Year 12 Physics Remote Learning Phase 3 - Ark Globe Academy

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Year12PhysicsRemoteLearningPhase

3

Name:________________________________________________________________

Resources:

• Allvideos,powerpoints,examquestionsandanswercanbefoundonthesharepoint:https://arkschools.sharepoint.com/:f:/r/sites/A-LevelPhysicsREsources/Shared%20Documents/A-Level%20Physics%20resources?csf=1&e=1FMnCr

• RunPHETsimulationstoaddresstheidentifiedgaps.https://phet.colorado.edu/en/simulations/category/physics

• UsetheYouTubechannel“OnlinePhysics“tohelpaddressthegaps.https://www.youtube.com/channel/UCZzatyx-xC-Dl_VVUVHYDYw/playlists?view=50&sort=dd&shelf_id=15

• Notesandexamquestionfromotherexamboardscanbefoundhere.https://www.physicsandmathstutor.com/

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InstructionforWeeksbeginning11thto22ndMay2020

Session Title Worktobecompleted

1 CircularmotionAcceleration,workdoneandequations1

-WatchvideoL1.1andL1.2.Takesnotes,whichgointoyourfolder.-Completequestions1-8oftheself-diagnosticandsendMrNagatheanswersbyemail.-Completeexamquestions1-17

2 CircularmotionAcceleration,workdoneandequations2

-Read17.1Uniformcircularmotion(Pages274-275)-Completeandmarkthesummaryquestions

3 CircularmotionForces1 -WatchvideoL1.3andL1.4.Takesnotes,whichgointoyourfolder.-Completeexamquestions18-24

4 CircularmotionForces2 -Read17.2Centripetalacceleration(Pages276-277)(Donolearntheproofofa=v2/r,notrequiredbyspecification)-Completeandmarkthesummaryquestions

5 CircularmotionVertical1

-WatchvideoL1.5andL1.6.Takesnotes,whichgointoyourfolder.-Completequestions9-12oftheself-diagnosticandsendMrNagatheanswersbyemail.-Completeexamquestions25-31

6 CircularmotionVertical2

-Read17.3Ontheroad(Pages278-279)(Donotspendtolongon“onanakedtrack”thisisexplainedinthenextvideo)-Completeandmarkthesummaryquestions

7 CircularmotionatAngles -WatchvideoL1.7andL1.8.Takesnotes,whichgointoyourfolder.-Completequestions13-14oftheself-diagnosticandsendMrNagatheanswersbyemail.-Completeexamquestions32-35

8 CircularmotionatAngles -Read17.4Atthefairground(Pages280-281)-Completeandmarkthesummaryquestions

9 CircularmotionSummary

-Completetheexamquestionsonpages282and283-Markanswers

10 CircularmotionSummary

-Completetheexamquestionsonpages284and285-Markanswers

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Q1. An object moving at constant speed in a circle experiences a force that is

A in the direction of motion.

B outwards and at right angles to the direction of motion.

C inwards and at right angles to the direction of motion.

D opposite to the direction of motion. (Total 1 mark)

Q2. A particle moves in a circular path at constant speed. Which one of the following statements is correct?

A The velocity of the particle is directed towards the centre of the circle.

B There is no force acting on the particle.

C There is no change in the kinetic energy of the particle.

D The particle has an acceleration directed along a tangent to the circle. (Total 1 mark)

Q3. For a particle moving in a circle with uniform speed, which one of the following statements is correct?

A The kinetic energy of the particle is constant.

B The force on the particle is in the same direction as the direction of motion of the particle.

C The momentum of the particle is constant.

D The displacement of the particle is in the direction of the force. (Total 1 mark)

Q4. For a particle moving in a circle with uniform speed, which one of the following statements is correct?

A The displacement of the particle is in the direction of the force.

B The force on the particle is in the same direction as the direction of motion of the particle.

C The momentum of the particle is constant.

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D The kinetic energy of the particle is constant. (Total 1 mark)

Q5. For a particle moving in a circle with uniform speed, which one of the following statements is incorrect?

A The velocity of the particle is constant.

B The force on the particle is always perpendicular to the velocity of the particle.

C There is no displacement of the particle in the direction of the force.

D The kinetic energy of the particle is constant. (Total 1 mark)

Q6. The Earth moves around the Sun in a circular orbit with a radius of 1.5 × 108 km. What is the Earth’s approximate speed?

A 1.5 × 103ms–1

B 5.0 × 103ms–1

C 1.0 × 104ms–1

D 3.0 × 104ms–1

(Total 1 mark)

Q7.

A model car moves in a circular path of radius 0.80 m at an angular speed of rad s–1.

What is its displacement from point P 6.0 s after passing P?

A zero

B 0.4π m

C 1.6 m

D 1.6π m (Total 1 mark)

Q8. What is the angular speed of a car wheel of diameter 0.400 m when the speed of the car is

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108 km h–1?

A 75 rad s–1

B 150 rad s–1

C 270 rad s–1

D 540 rad s–1

(Total 1 mark)

Q9. The diagram shows a disc of diameter 120 mm that can turn about an axis through its centre.

The disc is turned through an angle of 30° in 20 ms. What is the average speed of a point on the edge of the disc during this time?

A 0.5π m s–1

B π m s–1

C 1.5π m s–1

D 2π m s–1

(Total 1 mark)

Q10. A disc of diameter D is turning at a steady angular speed at frequency f about an axis through its centre.

What is the centripetal force on a small object O of mass m on the perimeter of the disc?

A 2πmfD

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B 2πmf 2D

C 2π2mf 2D

D 2πmf 2D 2

(Total 1 mark)

Q11. A fairground roundabout makes nine revolutions in one minute. What is the angular speed of the roundabout?

A 0.15 rad s–1

B 0.34 rad s–1

C 0.94 rad s–1

D 2.1 rad s–1

(Total 1 mark)

Q12. What is the angular speed of a point on the Earth’s equator?

A 7.3 × 10–5 rad s–1

B 4.2 × 10–3 rad s–1

C 2.6 × 10–1 rad s–1

D 15 rad s–1

(Total 1 mark)

Q13.

A model car moves in a circular path of radius 0.8 m at an angular speed of rad s–1.

What is its displacement from point P, 6 s after passing P?

A zero

B 1.6 m

C 0.4π m

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D 1.6π m (Total 1 mark)

Q14. A helicopter circles continuously at a constant speed around a horizontal path of diameter 800 m, taking 5.0 minutes to complete each orbit of the path.

What are the speed v and the centripetal acceleration a of the helicopter?

v / m s−1 a / m s−2

A 0.021 0.18

B 8.4 0.088

C 8.4 0.18

D 17 0.35 (Total 1 mark)

Q15.

The figure shows a smooth thin tube T through which passes a string with masses m and M attached to its ends. Initially the tube is moved so that the mass, m, travels in a horizontal circle of constant radius r, at constant speed, v. Which one of the following expressions is equal to M?

A

B mv2rg

C

D (Total 1 mark)

Q16.

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A girl of mass 40 kg stands on a roundabout 2.0 m from the vertical axis as the roundabout rotates uniformly with a period of 3.0 s. The horizontal force acting on the girl is approximately

A zero.

B 3.5 × 102 N.

C 7.2 × 102 N.

D 2.8 × 104 N. (Total 1 mark)

Q17. A mass on the end of a string is whirled round in a horizontal circle at increasing speed until the string breaks. The subsequent path taken by the mass is

A a straight line along a radius of the circle.

B a horizontal circle.

C a parabola in a horizontal plane.

D a parabola in a vertical plane. (Total 1 mark)

Q18. A chemical centrifuge consists of two test-tube holders which can be spun round in a horizontal circular path at very high speed as shown. The centrifuge runs at a steady speed of 3000 revolutions per minute and the test-tube holders are horizontal.

(a) Calculate the angular speed of the centrifuge in rad s–1.

___________________________________________________________________

___________________________________________________________________

(b) Calculate the magnitude of the acceleration at a point on the centrifuge 95 mm from the axis of rotation.

___________________________________________________________________

___________________________________________________________________

(c) State the direction of the acceleration in part (ii).

___________________________________________________________________ (Total 5 marks)

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Q19. The figure below shows a dust particle at position D on a rotating vinyl disc. A combination of electrostatic and frictional forces act on the dust particle to keep it in the same position.

The dust particle is at a distance of 0.125 m from the centre of the disc. The disc rotates at 45 revolutions per minute.

(a) Calculate the linear speed of the dust particle at D.

(3)

(b) (i) Mark on the diagram above an arrow to show the direction of the resultant horizontal force on the dust particle.

(1)

(ii) Calculate the centripetal acceleration at position D.

(2)

(c) On looking closely at the rotating disc it can be seen that there is more dust concentrated on the inner part of the disc than the outer part. Suggest why this should be so.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

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___________________________________________________________________

___________________________________________________________________

___________________________________________________________________ (3)

(Total 9 marks)

Q20. A particle of mass m moves in a circle of radius r at uniform speed, taking time T for each revolution. What is the kinetic energy of the particle?

A

B

C

D (Total 1 mark)

Q21. A particle of mass m moves in a circle of radius r at a uniform speed with frequency f. What is the kinetic energy of the particle?

A

B

C 2π2mf 2r2

D 4π2mf 2r2

(Total 1 mark)

Q22.

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A 150 g mass is attached to one end of a light inextensible string and the other end of the string is fixed at a point P as shown in the diagram above. The mass is held at point A so that the string is taut and horizontal. The mass is released so that it moves freely along a circular arc of 250 mm radius.

When the string moves through the vertical position, the mass is at point B. Neglecting the effect of air resistance, calculate

(a) the kinetic energy of the mass,

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(b) the velocity of the mass,

___________________________________________________________________

___________________________________________________________________

(c) the centripetal force acting on the mass,

___________________________________________________________________

___________________________________________________________________

(d) the tension in the string.

___________________________________________________________________

___________________________________________________________________ (Total 6 marks)

Q23. The figure below shows a car on a rollercoaster track. The car is initially at rest at A and is lifted to the highest point of the track, B, 35 m above A.

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The car with its passengers has a total mass of 550 kg. It takes 25 s to lift the car from A to B. It then starts off with negligible velocity and moves unpowered along the track.

(a) Calculate the power used in lifting the car and its passengers from A to B. Include an appropriate unit in your answer.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

power______________________unit_____________ (3)

(b) The speed reached by the car at C, the bottom of the first dip, is 22 ms–1. The length of the track from B to the bottom of the first dip C is 63 m.

Calculate the average resistive force acting on the car during the descent.

Give your answer to a number of significant figures consistent with the data.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

resistive force ______________________ N (4)

(c) Explain why the resistive force is unlikely to remain constant as the car descends from B to C.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

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___________________________________________________________________

___________________________________________________________________ (3)

(d) At C, a passenger of mass 55 kg experiences an upward reaction force of 2160 N when the speed is 22 ms–1.

Calculate the radius of curvature of the track at C. Assume that the track is a circular arc at this point.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

radius of curvature of the track ______________________ m (3)

(Total 13 marks)

Q24. Figure 1 shows a fairground ride called a Rotor. Riders stand on a wooden floor and lean against the cylindrical wall.

Figure 1

The fairground ride is then rotated. When the ride is rotating sufficiently quickly the wooden floor is lowered. The riders remain pinned to the wall by the effects of the motion. When the speed of rotation is reduced, the riders slide down the wall and land on the floor.

(a) (i) At the instant shown in Figure 2 the ride is rotating quickly enough to hold a rider at a constant height when the floor has been lowered.

Figure 2

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Draw onto Figure 2 arrows representing all the forces on the rider when held in this position relative to the wall. Label the arrows clearly to identify all of the forces.

(3)

(ii) Explain why the riders slide down the wall as the ride slows down.

______________________________________________________________

______________________________________________________________

______________________________________________________________ (2)

A Rotor has a diameter of 4.5 m. It accelerates uniformly from rest to maximum angular velocity in 20 s. The total moment of inertia of the Rotor and the riders is 2.1 × 105 kg m2.

(b) (i) At the maximum speed the centripetal acceleration is 29 m s–2.

Show that the maximum angular velocity of a rider is 3.6 rad s–1.

(2)

(ii) Calculate the torque exerted on the Rotor so that it accelerates uniformly to its maximum angular velocity in 20 s. State the appropriate SI unit for torque.

torque ____________________ SI unit for torque __________ (3)

(iii) Calculate the centripetal force acting on a rider of mass 75 kg when the ride is

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moving at its maximum angular velocity. Give your answer to an appropriate number of significant figures.

centripetal force ____________________ N (1)

(c) Figure 3 shows the final section of a roller coaster which ends in a vertical loop. The roller coaster is designed to give the occupants a maximum acceleration of 3g. Cars on the roller coaster descend to the start of the loop and then travel around it, as shown.

Figure 3

(i) At which one of the positions marked A, B and C on Figure 3 would the passengers experience the maximum reaction force exerted by their seat? Circle your answer below.

A B C

(1)

(iii) Explain why the maximum acceleration is experienced at the position you have chosen.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________ (2)

(Total 14 marks)

Q25. A young child of mass 20 kg stands at the centre of a uniform horizontal platform which rotates at a constant angular speed of 3.0 rad s–1. The child begins to walk radially outwards towards the edge of the platform. The maximum frictional force between the child and the platform is 200 N. What is the maximum distance from the centre of the platform to which the child could walk without the risk of slipping?

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A 1.1 m

B 1.3 m

C 1.5 m

D 1.7 m (Total 1 mark)

Q26.

A ball of mass m, which is fixed to the end of a light string of length l, is released from rest at X. It swings in a circular path, passing through the lowest point Y at speed ʋ. If the tension in the string at Y is T, which one of the following equations represents a correct application of Newtonߣs laws of motion to the ball at Y?

A T = − mg

B T − mg =

C mg − T =

D T + = mg (Total 1 mark)

Q27. A bob of mass 0.50 kg is suspended from the end of a piece of string 0.45 m long. The bob is rotated in a vertical circle at a constant rate of 120 revolutions per minute.

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What is the tension in the string when the bob is at the bottom of the circle?

A 5.8 N

B 31 N

C 36 N

D 40 N (Total 1 mark)

Q28. A 0.20 kg mass is whirled round in a vertical circle on the end of a light string of length 0.90 m.

At the top point of the circle the speed of the mass is 8.2 m s–1. What is the tension in the string at this point?

A 10 N

B 13 N

C 17 N

D 20 N (Total 1 mark)

Q29.

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A simple pendulum consists of a bob of mass m on the end of a light string of length l. The bob is released from rest at X when the string is horizontal. When the bob passes through Y its velocity is ʋ and the tension in the string is T. Which one of the following equations gives the correct value of T?

A T = mg

B T =

C T + mg =

D T − mg = (Total 1 mark)

Q30. A ball of mass 0.30 kg is attached to a string and moves in a vertical circle of radius 0.60 m at a constant speed of 5.0 m s–1.

Which line, A to D, in the table gives the correct values of the minimum and maximum tension in the string?

Minimum tension / N

Maximum tension / N

A 2.5 5.4 B 6.7 9.6 C 13 13 D 9.6 15

(Total 1 mark)

Q31. Figure 1 shows a parcel on the floor of a delivery van that is passing over a hump-backed bridge on a straight section of road. The radius of curvature of the path of the parcel is r and the van is travelling at a constant speed v. The mass of the parcel is m.

Figure 1

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(a) (i) Draw arrows on Figure 2 below to show the forces that act on the parcel as it passes over the highest point of the bridge. Label these forces.

Figure 2

(1)

(ii) Write down an equation that relates the contact force, R, between the parcel and the floor of the van to m, v, r and the gravitational field strength, g.

______________________________________________________________

______________________________________________________________ (1)

(iii) Calculate R if m = 12 kg, r = 23 m, and v = 11ms–1.

answer = ______________________ N (2)

(b) Explain what would happen to the magnitude of R if the van passed over the bridge at a higher speed. What would be the significance of any van speed greater than 15ms–1? Support your answer with a calculation.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

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___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________ (3)

(Total 7 marks)

Q32. A gymnast does a hand-stand on a horizontal bar. The gymnast then rotates in a vertical circle with the bar as a pivot. The gymnast and bar remain rigid during the rotation and when friction and air resistance are negligible the gymnast returns to the original stationary position.

Figure 1 shows the gymnast’s position at the start and Figure 2 shows the position after completing half the circle.

Figure 1 Figure 2

(a) The gymnast has a mass of 70 kg and the centre of mass of the gymnast is 1.20 m from the axis of rotation.

acceleration of free fall, g = 9.8 m s–2

(i) Show clearly how the principle of conservation of energy predicts a speed of 6.9 m s–1 for the centre of mass when in the position shown in Figure 2.

(3)

(ii) The maximum force on the arms of the gymnast occurs when in the position shown in Figure 2.

Calculate the centripetal force required to produce circular motion of the gymnast when the centre of mass is moving at 6.9 m s–1.

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(2)

(iii) Determine the maximum tension in the arms of the gymnast when in the position shown in Figure 2.

(1)

(iv) Sketch a graph to show how the vertical component of the force on the bar varies with the angle rotated through by the gymnast during the manoeuvre. Assume that a downward force is positive.

Include the values for the initial force and the maximum force on the bar.

Only show the general shape between these values.

(2)

(b) The bones in each forearm have a length of 0.25 m. The total cross-sectional area of the bones in both forearms is 1.2 × 10–3 m2 . The Young modulus of bone in compression is 1.6 × 1010 Pa.

Assuming that the bones carry all the weight of the gymnast, calculate the reduction in length of the forearm bones when the gymnast is in the start position shown in Figure 1.

(3)

(Total 11 marks)

Q33. (a) Figure 1 and Figure 2 each show a car travelling in a horizontal circular path.

(i) Draw and label on Figure 1 and Figure 2 arrows to indicate the other forces acting

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on the cars.

Figure 1

Figure 2 (2)

(ii) State the possible origins of the centripetal force on the car in Figure 2.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________ (4)

(b) Figure 3 shows a motorcycle stunt rider travelling around a track in a vertical circle of radius 5.2 m. At position Q, when the speed is the minimum necessary to keep the motorcycle in contact with the track, the centripetal force is supplied by the weight of the motorcycle and rider. The combined mass of the motorcycle and rider is 220 kg.

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Figure 3

Calculate the minimum speed which will keep the motorcycle in contact with the track at position Q. The acceleration due to gravity, g, is 9.8 m s–2.

(3)

(Total 9 marks)

Q34. A lead ball of mass 0.25 kg is swung round on the end of a string so that the ball moves in a horizontal circle of radius 1.5 m. The ball travels at a constant speed of 8.6 m s–1.

(a) (i) Calculate the angle, in degrees, through which the string turns in 0.40 s.

angle ____________________ degree

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(3)

(ii) Calculate the tension in the string. You may assume that the string is horizontal.

tension ____________________ N (2)

(b) The string will break when the tension exceeds 60 N. Calculate the number of revolutions that the ball makes in one second when the tension is 60 N.

number of revolutions ____________________ (2)

(c) Discuss the motion of the ball in terms of the forces that act on it. In your answer you should:

• explain how Newton’s three laws of motion apply to its motion in a circle • explain why, in practice, the string will not be horizontal.

You may wish to draw a diagram to clarify your answer.

The quality of your written communication will be assessed in your answer.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

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___________________________________________________________________

___________________________________________________________________

___________________________________________________________________ (6)

(Total 13 marks)

Q35. Figure 1 shows a side view of an act performed by two acrobats. Figure 2 shows the view from above.

Figure 1 Figure 2

The acrobats, each of mass 85 kg, are suspended from ropes attached to opposite edges of a circular platform that is at the top of a vertical pole. The platform has a diameter of 2.0 m A motor rotates the platform so that the acrobats move at a constant speed in a horizontal circle, on opposite sides of the pole.

When the period of rotation of the platform is 5.2 s, the centre of mass of each acrobat is 5.0 m below the platform and the ropes are at an angle of 28.5° to the vertical as shown in Figure 1.

(a) Show that the linear speed of the acrobats is about 4.5 m s–1

(2)

(b) Determine the tension in each rope that supports the acrobats.

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tension = ____________________ N (3)

(c) Discuss the consequences for the forces acting on the pole when one acrobat has a much greater mass than the other.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________ (3)

(Total 8 marks)

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Mark schemes

Q1. C

[1]

Q2. C

[1]

Q3. A

[1]

Q4. D

[1]

Q5. A

[1]

Q6. D

[1]

Q7. C

[1]

Q8. B

[1]

Q9. A

[1]

Q10. C

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[1]

Q11. C

[1]

Q12. A

[1]

Q13. B

[1]

Q14. C

[1]

Q15. D

[1]

Q16. B

[1]

Q17. D

[1]

Q18.

(a) f = = 50 (Hz) (1)

ω (= 2πf) = 314 (rad s–1) (1)

(b) α = (rω2) = 95 × 10–3 × 3142 = 9.4 × 103 m s–2 (1)

(c) (inwards) towards axis of rotation (1) [5]

Q19.

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(a) v = ωr or v = or v = 2πrf C1

ω = 2π × 45 / 60 or correct substitutions for v C1

0.59 ms–1

A1

(b) (i) radial arrow from D towards centre of disc B1

(ii) a = or a = ω2r condone a = ω2x but not a = – (2πf)2x

2.78 m s–2 but not if shm equation clearly used A1

(c) recognition that closer toward centre particles need smaller centripetal force

B1

support for this: v ∝ r or ω = constant along disc B1

idea that friction / electrostatic forces are sufficient to meet the requirements of particles close to centre but not for those further away

B1 [9]

Q20. D

[1]

Q21. C

[1]

Q22. (a) kinetic energy = mgh (1) = 0.37 J (1)

(b) υ = (1) = 2.22 ms–1 (1)

(c) Fc = 2.9 N [or 3.0 N if g = 10 used] (1)

(d) T = Fc + W = 4.4 N (1)

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[6]

Q23. (a) attempt to use power = mgh/t or P = Fv and v = s/t

C1

7546/7550/7600

A1

W (allow J s–1 and condone N ms–1)

B1 3

(b) loss of GPE = 550 × 9.81 × 35 = 189 kJ

C1

gain in KE = 0.5 × 550 × 222 = 133 kJ

C1

resistance force = their difference/63 (890 N if correct)

A1

answer to 2 sf (allow if answer is from working even if incorrect)

B1 4

(c) air resistance varies/increases

B1

frictional force varies/increases

B1

further detail: air resistance increases with speed/v or normal reaction force varies with angle of the slope

B1 3

(d) use of F = mv2/r

C1

arrives at r = 12 m (ignoring the weight)

C1

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16.4 m

A1 3

[13]

Q24. (a) (i) Weight / W / mg − vertically downwards from some point

on the body

B1

Friction − vertically upwards and touching both the wall and the body

B1

Centripetal force / normal reaction / R − horizontally to the left from the body Each must be correct and correctly labelled Minus one for each additional inappropriate force

B1 3

(ii) Centripetal force / reaction / R is smaller

B1

Frictional force reduces Frictional force is less than weight Resultant force is downward Friction is proportional to (normal) reaction

B1 2

(b) (i) rω2 = 29 or v2 / r = 29

B1

Use of correct radius leading to 3.590 (rad s−1 ) to at least 3 sig figs

2.54 using wrong r = 1 mark

B1 2

(ii) Angular acceleration, α = 3.6 / 20 OR 3.59 / 20 or 0.18 or 0.1795

C1

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3.8 (3.77, 3.78) × 104 cao

A1

N m or kg m2 s−2

B1 3

(iii) 2200 N cao

B1 1

(c) (i) C

B1 1

(ii) Speed greatest (as all PE turned to KE)

B1

Total reaction force = mrω2 + mg or v2 / r + mg or R is largest or R = ma + mg OR Acceleration = v2 / r

B1 2

[14]

Q25. A

[1]

Q26. B

[1]

Q27. D

[1]

Q28. B

[1]

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Q29. D

[1]

Q30. D

[1]

Q31. (a) (i) arrows to show R (or N) vertically up and mg (or W)

vertically down and along the same line (within ± 2 mm) 1

(ii) mg – R = R = mg – 1

(iii) use of R = m gives R = 12(9.81 – )

= 55 (54.6) (N) 2

(b) R decreases (as v increases)

because mg is unchanged but is larger

at higher speeds R becomes = 0 [or package is not in contact with the floor]

supported by calculation eg when v = 15 m s–1, R = 0.33 N (or ≈ 0)

max 3 [7]

Q32. (a) (i) loss of PE = gain of KE or mgh = ½mv2

allow for statement of conservation of energy

(energy can not be destroyed but can be converted from one form to another)

B1

correct height used (2.4 m or 2 × 1.2 seen in an equation) B1

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correct substitution including values for h and g (no u.p.) B1

(3)

(ii) F = mv2 / r

(allow mrω2

C1

2800 N (2780 N) or

2700 N (2740 N) if using v = 6.86 m s–1

A1 (2)

(iii) (ii) + 690 (3500 N or 3460 N)

(3400 N or 3430 N if using v = 6.86 m s–1) B1

(1)

(iv) graph shape down up down up (condone linear); minima at 90° and 270°

M1

graph starts at 690 (N); this point labelled;

maximum labelled consistent with answer to (iii),

zero at 90 and 270 (allow any shape between these points) A1

(2)

(b) stress = F / A and strain = extension /original length and E = stress / strain

or

E = Fl / Ae C1

correct substitution using 690 N (condone 700 N)

or substitution with e.c.f. from graph C1

allow e.c.f. for use of g without substitution if penalised in (i)

8.9 × 10–6 – 9.1 × 10–6 m A1

allow only 1 mark if candidate divides by 2 at any stage (3)

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Q33. (a) (i) a normal reaction shown and labelled on either diagram

B1

a frictional force correctly shown and labelled on either diagram (may be outward on second diagram)

deduct 1 mark for each wrong force (condone poor friction / reaction)

B1 (2)

(ii) friction (between surface and wheel / tyre) B1

(normal) reaction (at the surface) B1

horizontal component of either force / component towards the centre

B1

sum of horizontal components B1

(4)

(b) use of mg = mv2 / r or g = v2r, centripetal force = mv2 / r C1

correct substitution v2 = 9.8 × 5.2 C1

7.1 m s–1

A1 (3)

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Q34.

(a) (i) ω ( = 5.73 rad s−1) ✓

θ( = ωt ) = 5.73 × 0.40 = 2.3 (2.29) (rad) ✓

= × 360 = 130 (131) (degrees) ✓

[or s(( = vt) = 8.6 × 0.40 ( = 3.44 m) ✓

θ = × 360 ✓ = 130 (131) (degrees) ✓ ]

Award full marks for any solution which arrives at the correct answer by valid physics.

3

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(ii) tension F(=mω2r) = 0.25 × 5.732 × 1.5 ✓ = 12(.3) (N) ✓

[or F = ✓ = 12(.3) (N) ✓ ]

Estimate because rope is not horizontal.

2

(b) maximum ω = (= 12.6) (rad s−1) ✓

maximum f = 2.01 (rev s−1) ✓

[or maximum v = = (= 19.0) (m s−1) ✓

maximum f = = 2.01 (rev s−1) ✓ ]

Allow 2 (rev s−1) for 2nd mark. Ignore any units given in final answer.

2

(c) The student’s writing should be legible and the spelling, punctuation and grammar should be sufficiently accurate for the meaning to be clear. The student’s answer will be assessed holistically. The answer will be assigned to one of three levels according to the following criteria. High Level (Good to excellent): 5 or 6 marks The information conveyed by the answer is clearly organised, logical and coherent, using appropriate specialist vocabulary correctly. The form and style of writing is appropriate to answer the question. The student appreciates that the velocity of the ball is not constant and that this implies that it is accelerating. There is a comprehensive and logical account of how Newton’s laws apply to the ball’s circular motion: how the first law indicates that an inward force must be acting, the second law shows that this force must cause an acceleration towards the centre and (if referred to) the third law shows that an equal outward force must act on the point of support at the centre. The student also understands that the rope is not horizontal and states that the weight of the ball is supported by the vertical component of the tension.

A high level answer must give a reasonable explanation of the application of at least two of Newton’s laws, and an appreciation of why the rope will not be horizontal.

Intermediate Level (Modest to adequate): 3 or 4 marks

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The information conveyed by the answer may be less well organised and not fully coherent. There is less use of specialist vocabulary, or specialist vocabulary may be used incorrectly. The form and style of writing is less appropriate. The student appreciates that the velocity of the ball is not constant. The answer indicates how at least one of Newton’s laws applies to the circular motion. The student’s understanding of how the weight of the ball is supported is more superficial, the student possibly failing to appreciate that the rope would not be horizontal and omitting any reference to components of the tension.

An intermediate level answer must show a reasonable understanding of how at least one of Newton’s laws applies to the swinging ball.

Low Level (Poor to limited): 1 or 2 marks The information conveyed by the answer is poorly organised and may not be relevant or coherent. There is little correct use of specialist vocabulary. The form and style of writing may be only partly appropriate. The student has a much weaker knowledge of how Newton’s laws apply, but shows some understanding of at least one of them in this situation. The answer coveys little understanding of how the ball is supported vertically.

A low level answer must show familiarity with at least one of Newton’s laws, but may not show good understanding of how it applies to this situation. References to the effects of air resistance, and/or the need to keep supplying energy to the system would increase the value of an answer.

The explanation expected in a competent answer should include a coherent selection of the following points concerning the physical principles involved and their consequences in this case.

• First law: ball does not travel in a straight line, so a force must be acting on it

• although the ball has a constant speed its velocity is not constant because its direction changes constantly

• because its velocity is changing it is accelerating • Second law: the force on the ball causes the ball to

accelerate (or changes the momentum of it) in the direction of the force

• the acceleration (or change in momentum) is in the same direction as the force

• the force is centripetal: it acts towards the centre of the circle

• Third law: the ball must pull on the central point of support with a force that is equal and opposite to the force pulling on the ball from the centre

• the force acting on the point of support acts outwards • Support of ball: the ball is supported because the rope is

not horizontal • there is equilibrium (or no resultant force) in the vertical

direction • the weight of the ball, mg, is supported by the vertical

component of the tension, F cos θ, where θ is the angle

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between the rope and the vertical and F is the tension • the horizontal component of the tension, F sin θ, provides

the centripetal force m ω2 r Credit may be given for any of these points which are described by reference to an appropriate labelled diagram.

A reference to Newton’s 3rd law is not essential in an answer considered to be a high level response. 6 marks may be awarded when there is no reference to the 3rd law.

max 6 [13]

Q35. (a) Radius of orbit = 5 tan 28.5 + 1 = 3.71 m "#

1

Speed = 2 × 3.14 × 3.71/5.2 = 4.49 () "# For second mark only allow Use of sin 28.5 gives orbit radius 3.39 m and speed = 4.1 m s–1

Or Forgets to add 1 giving radius 2.71 and speed 3.27 m s–1

1

(b) Centripetal force = 85 × 4.492/3.71 = 460 N "# 470 N if using 4.5 m s–1 leads to 1000 N

1

Centripetal force = T sin 28.5 "# Allow the following as ecf: Forgetting to add the 1 m ( using r = 2.71 m) leads to centripetal force = 630 N T = 1300 N) Using r =3.39 m as ecf from part (e) which leads to Centripetal force = 510 N giving T = 1070 N

1

T = 950 - 970 N "# 1

OR

Weight = 85 × 9.8(1) 0r 834 N seen "#

Weight = T cos 28.5 "#

T = 950 (949) (N) "#

OR

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Centripetal force = 85 × 4.52/3.71 = 464 N "#

Weight = 834 N "#

T = = 950 – 970 N "#

Allow ecf for incorrect weight or centripetal force Allow the following as ecf: Forgetting to add the 1 m (using r = 2.71 m) leads to Centripetal force = 630 N, T = 1050 N Using r = 3.39 m leads to Centripetal force = 510 N giving T = 980N

(c) Vertical (compressive) force on the pole increases "# 1

Increases mass increases weight and hence tension in the rope(for the same angle) 1 "#

1

Centripetal Force

on the acrobats/masses would be different/not equal

OR

Would be greater on the more massive acrobat(travelling at the same speed/same angle to vertical) 1 "#

Unbalanced (horizontal) forces/resultant force exists (on the pole) "#

OR

Unbalanced moments acting (on pole)/resultant torque acting (on pole) "#

Causing the pole to sway/bend/move/ or tilt/topple the platform toward more massive acrobat 1 "#

1 Max 3

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Year 12 Physics Remote Learning Phase 3 - Ark Globe Academy

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Transcript of Year 12 Physics Remote Learning Phase 3 - Ark Globe Academy