1. the diagram below shows water falling from a dam. each...

Bucklers Mead Community School 1

1. The diagram below shows water falling from a dam. Each minute 12 000 kg of water falls

vertically into the pool at the bottom.

The time taken for the water to fall is 2 s and the acceleration of the water is 10 m/s².

(a) Assume the speed of the water at the bottom of the dam is zero. Calculate the speed of

the water just before it hits the pool at the bottom.

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

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

(b) Use your answer to part (a) to calculate the average speed of the falling water.

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

(c) Calculate the height that the water falls.

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

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

(d) What weight of water falls into the pool each minute?

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


(e) How much work is done by gravity each minute as the water falls?

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

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

(f) A small electrical generator has been built at the foot of the waterfall. It uses the falling

water to produce electrical power.

(i) How much energy is available from the falling water each minute?

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

(ii) How much power is available from the falling water?

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

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

(iii) If the generator is 20% efficient, calculate the electrical power output of

the generator.

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

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

2. (a) 20 m/s 2

gets 2 marks

Else working

gets 1 mark

(b) 10 m/s 1

(c) 20 m 2

gets 2 marks

Else working

gets 1 mark

(d) 12 000 N 2

gets 2 marks

Else working

gets 1 mark

(e) 2 400 000 J 2

gets 2 marks

Else working

gets 1 mark


(f) (i) Ans to (e) 1

(ii) Ans to (e)/60 1

Else working 1

(iii) Ans to (ii)/5 1 [13]

3. A driver is driving along a road at 30 m/s. The driver suddenly sees a large truck parked across

the road and reacts to the situation by applying the brakes so that a constant braking force stops

the car. The reaction time of the driver is 0.67 seconds, it then takes another 5 seconds for the

brakes to bring the car to rest.

(a) Using the data above, draw a speed-time graph to show the speed of the car from the

instant the truck was seen by the driver until the car stopped.

speed(m/s)

time (s)

(5)


(b) Calculate the acceleration of the car whilst the brakes are applied.

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

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

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

Answer = .................................... m/s2 (3)

(c) The mass of the car is 1500 kg. Calculate the braking force applied to the car.

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

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

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

Answer = .................................... N (3)

(d) The diagrams below show what would happen to a driver in a car crash.

(i) Explain why the driver tends to be thrown towards the windscreen.

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(ii) During the collision the front end of the car becomes crumpled and buckled. Use

this information to explain why such a collision is described as “inelastic”.

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

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


(iii) The car was travelling at 30 m/s immediately before the crash. Calculate the

energy which has to be dissipated as the front of the car crumples.

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

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

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

.......................................................................................................................... (8)

4. (a) Each scale optimum 5

Else both half size

Straight line joining

30,0

to 30,0.67

to 0, 5.67

any 5 for 1 mark each

(b) 6 3

Else a = 30/5

gets 2 marks

Else a = v/t

gets 1 mark

(c) 9000 3

Else F = 6 × 1500

gets 2 marks

Else F = ma

gets 1 mark

(d) (i) Driver has forward momentum 3

Which is conserved

Giving drive relative forward speed to car

for one mark each

(ii) If inelastic ke lost 2

Here ke does work crumpling car

for 1 mark each


(iii) Car stops in 75m 3

gets 1 mark

W = F.d or 9000 × 75

gets 1 mark

W = 675 000 J

OR

ke = 1/2 mv2

gets 1 mark

ke = 1/2.1500.302

ke = 675 000 J [19]

5. A racing driver is driving his car along a straight and level road as shown in the diagram below.

NEAB

5

(a) The driver pushes the accelerator pedal as far down as possible. The car does not

accelerate above a certain maximum speed. Explain the reasons for this in terms of the

forces acting on the car.

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

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

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

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

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

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

(b) The racing car has a mass of 1250 kg. When the brake pedal is pushed down a constant

braking force of 10 000 N is exerted on the car.

(i) Calculate the acceleration of the car.

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

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

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

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

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

(ii) Calculate the kinetic energy of the car when it is travelling at a speed of 48 m/s.


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

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

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

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

(iii) When the brakes are applied with a constant force of 10 000 N the car travels a

distance of 144 m before it stops. Calculate the work done in stopping the car.

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

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

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

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

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

................................................................................................................. (12)

6. (a) there is a (maximum) forward force 4

drag/friction/resistance (opposes motion) (not pressure)

increases with speed

till forward and backward forces equal

so no net force/acceleration

any 4 for 1 mark each

(b) (i) F = ma 4

10 000 = 1250a

a = 8

m/s2

for 1 mark each

(ii) ke = 1/2 mv2 4

ke = 1/2 1250.482

ke = 1 440 000

J

for 1 mark each

(iii) W = Fd 4

W = 10 000.144

W = 1 440 000

J

for 1 mark each

[16]


7. A car driver sees a dog on the road ahead and has to make an emergency stop.

The graph shows how the speed of the car changes with time after the driver first sees the dog.

30

25

20

15

10

5

0

Speed of

car

(m/s)

0 1 2 3 4 5 6

Time (seconds)

A

B

C

(a) Which part of the graph represents the “reaction time” or “thinking time” of the driver?

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

(b) (i) What is the thinking time of the driver? Time ........................ seconds (1)

(ii) Calculate the distance travelled by the car in this thinking time.

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

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

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

Distance ..................................... m (3)


(c) Calculate the acceleration of the car after the brakes are applied.

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

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

(d) Calculate the distance travelled by the car during braking.

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

Distance ................................................ m (3)

(e) The mass of the car is 800 kg. Calculate the braking force.

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

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

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

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

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

Braking force ........................................ N (3)


8. The Highway Code gives tables of the shortest stopping distances for cars travelling at various

speeds. An extract from the Highway Code is given below.

total stopping distance

thinking distance braking distance

thinking distance + braking distance = total stopping distance

(a) A driver’s reaction time is 0.7 s.

(i) Write down two factors which could increase a driver’s reaction time.

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

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

(ii) What effect does an increase in reaction time have on:

A thinking distance; ........................................................................................

B braking distance; ..........................................................................................

C total stopping distance? ................................................................................ (3)

(b) Explain why the braking distance would change on a wet road.

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

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

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

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


(c) A car was travelling at 30 m/s. The driver braked. The graph below is a velocity-time

graph showing the velocity of the car during braking.

Velocity

(m/s)

Time (s)

30

20

10

00 1 2 3 4 5

Calculate:

(i) the rate at which the velocity decreases (deceleration);

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

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

Rate .......................... m/s² (2)

(ii) the braking force, if the mass of the car is 900 kg;

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

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

Braking force ............................... N (2)

(iii) the braking distance.

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

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

Braking distance .............................. m (2)


9. (a) (i) tiredness / boredom 2

drugs

alcohol

distraction

any two for 1 mark each

(ii) A greater / longer 3

B no effect

C greater / longer

each for 1 mark

(b) on a wet road: 2

there is less friction / grip

for 1 mark

braking distance is greater / takes longer to stop

or car skids / slides forward

for 1 mark

(c) (i) deceleration = gradient or 30 / 4.8 2

each for 1 mark

(ii) force = mass × acceleration or 900 × 6.25 2

each for 1 mark

(iii) distance = area under graph or 0.5 × 4.8 × 30 or average 2

speed × time or 15 × 4.8

Accept answer in terms of change in k.e. = work done

if incorrect unit given (eg 72km) then no mark

each for 1 mark

[13]

10. The diagram shows a high jumper.

In order to jump over the bar, the high jumper must raise his mass by 1.25m.

The high jumper has a mass of 65kg. The gravitational field strength is 10N/kg.


(a) The high jumper just clears the bar.

Use the following equations to calculate the gain in his gravitational potential energy.

weight

(newton, N)

mass

(kilogram, kg)

gravitational field strength

(newton/kilogram, N/kg)

= ×

weight

(newton, N)

change in vertical height

(metre, m)

change in gravitational potential energy

(joule, J)

= ×

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

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

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

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

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

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

Gain in gravitational potential energy .................... J (4)

(b) Use the following equation to calculate the minimum speed the high jumper must reach

for take-off in order to jump over the bar.

kinetic energy

(joule, J)

mass

(kilogram, kg)

[speed]

[(metre/second) , (m/s) ]

= ××1

22

2 2

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

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

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

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

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

Speed .................... m/s (3)


11. (a) W = 65 × 10 4

(allow a maximum of 3 marks if candidate uses

g=9.8N / Kg (as ecf))

gains 1 mark

but

W = 650 (N)

(allow use of p.e.= m × g × h)

gains 2 marks

but

PE change = 650 × 1.25 or 65 × 10 × 1.25

gains 3 marks

but PE change = 812.5 (J) (allow 813J or 812J)

gains 4 marks

(b) k.e. = p.e. 3

gains1 mark

but

(speed)² = 812.5 × 2 / 65 or 812.5 = ½ × 65 × (speed)² ecf

gains 2 marks

but

speed = 5 (m/s) (allow 4.99 5.002)

(if answer = 25mls check working: 812.5 = ½ m × v gains 1 mark for

KE=PE)

(but if 812.5 = ½m × v² = ½ × 65 × v2 or v2 = 65

5.8122 gains 2 marks)

25, with no working shown gains 0 marks

gains 3 marks

[7]

12. The diagram shows an orbiter, the reusable part of a space shuttle. The data refers to a typical

flight.

Orbiter data

Mass

Orbital speed

Orbital altitude

Landing speed

Flight time

78 000 kg

7.5 km/s

200 km

100 m/s

7 days

(a) (i) What name is given to the force which keeps the orbiter in orbit around the Earth?

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


(ii) Use the following equation to calculate the kinetic energy, in joules, of the orbiter

while it is in orbit.

kinetic energy = ½ mv2

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

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

Kinetic energy = ............................. joules (2)

(iii) What happens to most of this kinetic energy as the orbiter re-enters the Earth’s

atmosphere?

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

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

(b) After touchdown the orbiter decelerates uniformly coming to a halt in 50 s.

(i) Give the equation that links acceleration, time and velocity.

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

(ii) Calculate the deceleration of the orbiter. Show clearly how you work out your

answer and give the unit.

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

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

Deceleration = ............................... (2)

(c) (i) Give the equation that links acceleration, force and mass.

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

(ii) Calculate, in newtons, the force needed to bring the orbiter to a halt. Show clearly

how you work out your answer.

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

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

Force = ............................ newtons (1)

(Total 9 marks)

13. (a) (i) gravity/weight 1


(ii) 2193750000000 2

or 2.19 × 1012

not 2.1912

allow 1 mark for the correct conversion to 7500 (m/s)

allow one mark for answer 2193750(J)

transferred to heat 1

ignore extras of sound and light

accept changed to heat

accept lost due to friction

(b) (i) acceleration = (taken)time

velocityinchange 1

accept word speed instead of velocity

accept a = t

uv

or correct rearrangement

do not accept

V

a t

even if subsequent calculation correct

v-u

a t

can gain credit if subsequent calculation correct

(ii) 2 1

ignore + or signs

m/s2 1

accept m/s/s or ms 2

(c) (i) force = mass × acceleration 1

accept correct rearrangement

accept F = m × a

do not accept

f

m a

unless subsequent calculation correct

(ii) 156 000 1

accept 78 000 × their (b)(ii)(only if (b)(i) correct)

[9]


14. (a) The graph shows how the distance travelled by a car changes with time during a short

journey.

1800

1600

1400

1200

1000

800

600

400

200

0

Distance

in metres

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

Time in minutes

(i) Describe fully the motion of the car during the first two minutes of the journey.

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

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

(ii) During the last minute of the journey the velocity of the car changes although the

speed remains constant. How is this possible?

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

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


(iii) “During the journey the car engine is 22% efficient.” Explain what this statement

means.

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

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

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

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

(b) The diagram shows a spanner being used to undo a tight nut.

Force

30 cm

50 cm

40 cm

The nut was tightened using a moment of 120 newton metres.

Use the following equation to calculate the force needed to undo the nut. Show clearly

how you work out your answer.

moment = force × perpendicular distance from pivot

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

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

Force = ............................................. N (2)

(Total 9 marks)

15. (a) (i) first statement must be accelerated 1

if it just accelerated then

decelerates award 2 marks

final statement must be stationary 1

interim statement decelerates 1

(ii) direction is changing 1


(iii) only 22% of the (total) energy input 1

from the (chemical) energy of the fuel 1

is transferred to (useful) kinetic 1

energy

accept rest is wasted as heat (and sound) if second marking

point missing

(b) 300 2

allow 1 mark for rearranging equation or correct substitution

[9]

16. The table shows the braking distances for a car at different speeds and kinetic energy. The

braking distance is how far the car travels once the brakes have been applied.

Braking distance

in m

Speed of car in

m/s

Kinetic energy of

car in kJ

5 10 40

12 15 90

20 20 160

33 25 250

45 30 360

(a) A student suggests, “the braking distance is directly proportional to the kinetic energy.”

(i) Draw a line graph to test this suggestion.

Kinetic

energy

in

kilojoules

(kJ)

Braking distance in metres (m) (3)

(ii) Does the graph show that the student’s suggestion was correct or incorrect? Give a


reason for your answer.

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

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

(iii) Use your graph and the following equation to predict a braking distance for a speed

of 35 metres per second (m/s). The mass of the car is 800 kilograms (kg). Show

clearly how you obtain your answer.

kinetic energy = ½ mv2

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

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

Braking distance = ........................................ m (2)

(iv) State one factor, apart from speed, which would increase the car’s braking distance.

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

(b) The diagram shows a car before and during a crash test. The car hits the wall at

14 metres per second (m/s) and takes 0.25 seconds (s) to stop.

WallDummy

(i) Write down the equation which links acceleration, change in velocity and time

taken.

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

(ii) Calculate the deceleration of the car.

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

Deceleration = ........................................ m/s2 (1)


(iii) In an accident the crumple zone at the front of a car collapses progressively. This

increases the time it takes the car to stop. In a front end collision the injury to the

car passengers should be reduced. Explain why. The answer has been started for

you.

By increasing the time it takes for the car to stop, the ...................................

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

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

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

(Total 11 marks)

17. (a) (i) linear scales used 1

do not credit if less than half paper used

points plotted correctly 1

all of paper used

(straight) line of best fit drawn 1

allow a tolerance of half square

(ii) correct and straight line through 1

origin

all needed

e.c.f. if their (a)(i) is straight but not through the origin -

incorrect because line does not go through origin

credit a calculation that shows proportionality

(iii) 62 ± 0.5 (m) 2

credit 1 mark for KE = 490000 or 490kJ

credit 1 mark for correct use of graph clearly shown

(iv) any one from: 1

wet or icy or worn or smooth road

accept slippery slope

brakes worn

accept faulty brakes

car heavily loaded

worn tyres

downhill slope

do not accept anything to do with thinking distance e.g. driver

tired or drunk


(b) (i) acceleration = takentime

velocityinchange 1

accept correct transformation

accept t

uv= a

accept m/s2 = s

m/s

do not accept acceleration = t ime

velocity

(ii) 56 1

accept 56

(iii) deceleration is reduced 1

accept deceleration is slower

accept acceleration

force on car and or passengers is 1

reduced

accept an answer in terms of change in momentum for full credit

[11]

18. A student carries out an experiment with a steel ball bearing and a tube of thick oil.

The diagram shows the apparatus used.

The student releases the ball bearing and it falls through the oil.

steel ball bearing

glass tube

thick oil

The forces X and Y act on the ball bearing as it falls through the oil.

This is shown on the diagram.

force

force

Y

X

ball bearing

The graph shows how the speed of the ball bearing changes as it falls through the oil.


A

B

C D

0 2.0 4.0time (s)

speed

(cm/s)

2.4

0

(a) (i) What is happening to the speed of the ball bearing between points A and B?

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

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

Explain, in terms of forces X and Y, why this happens ...................................

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

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

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

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

(ii) What is happening to the speed of the ball bearing between points C and D?

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

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

Explain, in terms of forces X and Y, why this happens ...................................

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

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

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

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


(b) Use the graph to help you to calculate the distance travelled by the ball bearing between

points C and D.

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

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

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

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

(Total 8 marks)

19. (a) (i) speed increasing or ball bearing 1

accelerating

force X greater than force Y 1

(ii) speed constant

any three from:

force Y increases

force X constant

force Y becomes equal to force X

no net force

in absence of any other marks allow terminal velocity 1 mark

(b) Distance = area under graph 2

allow speed × time

gains 1 mark

But distance = 4.8cm

gains 2 marks

[8]


20. A car travelling along a straight road has to stop and wait at red traffic lights. The graph shows

how the velocity of the car changes after the traffic lights turn green.

15

10

5

0

0 1 2 3 4 5 6 7 8 9

Time in seconds (s)

Velocity in

metres/second

(m/s)

(a) Between the traffic lights changing to green and the car starting to move there is a time

delay. This is called the reaction time. Write down one factor that could affect the driver’s

reaction time.

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

(b) Calculate the distance the car travels while accelerating. Show clearly how you work out

your answer.

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

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

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

(c) Calculate the acceleration of the car. Show clearly how you work out your final answer

and give the units.

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

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

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

Acceleration = ................................................................... (4)

(d) The mass of the car is 900 kg.

(i) Write down the equation that links acceleration, force and mass.

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


(ii) Calculate the force used to accelerate the car. Show clearly how you work out your

final answer.

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

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

Force = ..................................................... newtons (2)

(Total 11 marks)

21. (a) concentration / tiredness / drugs / alcohol 1

accept any reasonable factor that could affect a driver’s

reactions

do not accept speed or any physical condition unrelated to the

driver

(b) 31.25 3

credit for 1 mark correct attempt to calculate the area under the

slope or for using the equation

distance = average velocity (speed) × time

credit for 1 mark use of correct velocity change (12.5) and

correct time (5) or answer of 62.5

(c) 2.5 3

credit for 1 mark triangle drawn on slope or correct equation or

two correct pairs of coordinates

credit for 1 mark use of correct velocity change (12.5) and

correct time (5)

accept time = between 4.8 and 5.2 if used in (b)

do not accept an attempt using one pair of coordinates taken

from the slope

metres / second / second or

metres / second / squared or

m/s2 or ms–2 1

(d) (i) force = mass × acceleration 1

accept correct transformation

accept F = m × a

accept F

m a provided

subsequent use of is correct

do not accept an equation in units

(ii) 2250 2

credit their (c) × 900 for 2 marks

credit 1 mark for correct substitution

[11]

1. the diagram below shows water falling from a dam. each...

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