As Physics ISP Mechanics

Subject: Physics Teacher:Student: Unit Unit 1

Study Task outline:Mechanics ISPThis ISP covers the Mechanics part of Physics Unit 1: Physics on the Go. While much of the topic contains mathematical concepts which you will be asked quantitatively (using numbers) you may be asked to explain these ideas qualitatively (using words). This ISP will help you build the skills required to answer both styles of questions and aid you to build your knowledge for your assessments. It is of imperative importance that you can answer both quantitative and qualitative Physics questions; without this skill you will struggle to do any better than a D grade. To achieve a grade A-C you will need to be able to bring together different areas of Physics you have learnt (e.g. linking forces with equilibrium). You must also be able to use an equation in a qualitative manner, explaining the implications of changing a variable in the equation. The specification points by this ISP are all of the specification points within the Mechanics module (see your introduction to AS Physics booklet).Task Task detail Resources

neededCompleted/Comments

1 Specification points covered : 2, 3, 5You should read through the notes from lessons and p10-15 from the course textbook. The areas to focus on are:

1. the differences between scalars and vectors2. a list of base units,3. information that can be deduced from

displacement, velocity and acceleration time graphs

Complete Explaining Everything 1Complete questions 1 to 5 from your Mechanics ISP booklet

AS Physics Text book

Explaining Everything number 1

Mechanics ISP booklet

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2 Specification points covered : 1Students should read through their notes from lessons and p16-17 from the course textbook. The areas to focus on are:

1. the equations of motion (uniform acceleration)

2. the experiment and demonstrations we performed to demonstrate the equations of motion and the value of g





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3 Specification points covered : 6,7,9,10,11You should read through the notes from lessons and p18-30 from the course textbook. The areas to focus on are:

1. trigonometry and resolving vectors2. Newton’s 3 laws and their applicationsComplete Explaining Everything 3Complete questions 10 to 14 from your Mechanics ISP booklet




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Individual Study Plan: Mechanics

4 Specification points covered : 8,You should read through the notes from lessons and p31-33 from the course textbook. The areas to focus on are:

1. Equilibrium (statics) and resolving forces to find horizontal and vertical components.





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4,6,7,12,13,14,15,16,17You should read through the notes from lessons and p34-44 from the course textbook. The areas to focus on are:

1. How to resolve velocities into their horizontal and vertical components.

2. How to find the time of flight and maximum range of a projectile

3. How energy, work and power are linked, and how energy transfers can be described.





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List of data, formulae and relationships

Data

Acceleration of free fall g = 9.81 m s–2 (close to Earth’s surface)

Boltzmann constant k = 1.38 × 10–23 J K–1

Coulomb’s law constant k = 1/4 oεΠ

= 8.99 × 109 N m2 C–2

Electron charge e = –1.60 × 10–19- C

Electron mass me = 9.11 × 10–31 kg

Electronvolt 1 eV = 1.60 × 10–19 J

Gravitational constantG = 6.67 × 10–11 N m–2 kg–2

Gravitational field strengthg = 9.81 N kg–1 (close to Earth’s surface)

Permittivity of free spaceoε = 8.85 × 10–12 F m–1

Planck constant h = 6.63 × 10–34 J s

Proton mass mp = 1.67 × 10–27 kg

Speed of light in a vacuumc = 3.00 × 108 m s–1

Stefan Boltzmann constant σ 5.67 × 10–8 W m–2 K–4

Unified atomic mass unit u = 1.66 × 10–27 kg

Mechanics

Kinematic equations of motion v = u + at

s = ut + ½at2

v2 = u2 + 2as

Forces ΣF = ma

g = F/m

W = mg

Work and energy ΔW = FΔs

Ek = ½ mv2

ΔEgrav = mgΔh

Materials

Stokes’ law F = 6πηrv

Hooke’s law F = kΔx

Density ρ = m/V

Pressure p = F/A

Young’s modulus E = σ/ε where

Stress σ = F/A

Strain ε =Δ x/x

Elastic strain energy Eel = ½ Fx

Explaining Everything 1 – Distance velocity and acceleration time graphs

What can we conclude from the shape of a Distance time graph, and a velocity time graph?

Key words to include: Area underneath Tangent Gradient Velocity Displacement Acceleration

Explain how we can use ‘strobe photography’ or ‘video analysis software’ to analyse the motion of moving objects.

Key phrases to include: Time Displacement Reference point

Explaining Everything 2 – Distance velocity and acceleration time graphs



What can we conclude from the shape of a Distance time graph, and a velocity time graph?

Key words to include: Area underneath Tangent Gradient Velocity Displacement Acceleration



Explaining Everything 3 – Newton’s Laws



Explain how you can use Newton’s three laws to describe a bouncing ball.

Key phrases to include: Balanced Unbalanced Equal Reaction Acceleration Line of action Opposite

Explaining Everything 4 - Statics

Explain how you can use Newton’s three laws to describe a bouncing ball.

Key phrases to include: Balanced Unbalanced Equal Reaction Acceleration Line of action Opposite

Explain how an object can be travelling at a constant velocity down a slope. Draw a diagram, a free body force diagram, and a vector triangle to explain how you could calculate the size of the forces acting on the object.

Key phrases to include: Balanced Equal Reaction Acceleration Perpendicular Opposite Component

Frictional Slope Trigonometry

Explaining Everything 5 – Projectiles and Energy

Explain how an object can be travelling at a constant velocity down a slope. Draw a diagram, a free body force diagram, and a vector triangle to explain how you could calculate the size of the forces acting on the object.

Key phrases to include: Balanced Equal Reaction Acceleration Perpendicular Opposite Component

Frictional Slope Trigonometry

Explain how you could calculate the range and maximum height of a ball thrown at an angle θ with a velocity of u.

Key phrases to include: Resolve Negligible Components Horizontal Vertical Air resistance

Explain how you could calculate power of an electric motor that is lifting a mass.

Key phrases to include: Gravitational potential energy Kinetic energy Efficiency Calculate Power

Wootton Upper School

ISP - Mechanics BookletTask 1

1. The graph below shows how the velocity of a motorbike varies with time during the final 10 s of a race.

9 0

8 0

7 0

6 0

5 0

4 0

3 0

2 0

1 0

00 1 2 3 4 5 6 7 8 9 1 0

/ m s – 1

t / s

(a) (i) Describe the motion shown by the graph.

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(ii) Show that during the final 10 s the motorbike travels a distance of approximately 800 m.

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Explain how you could calculate power of an electric motor that is lifting a mass.

Key phrases to include: Gravitational potential energy Kinetic energy Efficiency Calculate Power

(b) Using the axes below, sketch a graph showing how the distance of the motorbike from the finishing line varies with time during the final 10 s of the race.

8 0 0

7 0 0

6 0 0

5 0 0

4 0 0

3 0 0

2 0 0

1 0 0

00 1 2 3 4 5 6 7 8 9 1 0

t / s

D is tan ce fro mfin ish in glin e / m

(3)(Total 8 marks)

2. A hot-air balloon is rising vertically at a speed of 10 m s–1. An object is released from the balloon. The graph shows how the velocity of the object varies with time from when it leaves the balloon to when it reaches the ground four seconds later. It is assumed that the air resistance is negligible.

1 5

1 0

5

0

5

– 1 0

– 1 5

– 2 0

– 2 5

– 3 0

– 3 5

tim e /s

v e lo c ity/m s

0 .5 1 .0 1 .5 2 .0 2 .5 3 .0 3 .5 4 .0 4 .5

– 1

(a) Use the graph to

(i) show that the object continues to rise for a further 5 m after it is released.

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...........................................................................................................................(1)

(ii) determine the total distance travelled by the object from when it is released from the balloon to when it reaches the ground.

...........................................................................................................................

...........................................................................................................................

Total distance = .......................................(2)

(b) Hence determine the object’s final displacement from its point of release from the balloon. ......................................................................................................................................

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Displacement = .........................................................................(2)

(c) Using the axes below, sketch a graph showing how the acceleration of the object changes during the time from when it leaves the balloon to when it hits the ground.Mark any significant values on the axes.

- - - - - - - -acce le ra tio n

/ m s

tim e / s0

1 .0 1 .5 2 .0 2 .5 3 .0 3 .5 4 .0 4 .5

-

0 .5

– 1

(3)(Total 8 marks)

3. (a) State the difference between distance and displacement.

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(b) Figure 1 shows an idealised displacement-time graph for the journey of a train along a straight horizontal track, from the moment when it passes a point A on the track.Initially the train moves in an easterly direction away from A.

Figure 1

D isp la cem en tfro m A/m

Tim e /m in u te s1 2 3 4 5 6 7 8 9

7 0 0

6 0 0

5 0 0

4 0 0

3 0 0

2 0 0

1 0 0

0

-1 0 0

-2 0 0

-3 0 0

(i) Describe the position of the train relative to A at the end of the 8 minutes covered by the graph.

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...........................................................................................................................(2)

(ii) Use the grid, Figure 2, to plot a velocity against time graph of the journey shown in Figure 1. Do the calculations that are required on the lines below the grid.

Figure 2

Velo c ity/m s -1

01 2 3 4 5 6 7 8 9

T i m e / m i n u t e s

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...........................................................................................................................(4)

(Total 7 marks)

4. A cyclist and a car are both stationary at traffic lights. They are alongside each other with their front wheels in line. The lights change and they both move forward in the same direction along a straight flat road. The idealised graph shows the variation of velocity against time for both the cyclist and the car from the instant the lights change to green to the instant they are again level.

C y c lis t

C a r1 8

1 6

1 4

1 2

1 0

8

6

4

2

00 1 2 3 4 5 6 7 8 9 1 0 11 1 2

T im e /s

Ve lo c ity/m s – 1

(a) What does the time interval of 0.8 s at the beginning of the graph represent?

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.....................................................................................................................................(1)

(b) (i) How long does it take, from the instant the lights change to green, for the car to reach the same velocity as the cyclist?

...........................................................................................................................(1)

(ii) Determine the distance between the cyclist and the car at this time.

...........................................................................................................................

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...........................................................................................................................

Distance = ..........................................................(3)

(c) What is the relationship between the average velocity of the cyclist and the average velocity of the car for the time interval covered by the graph?

.....................................................................................................................................(1)

(Total 6 marks)

5. An athlete runs a 100 m race. The idealised graph below shows how the athlete’s velocity v changes with time t for a 100 m sprint.

u /

u

m s – 1

m a x

00

2 4 6 8 1 0 1 2t s/

By considering the area under the graph, calculate the maximum velocity vmax of the athlete.

...............................................................................................................................................

...............................................................................................................................................

...............................................................................................................................................

...............................................................................................................................................

Maximum velocity = ........................................(3)

Using the axes below, sketch a graph showing how the acceleration of this athlete changes with time during this race. Mark any significant values on the axes.

A cce le ra tio n

00 Tim e

(4)(Total 7 marks)

Task 2

6. A lorry is travelling at 25 m s–1 down a mountain road when the driver discovers that the brakes have failed. She notices that an escape lane covered with sand is ahead and stops her lorry by steering it on to the sand.

E scap e lan e

S an d

The lorry is brought to a halt in 40 m. Calculate the average deceleration of the lorry.

................................................................................................................................................

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................................................................................................................................................

Average deceleration =............................(3)

Suggest how the depth of the sand affects the stopping distance. Justify your answer.

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(Total 4 marks)

7. A student is working on a spreadsheet to model the fall of a golf ball from rest from the window of a tall building.

(a) He assumes that the acceleration remains constant at 9.81 m s–2 for the first two seconds of the fall. Comment on whether this is a reasonable assumption.

.....................................................................................................................................

.....................................................................................................................................(1)

A B C D

1

time fromstart / s

velocity reached/

m s–1

distance fallenduring 0.20 s time

interval / m

total distancefrom the start

/m23 0.00 0.00 0.00 0.004 0.20 1.96 0.20 0.205 0.40 3.92 0.59 0.786 0.60 5.89 0.98 1.777 0.80 7.85 1.37 3.148 1.00 9.81 1.77 4.919 1.20 11.77 2.16 7.0610 1.40 13.73 2.55 9.6111 1.60 15.70 2.94 12.5612 1.80 17.66 3.34 15.8913 2.00 19.62 3.73 19.62

(b) Cell B6 is calculated using the formula B6 = 9.81*A6. Explain why this is appropriate.

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(c) Cell C7 is calculated using the formula C7 = ((B6+B7)/2)*0.20.

(i) Explain what (B6+B7)/2 represents.

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...........................................................................................................................(1)

(ii) Why is this fraction multiplied by 0.20?

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(d) Give an appropriate spreadsheet formula that uses cell D9 to calculate cell D10.

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(e) You can check that this spreadsheet model is giving sensible answers for the total distance fallen by calculating the distance using an equation from the list of formulae at the back of the paper. Calculate the answer for cell D11.

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(Total 8 marks)

8. An astronaut on the moon drops a hammer. The gravitational acceleration is 1.6 m s–2.

(a) How long does the hammer take to fall 1.0 m from rest?

Time = .....................................(2)

(b) Calculate the velocity of the hammer just before it hits the ground.

Velocity = ................................(2)

(Total 4 marks)

9. A careless soldier shoots a bullet vertically into the air at 450 m s–1. Calculate the time the bullet takes to reach the top of its flight. State any assumption you have made.

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Time ………………………………(3)

Sketch and label fully a velocity-time graph for the bullet’s complete flight.

Explain the shape of your graph.

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Use your graph to calculate the distance travelled by the bullet before it hits the ground.

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(Total 10 marks)

Task 3

10. (a) Complete the following statement of Newton’s third law of motion.

“If body A exerts a force on body B, then body B .....................................................

....................................................................................................................................”(2)

(b) A man checks the weight of a bag of potatoes with a newtonmeter. Two of the forces acting are shown in the diagram.

P u sh o fg ro u n d o nm an

N ew to n m ete r

W eig h t o f p o ta to e s

The table below gives these forces. For each force there is a corresponding force, the ‘Newton’s third law pair force’. In each case state

the body that the Newton’s third law pair force acts upon

the type of force (one has been done for you)

the direction of the Newton’s third law pair force.

Force Body the Newton’sthird law pair forceacts upon

Type of force Direction of theNewton’s third lawpair force

Weight ofpotatoes

Push of groundon man

Normal contact force

(3)(Total 5 marks)

11. (a) The diagram below shows the forces acting on a shopping trolley at rest.

N o rm a lco n tac t

fo rce

N o rm a lco n tac t

fo rce

W eig h t

(i) State Newton’s first law of motion.

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(ii) In everyday situations, it does seem that a force is needed to keep an object, for example the shopping trolley, moving at constant speed in a straight line.Explain why.

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...........................................................................................................................(1)

(iii) The vertical forces acting on the trolley are in equilibrium. Explain what equilibrium means.

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(b) (i) The weight of the trolley is one of a Newton’s third law force pair. Identify what the other force in this pair acts upon and what type of force it is.

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(ii) Give two reasons why the two normal contact forces do not form a Newton’s third law pair.

1 ........................................................................................................................

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2 ........................................................................................................................

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(Total 7 marks)

12. The diagrams show a man standing on the Earth and two free-body force diagrams, one for the man and one for the Earth.

E arth

M a n E arth

A

B C

D

Force A can be described as ‘the Earth pulling the man down with a gravitational force’. Use a similar form of words to describe force C which forms a Newton third law pair with force A.

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Noting that forces A and C are a Newton third law pair, write down three similarities and two differences between these two forces.

Similarities:

(i) ...............................................................................................................................

(ii) ..............................................................................................................................

(iii) .............................................................................................................................

Differences:

(i) ...............................................................................................................................

(ii) ..............................................................................................................................(5)

Which two forces show whether or not the man is in equilibrium?

...............................................................................................................................................(1)

(Total 8 marks)

13. A child is crouching at rest on the ground

Below are free-body force diagrams for the child and the Earth.

A

B

D

C

E ar th

Complete the following table describing the forces A, B and C.

Force Description of force

Body which exerts force

Body the force acts on

A Gravitational Earth Child

B

C

(4)

All the forces A, B, C and D are of equal magnitude.

Why are forces A and B equal in magnitude?

..............................................................................................................................................

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Why must forces B and D be equal in magnitude?

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The child now jumps vertically upwards. With reference to the forces shown, explain what he must do to

jump, and why he then moves upwards.

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(Total 9 marks)

14. Figure 1 shows a box resting on the floor of a stationary lift. Figure 2 is a free-body force diagram showing the forces A and B that act on the box.

L ift f lo o r

B o x A

B

For each of the following situations, tick the appropriate boxes to show how the magnitude of the forces A and B change, if at all, compared with when the lift is stationary.

Situation Force A Force B

increases no change decreases increases no change decreases

Lift accelerating upwards

Lift moving with constant speed upwards

Lift accelerating downwards

Lift moving with constant speed downwards

(Total 4 marks)

Task 4

15. The diagram shows the two vertical forces acting on a helicopter hovering at a constant height. In this situation the two forces are equal in magnitude.

The mass of the helicopter is 1500 kg.

Calculate the magnitude of the lift force.

...............................................................................................................................................

...............................................................................................................................................

Lift force = .........................................................(1)

Forward thrust is obtained by tilting the helicopter forward by 17°. The speed of the rotor blades is increased so that the helicopter remains at the same height as it accelerates forwards.

Explain why the vertical component of the force produced by the rotor blades must still be equal in magnitude to the weight.

...............................................................................................................................................

...............................................................................................................................................(1)

The force produced by the rotor blades is now 15 400 N.

Show that the horizontal component of the force is about 4500 N.

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...............................................................................................................................................(2)

Calculate the forward acceleration of the helicopter.

...............................................................................................................................................

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Acceleration = ...................................................(2)

Calculate the horizontal distance the helicopter will have travelled from rest after 10 s assuming the acceleration is constant.

...............................................................................................................................................

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...............................................................................................................................................

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Distance = .........................................................(2)

Explain whether this is likely to be the actual distance travelled in this time.

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(Total 10 marks)

16. Two campers have to carry a heavy container of water between them. One way to make this easier is to pass a pole through the handle as shown.

The container weighs 400 N and the weight of the pole may be neglected. What force must each person apply?

(1)

An alternative method is for each person to hold a rope tied to the handle as shown below.

In the space below draw a free-body force diagram for the container when held by the ropes.

(2)

The weight of the container is 400 N and the two ropes are at 40° to the horizontal. Show that the force each rope applies to the container is about 300 N.

..............................................................................................................................................

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Force = .....................................................(3)

Suggest two reasons why the first method of carrying the container is easier.

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..............................................................................................................................................(2)

Two campers using the rope method find that the container keeps bumping on the ground. A bystander suggests that they move further apart so that the ropes are more nearly horizontal. Explain why this would not be a sensible solution to the problem.

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..............................................................................................................................................(1)

(Total 9 marks)

Task 5

17. A coin is flicked off a table so that it initially leaves the table travelling in a horizontal direction with a speed

of 1.5 m s–1. The diagram shows the coin at the instant it leaves the table. Air resistance can be assumed to have a negligible effect throughout this question.

F lo o r

C o in

D irec tio n o f m o v em en t

(a) Add to the diagram the path followed by the coin to the floor.(1)

(b) (i) The table is 0.70 m high. Show that the coin takes approximately 0.4 s to reach the floor.

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...........................................................................................................................(3)

(ii) Hence calculate the horizontal distance the coin travels in the time it takes to fall to the floor.

...........................................................................................................................

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...........................................................................................................................

Horizontal distance = ......................................................(2)

(c) A coin of greater mass is flicked with the same horizontal speed of 1.5 m s–1. Compare the path of this coin with that of the coin in the first part of the question. Explain your answer. You may be awarded a mark for the clarity of your answer.

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18. A weightlifter raised a bar of mass of 110 kg through a height of 2.22 m. The bar was then dropped and fell freely to the floor.

(i) Show that the work done in raising the bar was about 2400 J.

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.....................................................................................................................................(2)

(ii) It took 3.0 s to raise the bar. Calculate the average power used.

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Power = …………………………………….(2)

(iii) State the principle of conservation of energy.

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.....................................................................................................................................(2)

(iv) Describe how the principle of conservation of energy applies to

(1) lifting the bar,

(2) the bar falling to the floor. Do not include the impact with the floor.

(1) ...........................................................................................................................

...........................................................................................................................

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(2) ...........................................................................................................................

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...........................................................................................................................(3)

(v) Calculate the speed of the bar at the instant it reaches the floor.

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Speed = …………………………………….(3)

(Total 12 marks)

1. (a) (i) Describe motion

Constant / uniform acceleration or (acceleration of) 15 m s–2 (1)

(Followed by) constant / uniform speed / velocity (of

90 m s–1) (1) 2

(ii) Show that distance is approximately 800 mAny attempt to measure area under graph or select appropriateequations of motion required to determine total distance (1)

Correct expression or value for the area under the graph betweeneither 0-4 s [240 m] or 4-10 s [540 m] (1)

Answer : 780 (m) (1)

Eg distance = 60 m s–1 × 4 s + 90 m s–1 × 6 s= 240 m + 540 m= 780 (m)

Eg distance in first 4 s

s = 2

s m 30s m 90

2

uv 11

t

4 s = 240 mDistance in final 6 s

s = ut = 90 m s–1 × 6 s = 540 mTotal distance = 240 m + 540 m = 780 (m) 3

(b) Sketch graphGraph starts at 760 m – 800 m/their value and initially showsdistance from finishing line decreasing with time (1)The next two marks are consequent on this first mark being awarded

Curve with increasing negative gradient followed by straight line (1)

Graph shows a straight line beginning at coordinate (4 s, 540 m)and finishes at coordinate (10 s, 0 m) (1) 3

[8]

2. (a) (i) Additional heightAnswer [ 5 (m)] (1)

Eg distance = area of small triangle = 0.5 × 1 s × 10 m s–1 = 5 m 1

(ii) Total distance travelled [Allow ecf of their value]Distance travelled between 1 s and 4s [45 m] (1)Answer [ 50 m] (1)

Eg distance fallen = area of large triangle

= 0.5 × 3 s × 30 m s–1

= 45 m total distance =45m + 5m = 50m 2

(b) Objects displacement40 m (1)Below (point of release) or minus sign (1)[Ecf candidates answers for additional height and distance ieuse their distance – 2 × their additional height] 2

(c) Acceleration time graphLine drawn parallel to time axis extending from t = 0 (1)[Above or below the time axis]The line drawn parallel to the time axis extends from 0 s to 4 s (1)[If line continues beyond or stops short of 4 s do not give this mark]

Acceleration shown as minus 10 m s–2 (1)[This mark is consequent on the second mark being obtained] 3

[8]

3. (a) Displacement and distance?

Displacement has direction distance doesn’t or displacement isa vector, distance is a scalar or an explanation in terms of an example. (1) 1

[Candidates who describe displacement as “measured from a point”but do not mention direction or equivalent do not get this mark]

(b) (i) Position of train relative to A

300 m (1)

West (of) or a description[Do not accept backwards, behind or negative displacement] (1) 2

(ii) Velocity against time graph

Constant velocity shown extending from t = 0, positive / negative (1)[Above mark awarded even if graph does not reach or stop att = 4 min]

Constant velocity shown beginning at t = 4 min and ending at t =8 min, negative/positive (respectively) (1)

Values 2.5 (m s–1) or 3.75 (m s–1) or 3.8 (m s–1) seen[either calculated or on graph] (1)

Both values [allow their values] correctly plotted using a scale (1)[Only give this fourth mark if marking points 1 and 2 are correct.Also a clear scale must be seen eg 1, 2, 3, –1, –2, –3.The plot must be accurate to about half a small square.] 4

[7]

4. (a) Meaning of 0.8 sReaction time (of cyclist and car driver) (1) 1[Accept descriptions of reaction time eg ‘time it takes both to takein that the lights have changed to green’. Accept response time]

(b) (i) Same speed timeAnswer [6.8 s –6.9 s] [Accept any value in the range] (1) 1

(ii) How much further ahead?EitherFor measuring area under car graph at 6.8 s (1)

eg = 2

s m 9 s 6 1

= 27 m [27.5 m if 6.9 s used]For measuring area under cyclist graph at 6.8 s (1)

eg 2

s m 9 s 2 1

+ 4 s × 9 m s–1 = 45 m [45.9 m if 6.9 s used]

[For candidates who read the velocity 9 m s–1 as 8.5 m s–1 butotherwise do their calculation(s) correctly give 2/3][Allow one mark to candidates who attempt to measure anappropriate area]Answer [(45 m – 27 m =) 18 m] (1)

OrFor recognising the area enclosed by cyclist and car graphsas the difference in distance travelled (1)Using values from the graph to determine this area (1)Answer [(45 m – 27 m =) 18 m] (1)

eg distance = 2

1

× (6.8 – 2.8) s × 9 m s–1

= 18 m 3

(c) Relationship between average velocitiesThey are the same (1) 1

[6]

5. Maximum velocity

Area = 100 m (1)

Attempt to find area of trapezium by correct method (1)

= 10 m s–1 (1) 3

Sketch graph

Horizontal line parallel to x axis

Some indication that acceleration becomes 0 m s–2

The initial acceleration labelled to be max 2 [ initial a = 5 (m s–2) (1)(ecf)]

t = 2 (s) where graph shape changes (1) 4[7]

6. Average deceleration

Select 2 = u2 + 2ax, ½ m 2 = Fx and F = ma OR equations of motion (1)

Correct substitutions of 40 m and 25 m s–1 (1)

a = 7.8 m s–2 [If a = –7.8 m s–2 2/3] (1) 3

Depth of sand and stopping distance

More sand shorter stopping distance/stops more quickly/slowsdown faster Because lorry sinks further/ bigger resistingforce / bigger friction force (1) 1

[4]

7. (a) Comment on assumption

Yes – air resistance negligible OR still close to Earth (ignore upthrust)or No – air resistance becomes significant (1) 1

(b) Explanation of why formula for cell B6 is appropriate

Recall of v = u + at (accept ∆v = a ∆t or ∆v = at for 1st mark) (1)

(v is B6), u is zero, a is 9.81 [m s–2] and t is A6 (1) 2

(c) (i) Explanation of 2

B7) B6(

it is average speed (for that interval)

or 2

v)u (

(1) 1

(ii) Why 2

B7) B6(

is multiplied by 0.20

because dist = ave speed × time [accept s = vt]and 0.20 is the time (1) 1

(d) Formula for D10

= D9 + C10 (1) 1

(e) Calculation to check D11

Use of appropriate equation of motion (1)

Correct answer [12.557 m] [no ue] (1) 2

Example of calculation:

s = ut + ½ at2

= 0 + ½ × 9.81 m s–2 × (1.6 s)2

= 12.557 m

N.B. use of υ2 = u2 + 2as gives answer s = 12.563 m[8]

8. (a) Use of s = ut + ½ at2 (1)Correct answer [1.1 s] (1)


s 1.16.1

122

a

st

2

(b) Use of v = u + at (1)

Correct answer [1.8 m s–1] (1)


v = u + at = 1.6 × 1.1 = 1.8 m s–1 2[4]

9. Calculation of time bullet takes to reach top of its flight and statement of any assumption made:

– 9.8 m S–2 = (0 m s–1 – 450 m s–1)/t

t = 46 s

Assumption: air resistance is negligible, acceleration constant or equivalent 3

Sketch of velocity-time curve for bullet’s flight:

Label axes

Show the graph as a straight line inclined to axis

+ 450 m s–1 and 46 s shown correctly

– 450 m s–1 or 92 s for a correctly drawn line

4 5 0 m s

– 4 50 m s

4 6 s9 2 s t

–1

–1

Explanation of shape of graph:

Why the line is straight - acceleration constant or equivalent or why the velocity changes sign or why the gradient is negative

Calculate the distance travelled by bullet, using graph:

Identification of distance with area between graph and time axis or implied in calculation

20 700 m for g = 9.8 ms–2 or alternative answers from different but

acceptable “g” values. 7

[Allow e.c.f with wrong time value.][10]

10. (a) Complete statement of Newton’s Third Law of Motion....exerts an equal force on (body) A (1)

(but) in the opposite direction (to the force that A exerts on B) (1) 2[‘exerts an equal but opposite force on body A’ would get both marks]

(b) Complete the table

1 mark for each of the three columns (1) (1) (1) 3

[Accept from earth for up. Accept towards ground or towards earth for down]

Earth Gravitational. [Not ‘gravity’. Not

gravitational field strength]

Up(wards) /

Ground Down(wards)

/ [5]

11. (a) (i) Newton’s First law of Motion

An object will remain (at rest or) uniform/constant velocity/speed/motionin a straight line unless (an external/impressed) force acts upon it /provided resultant force is zero. (1) 1

(ii) Everyday situation

Reference to air resistance / friction / drag etc. (1) 1

(iii) Equilibrium

The resultant force is zero / no net force /sum of forces is zero /forces are balanced / acceleration is zero (1) 1[Accept moments in place of force]

(b) (i) Identify the other force

Earth (1)Gravitational [consequent on first mark] [Do not credit gravity.] (1) 2

(ii) Why normal contact forces are not a Newton’s third law pair

Do not act along the same (straight) line / do not act from the same point (1)They act on the same body (1)They act in the same direction / they are not opposite forces (1)They are of different magnitudes (1)

max 2[7]

12. Description of force C which forms a Newton’s third law pair with A

Man pulling Earth upwards

with a gravitational force 2

Similarities and differences

Similarities [any 3]:

Magnitudes or equalKind (or type) of force or gravitational forcesLine of action [but not same plane, or point, or parallel]Time interval or durationConstant [not true in general but true in this instance] Max 3

Differences:

On different bodies [must say “bodies” or equivalent]

Direction [again, it answers this particular question] or opposite 2

Two forces which show whether or not man is in equilibrium:

A and B 1[8]

13. Completion of table:

Force Description of force Body which exerts force

Body the force acts on

A Gravitational Earth Child

B (Normal) reaction OR contact OR E/M (1)

Earth child

C Gravitational [Not gravitational weight] (1)

Child Earth(1) for both

4

Why A and B are equal in magnitude:

Child is at rest/equilibrium OR otherwise child would move/accelerate (1)[NB use of N3 would contradict this]

Why must forces B and D be equal in magnitude:

Newton’s third law OR action + reaction equal and opposite (1)[NB use of N1 or N2 here would contradict this] [Not Newton pair] 2

What child must do to jump and why he moves upwards:

Push down, increasing D (1)

B increases [must be clearly B or description of B] (1)

and is > A OR there is a resultant upward force [clearly on child] (1)[Not “movement”] 3

[9]

14. How A and B changeForce BFor ticking ‘no change’ in all 4 boxes (1)

Force A4 ticks right (1)3 ticks right (1)2 ticks right (1)

increases no change decreases

[4]

15. Show that lift is about 14 700 N

Lift = weight = mg

= 1500 kg × 9.81 N kg–1

= 14 700 N (1) 1

Explanation of why vertical component equals weight

No vertical acceleration / resultant vertical force = zero / verticalforces balanced (1) 1

Show that horizontal component is about 4500 N

Horizontal component = Fsin

OR

= 15 400 N × sin 17° OR 15 400 N × cos (90° – 17°) (1)= 4503 N [no up] (1) 2

Calculation of forward acceleration

a=F / m (1)

= 4503 N ÷ 1500 kg

= 3.0 m s–2 (1) 2

Calculation of distance travelled after 10 s

s = ut + ½ at2

= 0 + ½ × 3.0 m s–2 × (10 s)2 [e.c.f.] (1)

= 150 m (1) 2

Explanation of whether likely to be actual distance

Distance likely to be less (1)

Air resistance / drag will decrease resultant force / acceleration (1) 2[10]

16. Force:

200 N (1) 1

Free body force diagram:

[Unlabelled arrows = 0; ignore point of application of force][Double arrows = 0] [Arrows required for marks]

W / weight / 400 N /mg/ pull of Earth / gravitational force (1)[Not gravity]

Two tensions OR Two × 300 N OR two × 311 N (1)[Accept T for tension OR any label that is not clearly wrong,e.g. R/W/N 200 N] 2

Applied force:

Attempt to resolve vertically (1)

2T sin 40 = 400 (1)

[400 ×cos 40 306 N(no marks)400 × sin 40 257 N (no marks)

200/cos 40° = 261 N gets 1 out of 3 (attempted to resolve)]

T = 310 (N) OR 311 (N) [No unit penalty] (1) 3

Two reasons why first method is easier:

Force applied is smaller/feels lighter/tension smaller [Not weighs less] (1)

They are not pulled sideways/forces only upwards/pulling against each other (1)[Answer must be in terms of forces] 2

Why solution is not sensible:

Because the tension (or description of tension) would be greater (1)OR bigger sideways force 1

[Do not accept bigger force][9]

17. (a) Path of coinCurved line that must begin to ‘fall’ towards the ground immediately (1) 1

(b) (i) Show that..

Selects s = (ut +) 2

1

at2 or selects two relevant equations (1)Substitution of physically correct values into equation or both (1)equations.Answer [0.37 s – 0.38 s] (1)

[Allow use of g = 10 m s–2. Must give answer to at least 2 sig. fig.,bald answer scores 0. No ue.] 3

eg 0.7 m = 2

1

(9.81 m s–2)t2

(ii) Horizontal distance [ecf their value of t]

Use of v = t

d

with correct value of time. [s = t

uv

2

is sometimes (1)used. In this case v and u must be given as 1.5 m s–1 and t must

be correct. Also s = ut + 0.5at2 OK if ‘a’ is set = 0.]Answer [0.55 m – 0.60 m] (1)

eg d = 1.5 (m s–1) × 0.38 (s)= 0.57 m 2

(c) A coin of greater mass?QWOC (1)It will follow the same path [accept ‘similar path’,do not accept ‘same distance’] (1)All objects have the same acceleration of free fall / gravity oracceleration of free fall / gravity is independent of mass / it will takethe same time to fall (to the floor) (1)Horizontal motion / velocity is unaffected by any force or (gravitational)force (acting on coin) has no horizontal component or horizontalmotion/velocity is the same/constant. (1) 4

[10]18. (i) Work done

Use of work done = force × distance (1)

Answer given to at least 3 sig fig. [2396 J, 2393 J if 9.8 m s–2 is used, (1) 2

2442 J if g = 10 m s–2 is used. No ue.]

Work done = 110 kg × 9.81 m s–2 × 2.22 m = 2395.6 J

(ii) Power exerted

Use of power = timedonework

or power = F × v (1)

Answer: [799 W. 800 W if 2400 J is used and 814 W if 2442 J is 2used. Ecf value from (i)] (1)

Power = 3sJ2396

= 798.6 W

(iii) Principle of Conservation of Energy

EitherEnergy can neither be created nor destroyed (1) (1)OREnergy cannot be created/destroyed or total energy is not lost/gained (1)(merely) transformed from one form to another or in a closed/isolatedsystem. (1) 2

[Simple statement ‘Energy is conserved’ gets no marks][Information that is not contradictory ignore. Q = U + W, withterms defined acceptable for 1st mark]

(iv) How principle applied to...

Lifting the bar: -Chemical energy (in the body of the weightlifter) or work done(lifting bar) = (gain in) g.p.e. (of bar) (1)[Reference to k.e. is acceptable]

The bar falling: -Transfer from g.p.e. to k.e. (1)(and that) g.p.e. lost = k.e. gained (1) 3

[‘g.p.e. converted to k.e.’ would get one mark][References to sound and thermal energy are OK, but gpe to sound orthermal energy on its own gets no marks]

(v) Speed of bar on reaching the floor

Setting ½ mv2 = m g h or ½ mv2 = work done or 2400 J (1)[ecf their value][Shown as formulae without substitution or as numbers substitutedinto formulae]Correct values substituted (1)

[allow this mark if the 110 kg omitted – substitution gives v2 = (1)

43.55(6) m2 s–2 or 44.4 m2 s–2 if g = 10 m s–2 is used]

Answer: [6.6 m s–1. 6.7 m s–1 if g = 10 m s–2 is used.]

½ 110 kg × v2 = 110 kg × 9.81 m s–2 × 2.22 m or = 2400 J / 2396 J

v = 6.6 m s–1 [6.66 m s–1 if 10 m s–2 used] (1)

OR

Selects v2 = u2 + 2as or selects 2 relevant equations (1)Correct substitution into equation (1)

Answer [6.6 m s–1] (1)

v2 = 0. + 2 × 9.81 ms–2 × 2.22m 3

v = 6.6 m s–1

[12]

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