2. length of spring when load is zero = … cm [2] [Total: 6]

6 marks

Mark scheme: 3(a) 2.5 (N) A2 (weight =) mass (in kg)  g OR (weight =) 0.25  9.8 (C1) 3(b)(i) (length of spring =) 48.5 (cm) A2 line from 7.0 on x-axis to line on graph OR line from graph to about 48.5 on y-axis (C1) (b)(ii) (length of spring =) 33.5 (cm) A2 graph line extended in straight line to meet y-axis (C1)

This question in 0625/31 Oct/Nov 2025

2 Fig. 2.1 shows the horizontal forces acting on an ice skater. The ice skater is moving forwards. ice skater backward force = 45 N forward force = 80 N skate edge of skate in contact with the ice Fig. 2.1 (a) Calculate the resultant horizontal force acting on the ice skater. Determine the direction of the resultant force. resultant horizontal force = … N direction = … [2] (b) The weight of the ice skater is 700 N. The area of the skate in contact with the ice is 6.2 cm2. Calculate the pressure on the surface of the ice exerted by the skate. Give your answer to two significant figures. pressure = … N / cm2 [3] (c) The weight of the ice skater is 700 N. Calculate the mass of the ice skater. Give your answer to two significant figures. mass of skater = … kg [3] [Total: 8]

8 marks

Mark scheme: 2(a) 35 (N) B1 forwards / to the right B1 2(b) 110 (N / cm2) A3 700 ÷ 6.2 C2 (pressure =) force / area C1 2(c) 71 (kg) A3 700 ÷ 9.8 C2 (mass =) weight / gravitational field strength or W / g or W / 9.8 C1

This question in 0625/32 Oct/Nov 2025

2 (a) A builder uses a metal bar to raise one end of the rock. builder rock metal bar 320 N pivot 1.2 m Fig. 2.1 (not to scale) Calculate the moment of the 320 N force about the pivot. moment = … Nm [3] (b) The builder lifts another rock using a truck as shown in Fig. 2.2. truck position 2 0.60 m position 1 Fig. 2.2 (not to scale) The truck lifts the rock through a vertical height of 0.60 m. The weight of the rock is 4800 N. Calculate the work done in lifting the rock. work done … J [3] (c) The work done by the truck in lifting a different rock is 5800 J. The truck lifts the rock in a time of 7.4 s. Calculate the power of the truck in lifting the rock. Include the unit. power = … unit … [4] [Total: 10]

10 marks

Mark scheme: 2(a) 380 (Nm) A3 1.2  320 C2 (clockwise moment =) force  (perpendicular) distance C1 2(b) 2900 (J) A3 4800  0.6 C2 (work done =) force  distance C1 2(c) 780 (W) A3 5800 / 7.4 C2 (power =) work(done) / time or energy / time C1 W or watts B1

This question in 0625/33 Oct/Nov 2025