Velocity/Distance/ Accceleration (inc Graphs)

Velocity/Distance/ Accceleration (inc Graphs)

Mark Scheme 1

Level Edexcel

Subject Physics

Exam Board GCSE(9-1)

Topic Motions and Forces

Sub Topic Velocity/Distance/Acceleration(inc Graphs)

Booklet Mark Scheme 1

57 minutes

/47

/100

Time Allowed:

Score:

Percentage:

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Question

Number

Answer Acceptable answers Mark

1(a) stopwatch /stopclock (1)

{trundle/measuring} wheel/measuring tape or tape measure (1)

ignore speedometer/speed

camera/radar

(electronic) timer

timing app (on ‘phone)

clock and watch on their own are

insufficient

any suitable length measuring device e.g. accept metre {rule(r)/stick}

but ruler on its own is

insufficient

Answers may be in either

order

(2)

Question Number


1(b)(i) white (car) (1) Allow the use of other columns that identify correct car

e.g. 5.6(s)

(1)

Question Number


1(b)(ii) (1) substitution 80 ÷ 4.3

evaluation (1)

19 (m/s)

Throughout the paper do not

penalise answers to many places of

decimal e.g. here 18.604651 gets bothmarks

Allow full marks for correct answer with no working seen.

accept 18.6 (m/s)

ignore 18 and 18.0 as incorrect

rounding

accept any power of 10 error for 1 mark

(2)

Question Number


1(b)(iii) 40 (miles per hour) (1) accept answers in range 39 – 43 (miles per hour)

ecf from b(ii) multiply bii by 2.222 range +/-

2.0

(1)


Question

Number


1(c) {steady/constant} speed (at

first) (1)

(then) slows down (1)

accept velocity for speed

ignore as time increases distance travelled increases

(then) slower/less speed/decelerates/negative

acceleration

(2)

Total for Question 1 = 8 marks


Question Number


2(a) 20(m) value between 18 and 22 (1)

Question Number


2(b) substitution (1) 100/9.8

evaluation (1) 10

unit (1) m/s

Accept 10.2

give 2 marks for correct answer, no working accept for 1 mark 9.65 or 9.7

mps (3)

Question Number


2(c) An explanation linking the following points

• speed changes (1)

• (because) slower to beginwith / faster at the end (1)

not the same speed throughout

slows down after 100 m

he speeds up=2 (2)

Question Number


2(d)(i) B slowing down (1)

Question Number


2(d)(ii) speed in a stated direction (1)


Question Number


(3(a)(1)

Question Number


3(b) distance travelled = area under graph (1)

substitution (1) ½ x 20 x 2

evaluation (1) 20 (m)

distance = average speed x time

= 10 × 2

20 (m)

allow (distance) = speed × time or 20 x 2 for 1 mark

give full marks for correct answer, no working (3)

Question Number



• velocity is a vector (1)

• (whereas) speed is not (1)

velocity has magnitude and direction velocity has direction

speed is a scalar speed has {no direction}/{magnitude only}

allow for 2 marks velocity is speed in a straight line velocity = displacement

time

NOTE answers in terms of momentum must still refer to vectors or direction to gain credit (2)


Question Number

Indicative Content Mark

QWC *5(d) An explanation linking some of the following

Forces acting • weight down• air resistance up (opposing motion)

Forces during fall • weight constant• air resistance increases• with speed• resultant force = W – R

Effect on shape of graph • at start, resultant force is large so acceleration large /

gradient steep• mid resultant force decreasing so acceleration decreasing /

gradient decreasing• terminal velocity, resultant force is zero so acceleration

zero / gradient zero (6)

Level 0 No rewardable content 11 -2 1 • a limited explanation linking a few facts from the indicative

content. E.g. at terminal velocity, forces are equal so constantspeed.

• the answer communicates ideas using simple language and useslimited scientific terminology

• spelling, punctuation and grammar are used with limited accuracy22 -4 3 • a simple explanation linking some of the indicative content to the

shape of the graph e.g At the start weight > air resistance soacceleration and at the end weight = air resistance so noacceleration.

• the answer communicates ideas showing some evidence of clarityand organisation and uses scientific terminology appropriately

• spelling, punctuation and grammar are used with some accuracy33 -6 5 • a detailed explanation linking most of the indicative content to the

complete shape of the graph e.g. At the start weight > airresistance so acceleration. Then air resistance increases (withspeed) so acceleration decreases. At the end weight = airresistance so no acceleration.

• the answer communicates ideas clearly and coherently uses a rangeof scientific terminology accurately

• spelling, punctuation and grammar are used with few errors


Question Number


4(a) D (1)

Question Number


4(b)(i) 12 (m/s) (1)

Range from 11(m/s) to 14 (m/s)

(1)

Question Number


4(b)(ii) Substitution (1) 20-0 5

evaluation (1) 4 (m/s2)

20 5

Full marks for correct answer with no working

Allow answers between 3.6 and 4.7 for 2 marks to reflect readings taken from the graph

(2)

Question Number


4b(iii) • velocity/ speed (measured

in) m/s (1)

• divided by time in s (1)

velocity/ speed (measured in) ms-1

acceleration is rate of change of velocity

m/s/s m per s per s [accept per for divide]

do not accept m/s times time

(2)


Question Number


4b(iv) at constant vel • distance = 60 (m) (1)

slowing down

• distance = ½×2×20 (1)

• = 20 (m) (1) correct answer scores 2 marks

(3)

Total for question 3=10 marks


Question Number


5 (a) (i) 16 (s) (1)

Sixteen/ sixteen seconds/ 16 s/ 16 seconds

(1)

Question Number


5 (a) (ii) Downward arrow starting at centre of the block

(1)

Mark by eye ie ruler not required. Accept freehand lines and gaps between dot and line less than half the distance between dot and bottom of block by eye. Accept lines that are not quite vertical

(1)

Question Number


5 (a) (iii) D zero (1)

Question Number


5 (a) (iv) Substitution 3 / 2

(1) Evaluation 1.5

(1)

Unit m/s2

(1)

ms-2 or m/s/s

bald 1.5 x 10n m/s2

gains 2 marks eg bald 150 = 1 mark (BOD for correct substitution) 150 m/s2 gains 2 marks

give full marks for correct numerical answer, 1.5 m/s2 even if no working

(3)


Question Number


5 (a) (v)

An explanation to include two of the following points

• (At first/in first 2 secondsBlock is) accelerating (1)

• Which requires a (resultant)force (1)

• In addition to the forceneeded to balance theweight of the block(1)

• (In next 4 seconds) forcesare balanced

(1) • (Because) velocity is

constant (1)

(block is) speeding up/increasing velocity

there is an unbalanced force/ forces are not balanced

(Because) speed is steady

(2)

Question Number


5 (b) An explanation to include

Information taken from the graph (1)

A valid conclusion (1)

Ignore air resistance

(Overall) time is less OR velocity/speed is greater OR acceleration is greater OR bigger/faster change in velocity/speed

So (same amount of) work is done more quickly/energy is transferred faster

(2)


Velocity/Distance/ Accceleration (inc

Graphs) Mark Scheme 2

Level Edexcel

Subject Physics





40 minutes

/33

/100

Time Allowed:

Score:

Percentage:



Question number

Answer Mark

1(a)(i) D (1)

Question number

Answer Additional guidance Mark

1(a)(ii) 16.0 (m/s) read from graph (1)

Substitution (1) (distance travelled =) 16 × 0.5

Answer (1) 8.0 (m) (1)

award full marks for correct numerical answer without working

ecf for substitution and answer using wrong speed value

(3)

Question number

Answer Mark

1(a)(iii) A (1)

Question number


1(a)(iv) Obtain readings from graph (1)

Substitution (1) 162.0

Answer (1) 8.0 (m/s2)


(3)

Question number


1(b) Any three improvements from:

• suitable instrument tomeasure distance (1)

• using a greater distance(to reduce effect of reactiontimes) (1)

• suitable instrument tomeasure time (1)

• use of one student at the{first/second} lamp post tosignal when to {start/stop}timing (1)

allow tape measure, trundle wheel

allow stop watch/clock or timing app. on phone

(3)


• two of three sets of studentstaking readings for the samecar (1)


Question number


1(a) Rearrangement (1)

m = fa

Substitution and conversion (1)

m =18701.83

Answer and rounding to 3 s.f. (1) 1020 (kg)

maximum 2 marks if kN not converted to N


(3)

Question number


1(b) ( −v u)= a

t(1)Rearrangement of

v = u + at

Substitution (1) v = 0 + 1.83 × 16

Answer (1) 29.3 (m/s)


(3)

Question number

Answer Mark

1(c) Correctly identifies data points from the graph to calculate areas (1)

Calculates area under AB (1) 240  m

Calculates area under CD (1) 135  m

distance travelled at constant speed = 240  m is greater than distance travelled when slowing down = 135  m (1) (4)


Question

Number


2(a)(i) force

(1)

If than one word given then 0

marks.

(1)

Question Number


2 (a)(ii) B 0.07kg (1)

Question

Number


2 (a)(iii) Arrow pointing (vertically)

upwards (1)

Value of 1.2 (N) (written near to

arrow)(1)

Marks are independent of each

other

(2)

Question

Number


2(b)(i) Substitution

90 x 3.3 (1)

1000

evaluation

0.30 (N) (1)

A value which rounds to 0.30 eg

0.297

Give full marks for correct answer with no working

Ignore power of ten error until evaluation

Allow 1 mark for 297 even with no working shown

(2)


Question Number


QWC *2(b)(ii) An explanation demonstrating some of the following:

Descriptions of the graph

Accelerates upwards during stage1

Maximum velocity is reached at the end of stage 1

Accelerates downwards / decelerates during stage 2

Accelerates during stage 3

Comes to rest during stage 4.

Interpretations of the shape of the graph

Fuel is burnt creating thrust in stage

Thrust is upwards in stage 1/

Gravity/weight (is always) a downward force

Fuel runs out at end of stage 1/ has ran out by stage

2

Still going up during/ max height at end of stage 2

Starts to fall at start of stage 3

Negative velocity during stage 3 because it is falling.

Rapid deceleration / collision with the ground during

stage 4/end of stage 3

Explanations for changes in velocity

Resultant force upwards/ thrust greater than gravity

force during stage 1

Acceleration non-linear because mass is decreasing /

resultant force is increasing

Linear deceleration in stage 2/3 because force of

gravity is constant

Resultant downward force/only gravity/ weight is

acting during stage 2 and 3

Large resultant force of impact during stage 4

(6)


Total for Question 5 = 12 marks

Level 0 No rewardable content

1 1 - 2 A limited explanation involving descriptions of the graph.

E.g. The rocket gets faster as it goes up during stage 1. Therocket slows down during stage 2

the answer communicates ideas using simple language and useslimited scientific terminology

spelling, punctuation and grammar are used with limitedaccuracy

2 3 - 4 A simple explanation involving interpretations of the shape of

the graph e.g. The rocket’s velocity increases during stage 1because the burning fuel provides a force. The rocket

accelerates downwards during stage 3 the answer communicates ideas showing some evidence of

clarity and organisation and uses scientific terminologyappropriately

spelling, punctuation and grammar are used with some accuracy

3 5 - 6 A detailed explanation which includes descriptions andinterpretations for the shape of the graph including an

explanation.E.g. The rocket’s acceleration during stage 1 is increasing

because it is losing mass as the fuel is burnt. It then slows downuntil it reaches maximum height at the end of stage 2

the answer communicates ideas clearly and coherently uses a

range of scientific terminology accurately spelling, punctuation and grammar are used with few errors


Question Number


3(ai) D 150 m (1)

(1)

Question Number


3(aii) B at 7 s (1)

(1)

Question Number


3(aiii) 6 (s) (1) (1)

Question Number


3(aiv) Substitution: 15 ÷ 6(1)

Evaluation 2.5 (m/s2) (1)

Allow ecf from 4(aiii) Must be 15 divided by their 4(aiii)

ECF allowed from first marking point ie evaluation of 15 divided by their answer from 4(aiii)

Award 2 marks for correct answer, no working (2)


Question Number


3(bi) 100 - 30 (1)

70 (N) (1)

100 + 30 or 130 gains 1 mark

Award 2 marks for correct answer, no working (2)

Question Number


3(bii) 550 (N) (1)

539 (N) allow use of g = 9.8 N/kg 539.55 (N) for use of g = 9.81N/kg Award mark for correct answer, no working

(1)

Question Number


3(c) An explanation linking

(combined) mass is less (1)

smaller force required for same acceleration OR more acceleration from same force (1)

ignore references to weight, friction or backwards force

ignore “easier to accelerate” as in stem

less force needed (to accelerate) (2)

(Total for Question 4 =10 marks)


Velocity/Distance/ Accceleration (inc

Graphs) Mark Scheme 3

Level Edexcel

Subject Physics





27 minutes

/22

/100

Time Allowed:

Score:

Percentage:



Question Number


1 (a) (i) 8 – 0 (m/s) 8 (1)

Question Number


1(a)(ii) substitution 8 / 5 (1)

evaluation (1) 1.6 (m/s2)

ecf from (i)

full marks for correct answer (or ecf) with no working shown.

(2)

Question Number


1(a)(iii) 0 Nil / nothing / zero / none (no mark for no response)

(1)

Question Number


1(b) (1)

(1)

substitution F = 1200 x 0.8

evaluation 960 (N)

full marks for correct answer with no working shown. (2)


Question Number


QWCQWC * ) an explanation linking some of the following points:

compared to a car with just the driver, a fully loaded car will

• have a greater mass / be heavier• greater kinetic energy / momentum• experience the same braking force (when brakes are

applied)• require a greater braking force (than available) to stop

(in the same distance)• have a smaller acceleration / deceleration• take a longer time to come to rest (from given speed)• travel greater distance in this time• needs to do more work with same amount of force• use of relevant equations such as F = ma, work done

= F x d• consequence of driver distractions

(6)

Level 0 No rewardable content 1 1 - 2 • a limited explanation using one idea from the indicative content

eg fully loaded car is heavier.• in answer communicates ideas using simple language and uses

limited scientific terminology• spelling, punctuation and grammar are used with limited

accuracy2 3 - 4 • a simple explanation which links ideas from the indicative content

eg it is heavier and so it takes a longer distance to stop• the answer communicates ideas showing some evidence of clarity

and organisation and uses scientific terminology appropriately• spelling, punctuation and grammar are used with some accuracy

3 5 - 6 • a detailed explanation which links several ideas from theindicative content e.g. It has more momentum and so it will takea longer time to stop. This means that it will travel a furtherdistance. The answer communicates ideas clearly and coherentlyuses a range of scientific terminology accurately



Question number



m = fa


m =18701.83




(3)

Question number


2(b) ( −v u)= a


v = u + at

Substitution (1) v = 0 + 1.83 × 16

Answer (1) 29.3 (m/s)


(3)

Question number

Answer Mark






Newton’s Laws &Weight Mark Scheme 1

Level Edexcel

Subject Physics



Sub Topic Newton’s Laws & Weight


47 minutes

/39

/100

Time Allowed:

Score:

Percentage:



Question Number


1(a)(i) (force of) water (on ski) air resistance/drag

ignore wind/unqualified friction

(1)

Question Number


1(a)(ii) substitution (1) 500 – 300

evaluation (1) 200 (N)


Question Number


1(a)(iii) to the right forward/direction skier is travelling/towards the boat

(1)

Question Number


1(b)(i) B J (1)

Question Number


1(b)(ii) substitution (1) 54 × 10 × 5

evaluation (1) 2700

Ignore unit (J) even incorrect

give full marks for correct answer, no working

(2)

Question Number


1(b)(iii) A description including two of the following points

• (some) KE at the ramp (1)

• is transferred to GPE at top(1)

• still has some KE at top (1)

• some energy lost due to airresistance (1)

KE to GPE for 1 mark

air friction (2)


Question Number


2(a)(i) B to the left (1) l

Question Number


2(a)(ii) A accelerating (1)

Question Number


2(a)(iii) substitution 625x 10 (1) Evaluation 6250 (N) (1)

625 x 9.8

6125 (N)


(2)

Question Number


2(b)(i) (1)

air resistance (1)

upward arrow on any part of line

vertical line from any point on the diagram

air friction, upthrust, drag Ignore any downward arrow labelled weight or gravity

(2)

Question Number


2(b)(ii) Balanced (1)

Zero (1)

(2)

Total for marks for question 1 = 8


Question Number


3 (a) (i) C (1)

Question Number


3 (a) (ii) acceleration Recognisab mis-spellingle sMore than one word written scores zero EXCEPT for the phrase Acceleration due to gravity which scores 1 mark

(1 )

Question Number


3 (b) Substitution weight = 0.00008 x 10

(1) evaluation

0.0008 (N) (1)

8 x 10-4

1/1250

(2)

Question Number


3 (c)

(1)

Substitution speed = 13 / 1.7

evaluation

7.6 (m/s) (1)

An answer which rounds to 7.6 eg 7.647 7.65

7.7

(2)


Question Number


QWCQWC * ) A explanation including some of the following points • drops near the top are accelerating• due to force of gravity• travel a greater distance in given time• there is air resistance on the drops as they fall• this increases with velocity• resultant force is downward• this reduces resultant force• eventually resultant force is zero• drops have reached terminal/ maximum velocity• drops near bottom are all travelling at terminal velocity• so travel same distance in given time

(6)

Level

0 No rewardable content

1 1 - 2 • a limited explanation such as one which correctly addresses eitherwhy the drops at the bottom are evenly spaced or why the dropsat the top are note.e.

OR drops at bottom are all going at the same speed

drops at top are speeding up • the answer communicates ideas using simple language and uses

limited scientific terminology• spelling, punctuation and grammar are used with limited accuracy

2 3 - 4 • a simple explanation such asa correct comparison of the motion of the drops at top and

bottom e.g. drops at bottom are travelling at terminal velocitywhereas drops at top are still accelerating.

Or a complete explanation of motion at either top or bottom e.g.at the bottom, air resistance and gravity forces are balanced so they travel at constant speed


• spelling, punctuation and grammar are used with some accuracy3 5 - 6 • a detailed explanation such as one which explains why the motion

of the drops at top and bottom are different e.g.The drops were initially accelerating due to a resultant forcedownwards. The acceleration decreased as they fell andeventually reached zero. With no acceleration their velocity wasconstant and so equal distance travelled in given time at thebottom.

• the answer communicates ideas clearly and coherently uses arange of scientific terminology accurately



Question number



m = fa


m =18701.83




(3)

Question number


4(b) ( −v u)= a


v = u + at

Substitution (1) v = 0 + 1.83 × 16

Answer (1) 29.3 (m/s)


(3)

Question number

Answer Mark







Level Edexcel

Subject Physics





42 minutes

/35

/100

Time Allowed:

Score:

Percentage:



Question number


1(a) An answer that combines the following points of understanding to provide a logical description:

• measurement of timebetween(or at) twopositions usingsuitable timingequipment (1)

• measurement ofsuitable distancealong the runwaywith metre rule (1)

• measurement of verticalheight to starting position(1)

• repeats AND averagesAND use of a correctequation (1)

allow

stopwatch, light gates

minimum is 0.5  m metal tape measure

average speed = distance/time OR average speed = (speed at A − speed at B)/2

(4)

Question number


1(b)(i) Substitution of correct data from graph and mass conversion (1)

0.5 × 0.65 × (0.61)2

Answer (1)

0.12 (J)

maximum of 1 mark if mass in g used

allow tolerance of ±0.2 for speed

(2)

Question number


1(b)(ii) • Tangent to the graph ath = 0.1 (1)

• Answer in the region 3.5 to3.6

either seen on graph or suitable pairs of values of Δv and Δh

(2)


Question number

Answer Mark

1(b)(iii) An answer that combines points of interpretation/evaluation to provide a logical description:

• for each change in height, as the height increases thespeed of the trolley increases

• the greatest change in speed is between the change inheight from 0.04  m to 0.9  m

(2)

Question number


1(c) An answer that combines the following points to provide a logical description of the plan/method/experiment:

• identifies controlvariables (1)

• uses at least 3 differentsurfaces (1)

• calculates average speedfor each surface andrepeats (1)

constant height, constant slope, constant starting points and same length of surface

(3)


Question Number


2(a) A(1)

Question Number


2(b) distance travelled = area under graph (1)

substitution (1) ½ x 20 x 2

evaluation (1) 20 (m)

distance = average speed x time

= 10 × 2

20 (m)

allow (distance) = speed × time or 20 x 2 for 1 mark


Question Number



• velocity is a vector (1)

• (whereas) speed is not (1)

velocity has magnitude and direction velocity has direction

speed is a scalar speed has {no direction}/{magnitude only}

allow for 2 marks velocity is speed in a straight line velocity = displacement

time

NOTE answers in terms of momentum must still refer to vectors or direction to gain credit (2)


Question Number


QWC *2(d) An explanation linking some of the following

Forces acting • weight down• air resistance up (opposing motion)

Forces during fall • weight constant• air resistance increases• with speed• resultant force = W – R

Effect on shape of graph • at start, resultant force is large so acceleration large /

gradient steep• mid resultant force decreasing so acceleration decreasing /

gradient decreasing• terminal velocity, resultant force is zero so acceleration

zero / gradient zero (6)

Level 0 No rewardable content 11 -2 1 • a limited explanation linking a few facts from the indicative

content. E.g. at terminal velocity, forces are equal so constantspeed.


• spelling, punctuation and grammar are used with limited accuracy22 -4 3 • a simple explanation linking some of the indicative content to the

shape of the graph e.g At the start weight > air resistance soacceleration and at the end weight = air resistance so noacceleration.


• spelling, punctuation and grammar are used with some accuracy33 -6 5 • a detailed explanation linking most of the indicative content to the

complete shape of the graph e.g. At the start weight > airresistance so acceleration. Then air resistance increases (withspeed) so acceleration decreases. At the end weight = airresistance so no acceleration.

• the answer communicates ideas clearly and coherently uses a rangeof scientific terminology accurately



Question Number


3 (a) (i) 16 (s) (1)

Sixteen/ sixteen seconds/ 16 s/ 16 seconds

(1)

Question Number


3 (a) (ii) Downward arrow starting at centre of the block

(1)

Mark by eye ie ruler not required. Accept freehand lines and gaps between dot and line less than half the distance between dot and bottom of block by eye. Accept lines that are not quite vertical

(1)

Question Number


3 (a) (iii) D zero (1)

Question Number


3 (a) (iv) Substitution 3 / 2

(1) Evaluation 1.5

(1)

Unit m/s2

(1)

ms-2 or m/s/s

bald 1.5 x 10n m/s2

gains 2 marks eg bald 150 = 1 mark (BOD for correct substitution) 150 m/s2 gains 2 marks

give full marks for correct numerical answer, 1.5 m/s2 even if no working

(3)


Question Number


3 (a) (v)

An explanation to include two of the following points

• (At first/in first 2 secondsBlock is) accelerating (1)

• Which requires a (resultant)force (1)

• In addition to the forceneeded to balance theweight of the block(1)

• (In next 4 seconds) forcesare balanced

(1) • (Because) velocity is

constant (1)

(block is) speeding up/increasing velocity

there is an unbalanced force/ forces are not balanced

(Because) speed is steady

(2)

Question Number


3 (b) An explanation to include

Information taken from the graph (1)

A valid conclusion (1)

Ignore air resistance

(Overall) time is less OR velocity/speed is greater OR acceleration is greater OR bigger/faster change in velocity/speed

So (same amount of) work is done more quickly/energy is transferred faster

(2)



Level Edexcel

Subject Physics





28 minutes

/23

/100

Time Allowed:

Score:

Percentage:



Question number

Answer Mark

1(a)(i) D (1)

Question number


1(a)(ii) 16.0 (m/s) read from graph (1)

Substitution (1) (distance travelled =) 16 × 0.5

Answer (1) 8.0 (m) (1)


ecf for substitution and answer using wrong speed value

(3)

Question number

Answer Mark

1(a)(iii) A (1)

Question number


1(a)(iv) Obtain readings from graph (1)

Substitution (1) 162.0

Answer (1) 8.0 (m/s2)


(3)

Question number


1(b) Any three improvements from:

• suitable instrument tomeasure distance (1)

• using a greater distance(to reduce effect of reactiontimes) (1)

• suitable instrument tomeasure time (1)

• use of one student at the{first/second} lamp post tosignal when to {start/stop}timing (1)

allow tape measure, trundle wheel

allow stop watch/clock or timing app. on phone

(3)


• two of three sets of studentstaking readings for the samecar (1)


Question number


2(a) rearrangement (1)

m = fa

substitution and conversion (1)

1.83m =

1870

answer and rounding to 3 s.f. (1) 1020 (kg)



(3)

Question number


2(b) ( −v u)= a

t(1)rearrangement of

v = u + at

substitution (1) v = 0 + 1.83 × 16

answer (1) 29.3 (m/s)


(3)


Level Mark Descriptor

0 No awardable content.

1 1–2 • Interpretation and evaluation of the information attempted butwill be limited with a focus on mainly just one variable.Demonstrates limited synthesis of understanding. (AO3)

• The description attempts to link and apply knowledge andunderstanding of scientific ideas, flawed or simplisticconnections made between elements in the context of thequestion. (AO2)

2 3–4 • Interpretation and evaluation of the information on bothvariables, synthesising mostly relevant understanding. (AO3)

• The description is mostly supported through linkage andapplication of knowledge and understanding of scientific ideas,some logical connections made between elements in thecontext of the question. (AO2)

Question number

Indicative content Mark

*9(c) Answers will be credited according to candidate's deployment of knowledge and understanding of the material in relation to the qualities and skills outlined in the generic mark scheme.

The indicative content below is not prescriptive and candidates are not required to include all the material which is indicated as relevant. Additional content included in the response must be scientific and relevant.

AO2

• fuel forms a store of chemical (potential) energy• chemical energy is transferred to kinetic energy and

thermal energy when the car moves• kinetic energy transferred to thermal energy as the car

slows down

AO3 • during X, kinetic energy increases as the car’s speed

increases/car accelerates and the increase in kinetic energyis provided by the chemical energy store

• during all three sections, work is done against frictionalforces in the moving parts of the car and against the dragfrom the air

• during Y, kinetic energy stays constant when the car movesat constant speed but energy is still transferred to thermalenergy

• during Z, kinetic energy decreases as the car slows down (6)


3 5–6 • Interpretation and evaluation of the information,demonstrating throughout the skills of synthesising relevantunderstanding. (AO3)

• The description is supported throughout by linkage andapplication of knowledge and understanding of scientific ideas,logical connections made between elements in the context ofthe question. (AO2)


Save My Exams! – The Home of Revision For more awesome GCSE and A level resources, visit us at www.savemyexams.co.uk

Page 1

Time Allowed: 62 minutes

Score: /62

Percentage: /100

Circular Motion Mark Scheme 1

Level GCSE(9-1)Subject Physics

Exam Board Edexcel


Topic Circular Motion Sub-Topic

http://www.savemyexams.co.uk/


Page 2

M1.(a) A 1

(b) (i) 9000

an answer of 9 k(N) gains 1 mark 1

(ii) increase

accept other comparative terms, eg give a bigger

affect / change is insufficient 1

(iii) smaller

accept other comparative terms, eg less 1

(c) Q N M

all three in correct boxes

one statement in correct box gains 1 mark 2

(d) any two from:

• increase the current / p.d. (supplied to the coil)

accept reduce the resistance of the coil or increase cross sectional area of wire

accept more cells / batteries or turn up the power supply

increase power is insufficient

• increase number of turns (on the coil)

• increase the area (of the coil)

accept increase the width of the coil

increase width / size is insufficient

• increase the (strength of the permanent) magnetic field



Page 3

accept move the magnets closer to the coil

accept use stronger magnets

do not accept use larger magnets 2

(e) an economic 1

[9]

M2.(a) increases 1

(b) (i) B 1

(ii) tension in the wire 1

(iii) C 1

[4]

M3. (a) (i) centripetal force

accept any unambiguous correct indication 1

(ii) B 1

(b) (i) decrease

accept any unambiguous correct indication



Page 4

1

(ii) increase

accept any unambiguous correct indication 1

(c) Bull pit 1

smallest (inside) radius

accept smallest diameter / circumference 1

[6]

M4. (a) increases 1

(b) (i) B 1

(ii) tension in the wire 1

(iii) C 1

[4]



Page 5

M5. (a) centripetal 1

(b) B 1

(c) decreases 1

[3]

M6. (a) either its direction or its speed 1

(b) (i) friction 1

(ii) centripetal 1

(iii) increase 1

(iv) increase 1

(c) examples

(yes) noisy (1)

disturbs people living nearby (1)

(yes) encourages people to drive fast (1) which makes (road) accidents more serious/likely (1)

(no) leads to improvements in safety features (1) such as better brakes (1)

(don’t know) noisy (1) but new tyres have a better grip (1)

whichever box has been ticked, the mark(s) is/are for an appropriate response

note, accept responses which assume that the public may use the racetrack

2

[7]



Page 6

M7. (a) (i) direction

accept any unambiguous indication 1

(ii) centripetal


(b) A


(c) mass of the passengers is greater


[4]

M8. (a) tension

accept any unambiguous method of indication eg it’s underlined or ticked

1

(b) (i) speed of the ball is increased 1

(ii) the direction of the ball



Page 7

1

(c) centripetal

accept any unambiguous method of indication eg it’s underlined or ticked

1

[4]

M9. (a) (i) P 1

(ii) the child’s grip / hold / pull (on the roundabout / bar / rail)

or ‘the tension in the child’s arms’

accept ‘the child’s muscles’

accept ‘friction between the child and the roundabout’

do not accept just ‘friction’ 1

(iii) increases

accept any unambiguous indication that this ending has been selected

1

(b) (i) 360 (Nm)

credit either 240 × 1 ½

or 240 × 1.5 with 1 mark 2

(ii) move to(wards) the left / to(wards) the / his end

or move away from the centre / pivot / axis (of rotation)

or move away from the girl / the child / his daughter 1



Page 8

(c) (i) C 1

(ii) friction / grip of the car / tyres / wheels (on the road)

do not accept just ‘friction / grip’ 1

[8]

M10. (a) converging 1

image 1

object 1

image 1

object 1

shorter 1

(b) (i) (Earth’s) gravity

accept centripetal accept minor misspellings, however, do not credit any response which could be ‘centrifugal’

1

(ii) to(wards) (the centre of the) Earth

allow inwards do not accept downwards 1

(c) (i) either

friction (force) or centripetal force

allow grip 1



Page 9

between the tyres / wheels and (the surface of) the road

allow on the tyres / wheels or towards the centre of the bend / arc / circle

1

(ii) mass or speed or momentum

allow weight allow velocity 1

radius / diameter

do not credit ‘curvature’ or ‘circumference’ 1

(d) centripetal

accept minor misspellings (see above) 1

[13]



Page 1


Score: /38

Percentage: /100



Exam Board Edexcel





Page 2

M1.(a) (i) D 1

(ii) friction 1

(iii) any two from: • the speed / velocity • the radius of the bend

the radius is insufficient

accept curvature of the road

size of the bend is insufficient

accept distance of car from centre (of bend) • the mass (of the car).

accept weight for mass 2

(b) the car has a wide base

accept any description of a wide base e.g. the wheels are far apart

accept wide wheel base

do not accept long wheel base

a large surface area is insufficient

wide tyre(s) is insufficient 1

the car has a low centre of mass / gravity

accept any description of low centre of mass e.g. mass is close to the ground

a down force is insufficient 1

[6]

M2.(a) gravitational attraction (between the satellite and the Earth)

allow gravity

allow weight of the satellite 1

(b) any two from: • mass of satellite • speed / velocity (of satellite)



Page 3

• radius of orbit / circle

allow height above the Earth

radius / height alone is insufficient 2

(c) (i) increasing the height (above the Earth’s surface) increases the time (for one orbit)

allow a positive correlation

allow as one gets bigger, the other gets bigger, or vice versa

ignore they are directly proportional 1

(ii) there is no relationship / correlation 1

(d) Isaac Newton was a respected scientist who had made new discoveries before

1

[6]

M3. (a) (i) towards the centre of the circle

accept inwards

accept a correct description

‘along the string’ is insufficient 1

(ii) tension (in the string)

accept pull of the string

‘the string’ is insufficient orweight (on the end of the string)

‘the student’ is insufficient

‘turning action’ is insufficient 1

(b) (i) each may (also) affect the speed

accept results for speed 1

so only one independent variable

accept only one variable affects dependent variable



Page 4

‘fair test’ is insufficient

‘they are control variables’ is insufficient 1

(ii) continuous

both required

dependent 1

(iii) reduces (absolute) timing error (for one rotation)

accept too fast to time one orincreases / improves reliability / accuracy (for one rotation)

ignore checking for anomalous results

to work out an average is insufficient 1

(c) speed increases with centripetal force

accept positive correlation

do not accept proportional 1

(d) (i) gravitational pull (of the Earth)

accept gravity 1

(ii) No

both parts required – however this may have been subsumed within the reason

geostationary orbits once every 24 hours

accept a correct comparative description 1

[9]



Page 5

M4. (a) (i) arrow from centre of the ball and at right angles to the string and in the correct direction

arrow should point to the student’s belt accept free-hand ‘straight’ line do not accept curved line

1

(ii) increase

accept ‘be stronger / bigger’ 1

increase


increase


(b) speed velocity direction

all three correct any two correct for 1 mark otherwise 0 marks

2

(c) (i) centripetal

accept ‘centripedal’ and other minor misspellings do not accept anything which could be ‘centrifugal’

1

(ii) gravity

accept ‘weight’ accept ‘force of attraction due to mass(es) (of the Moon and the Earth)’

1

(iii) electron(s) 1



Page 6

(iv) electrostatic

accept ‘electrical’ do not accept just ‘centripetal’

1

[10]

M5. (i) towards Earth

for 1 mark 1

(ii) gravity

for 1 mark 1

(iii) changes direction

for 1 mark 1

(iv) polar orbit; closer

for 1 mark each 2

(v) speed constant (1) mass constant (1)

for 1 mark each 2

[7]



Page 7



Page 1


Score: /33

Percentage: /100



Exam Board Edexcel





Page 2

M1.(a) centripetal (force)

allow tension (between astronaut and seatbelt) 1

towards the centre (of the G-machine / circle)

do not accept towards the centre of the Earth

allow inwards 1

(b) (i) the greater the speed (of a centrifuge), the greater the force

answers must be comparative

accept velocity for speed

accept positive correlation between speed and force

speed and force are not proportional – treat as neutral 1

the smaller the radius, the greater the force (at a given speed)

allow (G machine) 1 has / produces a greater force (than

G machine 2 ) at the same speed

must be comparative, eg a small radius produces a large force = 0 marks on own

1

as the speed increases the rate of change in force increases

accept force is proportional to the square of the speed

or

doubling speed, quadruples the force

accept any clearly correct conclusion 1

(ii) 12000 (N)

or

12 k(N) 1



Page 3

(c) (i) the current (in the coil) creates a magnetic field (around the coil)

accept the coil is an electromagnet 1

so the magnetic field of the coil interacts with the (permanent) magnetic field of the magnets (producing a force)

accept the two magnetic fields interact (producing a force)

if no marks scored an answer in terms of current is perpendicular to the (permanent) magnetic field is worth max 1 mark

1

(ii) vertically downwards arrow on side A

one arrow insufficient

and

vertically upwards arrow on side C 1

(iii) the current is parallel to the magnetic field

allow the current and magnetic field are in the same direction

allow it / the wire is parallel to the magnetic field 1

(d) increase the current / p.d. (of the coil)

accept decrease resistance

accept voltage for p.d.

accept increase strength of magnetic field / electromagnet 1

(e) yes with suitable reasonorno with suitable reason

eg

yes – it has increased our knowledge

yes – It has led to more (rapid) developments / discoveries (in technology / materials / transport) accept specific



Page 4

examples

no – the money would have been better spent elsewhere on such things as hospitals (must quote where, other things not enough)

no mark for just yes / no

reason must match yes / no 1

[12]

M2.(a) any two from:

• (acceleration occurs when) the direction (of each capsule) changes

• velocity has direction

• acceleration is (rate of) change of velocity 2

(b) to(wards) the centre (of the wheel) 1

(c) the greater the radius / diameter / circumference (of the wheel) the smaller the (resultant) force (required)

accept ‘the size’ for radiusboth parts required for the mark 1

[4]

M3. (a) the direction of the riders is constantly changing 1

therefore the velocity of the riders is changing 1

and because acceleration is the rate of change of velocity the acceleration is changing

1

(b) to(wards) the centre (of the cylinder / rotor)



Page 5

1

(b) centripetal 1

(b) it is reduced 1

[6]

M4. (a) (from present/recent) data/evidence/observations of (the rate of change in) Phobos’/the moon’s orbit (1)

or appropriate example of data (1)

and its correct use (1)

(and) continued/extended/extrapolated (the pattern/trend for the next 100 million years) (1)

example (present) distance from Phobos to Mars (1) ÷ (average) rate of approach (1)

2

(b) (it is) increasing (1)

Phobos/the moon will be nearer (to Mars) (1)

or the radius/circumference/diameter of the orbit of Phobos/the moon will decrease/be less

only credit 2nd mark if the first mark is correct 2

(c) it will increase/be more (1)

(because) Phobos/the moon will get/be closer to Mars/the planet (1)

only credit 2nd mark if the first mark is correct

note part(s) of this response may be included as the answer to part (b)



Page 6

read both before marks are awarded 2

[6]

M5. (a) any two ideas:

• (acceleration occurs when) the direction (of each capsule) changes

• velocity has direction

• acceleration is (rate of) change of velocity 2

(b) to(wards) the centre (of the wheel) 1

(c) centripetal

allow minor misspellings but do not credit a response which could be ‘centrifugal’

1

(d) the greater the radius / diameter / circumference (of the wheel) the smaller the (resultant) force (required)

accept ‘the size’

both parts required for the mark

accept converse 1

[5]


Momentum Mark Scheme 1

Level Edexcel

Subject Physics



Sub Topic Momentum


60 minutes

/50

/100

Time Allowed:

Score:

Percentage:



Question Number


1(ai) B momentum (1)

(1)

Question Number


1(aii) power (1) (1)

Question Number


1 (bi) Substitution: ½ x0.8 x 252

(1)

Evaluation 250 (1)

0.25 kJ scores 3 marks

J bod j (1)

Allow both marks for correct answer with no method shown.

Ignore power of 10 until evaluation e.g. 2 marks for 25 J 1mark for25 W

Nm ignore kg (m/s)2 Unit mark is independent of numerical answer. (3)

Question Number


1 (bii) 250 (1)

Ignore any unit given by the candidate

Allow ecf from 1(bi)

(1)


Question Number


1 (biii) A suggestion to include:

work done = force x distance (1)

(force) used over a longer distance (1)

ignore references to more power, greater speed, longer time, larger force, momentum and how far javelin travels.

the longer they are pushing (it/the javelin) [bod distance]

they can push the javelin (forward) for longer [bod distance]

the arm can move further (2)

(Total for Question 2 =8 marks)


Question Number


2 (a)(i) C - power (1)

Question Number


2 (a)(ii) energy work Must be in correct order (1)

Question Number


2 a(iv) Substitution 450 / 1.5 (1)

Evaluation 300 (N) (1)

Allow full marks for correct answer with no working shown Allow (1) for 167 (N) obtained by 450-200 / 1.5

(2)

Question Number


2 (a)(v) An explanation to include

(quantity has) a size and a direction

ignore any named examples

(1)

Question Number


2 a(iii) Substitution 50 x 4 (1)

Evaluation 200 (kg m/s) (1)

Allow full marks for correct answer with no working shown

(2)


Question Number


2 (b) An explanation which uses conservation of momentum to link three from

Mother and daughter have different mass (1)

Momentum is conserved / is zero to start with (1)

Both have same size momentum (after the push) (1)

so speed of the daughter is greater than that of the mother (1)

An explanation based on Newton’s laws and linking three from

Each have a different mass (1)

Each experience the same size force / action and reaction are equal (1)

Each experiences a different acceleration (1)

so speed of the daughter is greater than that of the mother (1) (3)

(Total for Question 3 = 10 marks)


Question Number


3 (a) (i) D the same size as the driving force

(1)

Question Number


3 (a) (ii) transposition: (1) {change in) speed= accelerationxtime

substitution: (1)

speed = 12 x 4

evaluation: (1)

48 (m/s) (1)

transposition and substitution can be in either order substitution mark can be scored when incorrectly transposed word/symbol equation is given

Give full marks for correct answer no working

(3)

Question Number


3 (b) An explanation linking

• {acceleration of sports is2x / time to reach 30 m/sis ½} that of family car /RA (1)

• mass of sports car LESSthan ½ that of family caror RA (1)

(so resultant force required is less)

Attempt to use f = m x a scores one mark e.g. 4200 OR 3600 scores 1

Correct numerical comparison scores both marks e.g. 4200:3600 numerically or in words scores 2 marks

(2)


Question Number


QWCQWC * ) An explanation including some of the following ideas • brakes apply a force to the car

• this force from brakes makes the car decelerate/ losevelocity

• a force also acts on the driver

• driver decelerates at same rate as the car

• does not move with respect to car/ stays in the drivingseat

• moves slightly because belt stretches

• small/ no horizontal force acts on the shopping bag

• shopping bag continues at similar/ same velocity

• until shopping bag falls off seat / hits dashboard

• ideas can be expressed in terms of energy, momentumand/or by reference to Newton’s laws

(6)

Level 0 No rewardable content 1 1 - 2 • A limited explanation of the difference in decelerations of at least

two of the objects Car (C), Shopping (S) and Passenger (P)mainly describing the effects.

E.g. (at start) C stops (very quickly) while {P / S} carries onmoving (for a longer time) OR S {carries on at same speed / hits the dashboard} while P is

{held back / slowed down} (by the seatbelt) • the answer communicates ideas using simple language and uses


accuracy2 3 - 4 • A simple explanation of the difference in decelerations of at least

two of the objects Car, Shopping and Passenger, including areason for at least one of the decelerations.

E.g.(at start) C stops (very quickly) because of friction at thebrakes and at the road while {P / S} carries on moving (for alonger time)

OR S {carries on moving (at same speed) / hits the dashboard}while P is {held back / slowed down} because of stretching force from the seatbelt)


• spelling, punctuation and grammar are used with some accuracy


3 5 - 6 • A detailed explanation of the relative decelerations of C, S and Pincluding mention of the physical principles involved in any twosuch as that named forces are needed to change given motions.

E.g. (The force of) friction is large for C to {slow down / stop}quickly but is low for P and S. {So / thus / therefore etc} Por S carry on at the same speed (initially). P decelerates moreslowly than C {because / as a result etc} of the stretching(force) of the seatbelt.

OR The idea of {Newton’s first law / inertia / need for a force to change motion} and the role of friction and {elastic / tension / stretching} force in producing the three named decelerations.

OR Named force needed for a described change in {momentum/kinetic energy} to {stop / slow down} each of the three objects.




Question Number


4 (a)(i) 2.5 (m) Allow answers between (and including) 2.45 & 2.55

(1)

Question Number


4 (a)(ii) 0.7 (s) Allow answers between (and including) 0.68 & 0.72

(1)

Question Number


4 (a)(iii)

line:

same shape as original (1)

peak at 1.9 m (1)

time taken < 0.7 s (1)

Ignore any part of the graph after the peak

(3)

Question Number


4 (a)(iv) An explanation linking:

energy lost (1)

in collision with ground / air resistance (1)

Inelastic collision worth (2)

as sound or heat (2)


Question Number


4 (b)(i) shown using data Any two from kinetic energy before = 12.5 + 0 (=12.5) (1) kinetic energy after = 4.5 + 8 (=12.5) (1)

Kinetic energy is the same before and after the collision (1)

Kinetic energy is conserved/no energy lost (2)

Question Number


4 (b)(ii) cyclotron (1) named particle accelerator accept CERN (1)

Total mark for question 4 = 10


Question Number


5(a)(ii) momentum before = momentum after (1)

5.1 = 0.83 × v (1)

v = 6.1 (m/s) (1)

allow 5.0 = 0.80 x v for 1 mark max

5.0 = 0.83 x v

v = 6.0 (m/s) allow ecf from (a)(i) give full marks for correct answer, no working

(3)

Question Number


5(a)(iii) Statement to include any two from

• kinetic energy is notconserved (1)

• (lost ke) appears asheat/sound (1)

• momentum is conserved(1)

ke not conserved / some ke lost

no momentum lost (2)

Question Number


5(b)(i) an explanation linking • momentum (must be)

conserved (1)

• so must have positive andnegative momentum (1)

photons move in opposite directions

indication of movement in opposite directions (e.g. opposite velocities)

(2)

Question Number


5(b)(ii) E = (2 ×) 9.1 × 10-31 × [3 × 108]2 (1)

= 1.6 × 10-13 (J) (1)

8.2 × 10-14 (0.82 × 10-13) for 1 mark


(2)

Question Number


5(a)(i) momentum = 0.03 × 170 (1) Accept 5.1 seen (1)



Level Edexcel

Subject Physics



Sub Topic Momentum


58 minutes

/48

/100

Time Allowed:

Score:

Percentage:



Question Number


1(a)(i)1( 1260 W (1)

Question Number


1(a)(ii) substitution (1) 5040 = 240 × 10 × height

transposition (1) height = 5040

240 × 10

evaluation (1) 2.1 (m)

substitution and transposition in either order


(3)

Question Number


1(b) no movement (in direction of force) / (work done=) weight x 0 = 0

stationary it is not changing height is in same position

ignore ref to terminal velocity, force and acceleration (1)

Question Number


1(c) substitution (1) 240 × 6.4

evaluation (1) 1500

Unit (1) kg m/s independent mark

1536

give (2) marks for correct answer, no working

Ns (3)


Question Number


2(a) Description including 3 of the following:

• (Gravitational) potentialenergy (transferred) to KE(1)

• Idea of energy transfer toheat/sound whilst descending(1)

• Chemical energy is transferredto heat energy in Andrew (1)

• Idea of energy dissipated onstopping (1)

(G)PE (transferred) to KEAllow gravitational energy forGPE

Energy transferred to heat because of air resistance/ friction

The energy goes to heat as he stops. Energy is transferred to the surroundings

(3)

Question Number


2(b)(i) substitution (1) 67 × 31

evaluation (1) 2077 (kg m/s)

2080, 2100

working backwards using 2000 (v=) 29.85, 30 (m=) 64.52, 65

67 X 31=2000 scores only one mark

(2)

Question Number


2(b)(ii) substitution (1) 2000 ÷ 2.3 evaluation (1) 870 (N)

answer to (b)(i)) ÷ 2.3

900, 869.6, 869.5 903

(2)


Question Number


2(b)(iii) an explanation linking two of the following

• Force on Andrew is quitesmall (1)

• Because impact time islong (1)

• Theacceleration/deceleration isquite small (1)

• Because impact distance isfar (1)

force is reduced/ less /not as strong

slows down/changes momentum gradually

acceleration = 1.35 ‘g’ or 13.5 m/s2

slows down (rate of) change of momentum scores 2 marks

(2)

Total question 2 = 8 marks


Question Number


3(a)(i) D 23 m (1)

Question Number


3(a)(ii) A the driver is tired (1)

Question Number


3(b)

(1)

substitution (1) 800 x 3

evaluation 2400 (kg m/s)

Give full marks for correct numerical answer, even if no working

bald 2.4 x 10n gains 1 mark (BOD for correct substitution) eg bald 240 = 1 mark

In all calculations if the candidate gives two different methods and writes the wrong answer in the answer space award no marks If the candidate writes correct answer they will gain full marks.

(2)

Question Number


3(c)(i) substitution (1) 600 x 15

evaluation (1) 9000 (J)

bald 9.0 x 10n gains 1 mark eg bald 900 = 1 mark (BOD for correct substitution)

give full marks for correct numerical answer, 9000 (J) even if no working

(2)

Question Number


3(c)(ii) A the energy transferred (1)


Question Number


4(a)(i) D towards the centre of the circle

(1)

Question Number


4(a)(ii) centripetal (force) reject centrifugal force accept misspellings where meaning is clear e.g. centripedal

(1)

Question Number


4(a)(iii) Any two of the following :-

ball slows down (1)

ball / it drops (down) / circles at a lower height (1)

go in smaller circles (1)

less kinetic energy / momentum

any lowering / less potential energy

stops going in circles the ball/it would not make complete circles (not just ‘stops’)

(2)

Question Number


4(a)(iv) An explanation linking:

the idea that momentum(of the closed system)would stay the same (1)

the idea that kineticenergy would not beconserved (1)

momentum of the ball decreases / changes (direction) / passed to wall

must specify which momentum; do not credit ‘momentum decreases’ by itself

kinetic energy heat/sound/wall

ignore ‘KE decreases / is lost’ without qualification allow ‘KE is lost because it’s not elastic’ (i.e. qualified)

(2)


Question Number


QWCC 4(b) A description including some of the following points :- Cyclotron

two D-shaped halves gap between the Dees (alternating) voltage across the gap magnetic field (at right angles to the moving particles) vacuum enables free movement of particles

Particle movement accelerate start at the centre move in a circular path spiral outwards exit in a straight line

Examples of labelled diagrams which would give Level 3 by themselves (not all labels / details needed)

Level 2 if no labels but Dees AND particle path shown. Level 1 if no labels but either Dees OR spiral of particle shown Ignore uses of cyclotron

(6)


(Total for Question 5 = 12 marks)

evLevelL No rewardable content 1 1 - 2 a limited description of either particle movement OR cyclotron

e.g. The particles move in a circle OR Cyclotrons have two DeesOR Cyclotrons are particle accelerators OR there’s a vacuum

the answer communicates ideas using simple language and useslimited scientific terminology


2 3 - 4 a simple description of particle movement AND cyclotron OR amore detailed description of one e.g. A cyclotron has two D-shaped halves and the particles inside accelerate OR A cyclotronhas a magnetic field and a voltage across the gap OR Chargedparticles increase in speed as they spiral outwards OR vacuumallows free movement of particles

the answer communicates ideas showing some evidence of clarityand organisation and uses scientific terminology appropriately

spelling, punctuation and grammar are used with some accuracy3 5 - 6 a description of particle movement AND cyclotron with a detailed

description of one of them e.g. the charged particles get faster asthey accelerate across the gap in the Dees OR the magnetic field(of the cyclotron) causes the particles to move in a circle

the answer communicates ideas clearly and coherently uses arange of scientific terminology accurately

spelling, punctuation and grammar are used with few errors


Question Number


5(a)(i) Circular/spiral/circle (1)

Question Number


5 (a)(ii) An explanation linking three of the following.

(fast moving) protons(1)

absorbed by(1)

nuclei(1)

(produces)unstable nuclei(1)

bombard / hit /strike / collide with

stable atoms / stable element

(3)

Question Number


5 (b)(i) B momentum (1)

Question Number


5 (b)(ii) (Momentum/it)equals mass x velocity

p = m x v kilograms / kg is the mass and metres per second / m/s is the velocity

Accept “times” for x

(1)


Question Number


QWC *5(b)(iii)

An explanation including some of the following points Diagram 1

Moving in opposite directions before collision inelastic collision stationary after collision momentum zero after collision (therefore) total momentum must have been zero

before collision (therefore) cars were moving at the same speed in

opposite directions (assuming cars have equal mass) both cars had kinetic energy before the collision KE zero after collision KE converted into heat, sound, elastic potential

energy etc.Diagram 2

Elastic collision / almost elastic collision Momentum conserved Momentum transferred from first to last sphere KE conserved / almost conserved (because)last sphere reaches same height as first

sphere Three spheres always have zero momentum Small amount of energy transferred to sound/heat

(6)


(Total marks for question 6 = 12 marks)

evLevelL No rewardable content 1 1 - 2 A limited analysis of ONE collision which is given by a correct

statement e.g. In collision 1, kinetic energy has been lost ORIn collision 2 momentum is transferred from the first to thelast sphere.

the answer communicates ideas using simple language anduses limited scientific terminology


2 3 - 4 a simple analysis of BOTH collisions considering BOTHmomentum AND kinetic energy correctly for each one e.g. Incollision 1, momentum is conserved and the kinetic energyof the cars changes. In collision 2, momentum and the kineticenergy is conserved.

answer communicates ideas showing some evidence of clarityand organisation and uses scientific terminologyappropriately

spelling, punctuation and grammar are used with someaccuracy

3 5 - 6 a detailed analysis of BOTH collisions considering momentumAND kinetic energy for each collision correctly for each ANDdetailed reference to EITHER diagram. e.g. In collision 1, themomentum before and after the collision is zero becausemomentum is always conserved, but the KE is lost. Incollision 2, all the momentum and KE is transferred to thelast sphere because it gets to the same height as the firstone.

the answer communicates ideas clearly and coherently uses arange of scientific terminology accurately

spelling, punctuation and grammar are used with few errors



Level Edexcel

Subject Physics



Sub Topic Momentum


37 minutes

/31

/100

Time Allowed:

Score:

Percentage:



Question Number


1(a)(i) B it decreases (1)

Question Number


1(a)(ii) C it does not change (1)

Question Number


1 (b)(i) horizontal arrow (judge by eye), pointing to the right anywhere on the diagram

(1)

Question Number


1(b)(ii) substitution: (1) 130 000 × 75

evaluation: (1) 9 750 000 (kgm/s) (Ns)


Ignore minus sign 9.75 x 106 (kgm/s) (Ns)

(2)

Question Number


1 (b)(iii) 9 750 000 kgm/s same value as answer to (b)(ii)

Ignore minus sign

(1)

Question Number


1(c)(i) An explanation linking two of the following:

• force is smaller/less (1)• momentum changes more

slowly (1)• lower deceleration (1)• use of the formula (1)

pressure is smaller/less

slower deceleration force is proportional to rate of change of momentum/F= (mv – mu)/t

(2)


Question Number


1(c)(ii) Any two from:

(for loaded aircraft) • has more mass (1)• has more momentum (1)• has more k.e. (1)• higher velocity• brakes need to do more work

(1)

accept reverse argument for empty aircraft

heavier/more passengers/more cargo

higher speed/moving faster

(2) expert

Total marks for question 4 = 10 marks


Question Number


2(a)(i) substitution (1) work done = 84 x 0.25

evaluation (1) 21(J) Full marks for correct answer

even if no working is evident (2)

Question Number


2(a)(ii) 21 J Ecf from (a)(i) (1)

Question Number


2(a)(iii) substitution (1) KE = ½ x 27 x (2.3)2 evaluation (1) = 71.4 (which is approx 71)

V=2.29 gains two marks

Reverse argument which shows that V = √5.3 gains two marks

(2)

Question Number


2 (a)(iv) B(1)


Question Number


QWC *2(b) An explanation linking some of the following points • kinetic energy varies during swing• kinetic energy maximum at bottom of swing• kinetic energy minimum at top of swing• gravitational potential energy(gpe) varies during swing• gpe maximum at top of swing• gpe minimum at bottom of swing• (continuous) interchange of KE and gpe• total amount of energy is constant during one swing• over a number of swings max KE and max PE decreases• energy is dissipated/’lost’ to surroundings• because of air resistance / friction• amplitude/size of swings decrease ( as energy ‘lost’ to

surroundings)

ignore references to momentum

(6)

Level

Descriptor Mark

0 No rewardable content1 1 - 2 • a limited explanation which states some facts e.g.

(max) Kinetic energy decreases over time. KE will transfer to GPE.orKE increases and decreases over one swing. The height which theswing reaches gets less over time.


• spelling, punctuation and grammar are used with limited accuracy2 3 - 4 • a simple explanation with links between facts; either over one

period of oscillation or over several periods of oscillations.Kinetic energy decreases as he gets higher and the GPE increases. There is a continuous interchange of KE and gpe as he swings. or KE is gradually transferred to heat so swing rises to a slightly lower height each time. • the answer communicates ideas showing some evidence of clarity


3 5 - 6 • a detailed explanation with links between facts over one period ofoscillation and over several periods of oscillations e.g.kinetic energy is at a maximum at bottom of swing There is acontinuous interchange of KE and gpe. KE (and gpe) reduce over anumber of swings as energy is dissipated to the surroundings dueto friction.




Question number

Answer Mark

3(a) D (1)

Question number

Answer Mark

3(b) C (1)

Question number


3(c)(i) An explanation that combines identification – application of knowledge (1 mark) and reasoning/justification – application of understanding (1 mark):

• frictional forces increase asmore trucks are added (1)

Plus one from:

• hence, in order to keepconstant speed, thestudent must increase theforce she applies to Z (1)

• when Y and Z separate,the frictional forces (to theleft) are more thanmagnetic attractionbetween Y and Z (1) (2)


Question number

Answer Mark

3(c)(ii) An answer that combines the following points to provide a plan:

• use of a Newton meter used horizontally (1)• record largest force observed (1)• repeat readings several times under same conditions (1)

(3)

Question number

Answer Mark

3(c)(iii) An explanation that combines identification – understanding (1 mark) and reasoning/justification – understanding (1 mark):

• the applied force must be resolved horizontally to determinethe force that separates the engine from the trucks

• and since the (size of) the resolved force is always less thanthe (size of) the actual force then a larger force (applied atan angle) is needed to separate the trucks from the engine (2)


Stopping Distances Mark Scheme

Level Edexcel

Subject Physics



Sub Topic Stopping Distances

Booklet Mark Scheme

34 minutes

/28

/100

Time Allowed:

Score:

Percentage:



Question Number


1 (a) (i) 8 – 0 (m/s) 8 (1)

Question Number


1(a)(ii) substitution 8 / 5 (1)

evaluation (1) 1.6 (m/s2)

ecf from (i)

full marks for correct answer (or ecf) with no working shown.

(2)

Question Number


1(a)(iii) 0 Nil / nothing / zero / none (no mark for no response)

(1)

Question Number


1(b) (1)

(1)

substitution F = 1200 x 0.8

evaluation 960 (N)

full marks for correct answer with no working shown. (2)


Question Number


QWCQWC * ) an explanation linking some of the following points:

compared to a car with just the driver, a fully loaded car will

• have a greater mass / be heavier• greater kinetic energy / momentum• experience the same braking force (when brakes are

applied)• require a greater braking force (than available) to stop

(in the same distance)• have a smaller acceleration / deceleration• take a longer time to come to rest (from given speed)• travel greater distance in this time• needs to do more work with same amount of force• use of relevant equations such as F = ma, work done

= F x d• consequence of driver distractions

(6)

Level 0 No rewardable content 1 1 - 2 • a limited explanation using one idea from the indicative content

eg fully loaded car is heavier.• in answer communicates ideas using simple language and uses


accuracy2 3 - 4 • a simple explanation which links ideas from the indicative content

eg it is heavier and so it takes a longer distance to stop• the answer communicates ideas showing some evidence of clarity


3 5 - 6 • a detailed explanation which links several ideas from theindicative content e.g. It has more momentum and so it will takea longer time to stop. This means that it will travel a furtherdistance. The answer communicates ideas clearly and coherentlyuses a range of scientific terminology accurately



Question Number


2(a) Description including 3 of the following:

• (Gravitational) potentialenergy (transferred) to KE(1)

• Idea of energy transfer toheat/sound whilst descending(1)

• Chemical energy is transferredto heat energy in Andrew (1)

• Idea of energy dissipated onstopping (1)

(G)PE (transferred) to KEAllow gravitational energy forGPE

Energy transferred to heat because of air resistance/ friction

The energy goes to heat as he stops. Energy is transferred to the surroundings

(3)

Question Number


2(b)(i) substitution (1) 67 × 31

evaluation (1) 2077 (kg m/s)

2080, 2100

working backwards using 2000 (v=) 29.85, 30 (m=) 64.52, 65

67 X 31=2000 scores only one mark

(2)

Question Number


2(b)(ii) substitution (1) 2000 ÷ 2.3 evaluation (1) 870 (N)

answer to (b)(i)) ÷ 2.3

900, 869.6, 869.5 903

(2)


Question Number


2(b)(iii) an explanation linking two of the following

• Force on Andrew is quitesmall (1)

• Because impact time islong (1)

• Theacceleration/deceleration isquite small (1)

• Because impact distance isfar (1)

force is reduced/ less /not as strong

slows down/changes momentum gradually

acceleration = 1.35 ‘g’ or 13.5 m/s2

slows down (rate of) change of momentum scores 2 marks

(2)

Total question 2 = 8 marks


Question Number


3(a)(i) D 23 m (1)

Question Number


3(a)(ii) A the driver is tired (1)

Question Number


3(b)

(1)

substitution (1) 800 x 3

evaluation 2400 (kg m/s)

Give full marks for correct numerical answer, even if no working

bald 2.4 x 10n gains 1 mark (BOD for correct substitution) eg bald 240 = 1 mark

In all calculations if the candidate gives two different methods and writes the wrong answer in the answer space award no marks If the candidate writes correct answer they will gain full marks.

(2)

Question Number


3(c)(i) substitution (1) 600 x 15

evaluation (1) 9000 (J)

bald 9.0 x 10n gains 1 mark eg bald 900 = 1 mark (BOD for correct substitution)

give full marks for correct numerical answer, 9000 (J) even if no working

(2)

Question Number


3(c)(ii) A the energy transferred (1)


Documents

Velocity/Distance/ Accceleration (inc Graphs)