Edexcel GCSE Additonal Science P2 Topic 4 test 13_14 with marks scheme

8/12/2019 Edexcel GCSE Additonal Science P2 Topic 4 test 13_14 with marks scheme

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PART A

PART A

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FORMULAE

You may find theses formulae useful

charge = current time Q= I t

potential difference = current resistance V= I R

electrical power = current potential difference P= I V

energy transferred = current potential difference time E= I V t

force = mass acceleration F= m a

weight = mass gravitational field strength W= m g

momentum = mass velocity p= m v

work done = force distance moved in the direction of the force E= F d

GPE = m g h

Do not forget to include units in all your answers.



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PART A

PART A

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Part A

Part A

Surname Name

American Academy LarnacaYear 5 Additional Science

Semester 2 Test 1

Unit P2 Physics for your future

Topic P2.4 Momentum, energy, work and power

Part A Time: 20 minutes

The total marks for this part is 20.

The total marks for the paper is 40.The marks for each question are shown in square brackets.Use this as a guide as to how much time to spend on each question.

You should take particular care with your spelling and grammar, as well as theclarity of expression, on these questions.

Answer the questions in the spaces provided.There may be more space than you need.


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1. The overall stopping distance for a car is given by the equation below:

Overall stopping distance = Thinking distance + Braking distance

The graph below is the speed-time graph for a car initially travelling at a constant speed.The time starts from the moment the driver sees an obstacle on the road.

(i) What is the reaction (thinking) time of the driver of this car? [1]

time =__________(ii) Use the equation

distance = speed time

to calculate the distance travelled by the car during the drivers thinking time. [2]

distance =__________ m

(iii) Calculate the total stopping distance for the car if the braking distance is 52.5m. [1]

total stopping distance =__________m

(iv) How long does it take the car to come to a stop after the brakes have been applied? [1]

time = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . s

[Total marks Q1 = 5]


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2. Dinos skis down a hill.

(a) Dinos starts from the top of the hill and his speed increases as he goes downhill.He controls his speed and direction by using his skis.He brings himself to a stop at the bottom of the hill.Describe the energy changes that happen between starting and stopping. [2]

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

(b) Dinos returns to the top of the hill and starts again.

(i) His mass is 67 kg.Show that his momentum is about 2000 kg m/s when his velocity is 31 m/s. [2]

momentum =__________kgm/s

(ii) He falls over when his momentum is 2000 kg m/s.After he falls over, he slows down by sliding across the snow.

It takes 2.3 s for his momentum to reduce to zero.Calculate the average force on Dinos as he slows down. [2]

average force =__________ N


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(iii) Dinos is not injured by the fall even though he was moving quickly.

Use ideas about force and momentum to explain why he is not injured. [2]

_____________________________________________________________________

_____________________________________________________________________

_____________________________________________________________________

[Total marks for Q2 = 8]

3. A car of mass 1 500 kg, travelling at 15 m/s has its speed reduced to 5 m/s when it travels 7.5 mthrough a pile of sand in the road.

(a) Calculate the loss of kinetic energy of the car. [3]

loss in kinetic energy =__________ J

(b) Use your answer to part (a) along with the equation

force x distance = [change in kinetic energy]

to find the (mean) resistive force produced by the sand during the collision. [3]

resistive force = ........................................ N

(c) Write down the value of the horizontal force that acts on the sand in this collision. [1]

Force on the sand = ........................................ N


[Total marks for Part A = 20]


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PART B

PART B


FORMULAE

You may find theses formulae useful

charge = current time Q= I t

potential difference = current resistance V= I R

electrical power = current potential difference P= I V

energy transferred = current potential difference time E= I V t

force = mass acceleration F= m a

weight = mass gravitational field strength W= m g

momentum = mass velocity p= m v

work done = force distance moved in the direction of the force E= F d

GPE = m g h

Do not forget to include units in all your answers.



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PART B

PART B

BLANK PAGE


9/15

Part B

Part B

Surname Name

American Academy LarnacaYear 5 Additional Science

Semester 2 Test 1

Unit P2 Physics for your future

Topic P2.4 Momentum, energy, work and power

Part B Time: 20 minutes

The total marks for this part is 20.

The total marks for the paper is 40.The marks for each question are shown in square brackets.Use this as a guide as to how much time to spend on each question.

You should take particular care with your spelling and grammar, as well as theclarity of expression, on these questions.

Answer the questions in the spaces provided.There may be more space than you need.


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The shelf is 1.8 m above the floor.

(i) The box has gained gravitational potential energy.Calculate the gain in gravitational potential energy.Gravitational field strength = 10 N/kg [2]

gain in gravitational potential energy =__________J

(ii) The box falls off the shelf.State the kinetic energy of the box just before it hits the floor. [1]

kinetic energy =__________J

(iii) Just before the box hits the floor it has a momentum of 4.8 kg m/s.Calculate the velocity of the box just before it hits the floor. [3]

velocity = ........................................ m/s



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3. A child is stationary on a swing.

(a) The child is given a push by his brother to start him swinging.

His brother applies a steady force of 84 N over a distance of 0.25 m.

(i) Calculate the work done by this force. [2]

work done =__________J

(ii) State how much energy is transferred by this force. [1]

_____________________________________________________________________

(iii) After several more pushes, the child has a kinetic energy of 71 J.The mass of the child is 27 kg.Show that the velocity of the child at this point is about 2.3 m/s. [2]

Velocity =__________ m/s

(iv)Which one of these quantities changes in both size and direction while he is swinging?

Put a cross ( ) in the box next to your answer. [1]

A his gravitational potential energyB his momentumC the force of gravity acting on himD his kinetic energy


[Total marks for Part B = 20]


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Mark scheme

Part A

QuestionNumber

Answer Acceptableanswers

Mark

1(i) 0.6s (1)

1(ii) 25 x 0.6 (1)= 15 (1)

(2)

1(iii) 15 + 52.5= 67.5 (1)

Allow ecf from 1(ii) (1)

1(iv) 4.80.6= 4.2 (1)

(1)

2(a) 2 of the following:

GPEKE(1)

energyheat/sound whilst

descending (1)Chemical energyheat energy inDinos (1)

energy dissipatedon stopping (1)

PE KE

energy heatbecause of airresistance/

friction energy heat

as he stops

energy surroundings

(2)

2(b)(i) 67 31 (1)= 2077 (kg m/s) (1)

2080, 2100

workingbackwardsusing 2000(v=) 29.85, 30

(m=) 64.52, 6567 X 31=2000scores only onemark

(2)

2(b)(ii) 2000 2.3 (1)= 870 (N) (1)

answer to (b)(i)) 2.3 900, 869.6,869.5, 903

(2)

2(b)(iii) explanation linkingtwo of the following

force on Dinos is

quite small (1) because impacttime is long (1)

rate of change ofmomentum is quitesmall (1)

because impactdistance is far (1)

force is reduced/less /not as strongslows down/

changesmomentumgraduallyacceleration = 1.35'g' or 13.5 m/s2slows down (rateof) change ofmomentum scores2 marks

(2)

3(a) x 1500 x 15

= 168 750 (J) (1)

x 1500 x 52

= 18 750 (J) (1)168 75018 750= 150 000 (1)

(3)


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QuestionNumber


Mark

3(b) F x 7.5 = 150 000(1)F = 150 000 7.5(1)= 20 000 (1)

(3)

3(c) 20 000 (1) (1)


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Part B

QuestionNumber


Mark

1(a) C (1) (1)

1(b) A(1) (1)

1(c)(i) 60 10 50 or 600

50 (2)= 30 000 (1)

give two marks for

correct answer noworking

(3)

1(c)(ii) After falling 50 m /when the cordbecomesstraight/when cordfully stretched

tension starting toincrease atterminal velocityignore maximumvelocity/speed

(1)

1(c)(iii) An explanationlinking any two of

not all GPE is

transferred to KE(1)

some of the GPEthermal energy/work is done (1)

due to drag (1)

not all GPE goes toKE maximumenergy is same(value) as GPEbefore falling/speed does notreach the speed atwhich he should fallsome lost asheat/sound (of ropeor movementthrough air) (air)resistance / frictionignore wind

(2)

2(i) 0.8 10 1.8 (1)= 14.4 (1)

give full marks forcorrect answer, noworking

(2)

2(ii) 14.4 (J) e.c.f from part (i) (1)

2(iii) 4.8 = 0.8 v (1)v = 4.8 / 0.8 (1)= 6 (1)

allowsubstitutionand transposition ineither order give fullmarks for correctanswer, no working

(3)

3(a)(i) 84 0.25 (1)= 21 (1)

Full marks forcorrect answer

even if no workingis evident

(2)

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

3(a)(iii) 27 x v2 = 71.4

(1)v = 2.29 (1)

v = 2.29 gains (2).Reverse argumentwhich shows thatV = 5.3 gains (2)

(2)

3(a)(iv) B (1) (1)

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Edexcel GCSE Additonal Science P2 Topic 4 test 13_14 with marks scheme