Post on 28-Jul-2020
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.
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.................................................................................................................................... (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.
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.................................................................................................................................... (2)
(d) What weight of water falls into the pool each minute?
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.................................................................................................................................... (2)
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(e) How much work is done by gravity each minute as the water falls?
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.................................................................................................................................... (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?
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(ii) How much power is available from the falling water?
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(iii) If the generator is 20% efficient, calculate the electrical power output of
the generator.
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.......................................................................................................................... (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
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(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)
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(b) Calculate the acceleration of the car whilst the brakes are applied.
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Answer = .................................... m/s2 (3)
(c) The mass of the car is 1500 kg. Calculate the braking force applied to the car.
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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”.
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(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.
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.......................................................................................................................... (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
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(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.
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........................................................................................................................... (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.
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(ii) Calculate the kinetic energy of the car when it is travelling at a speed of 48 m/s.
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(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.
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................................................................................................................. (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]
Bucklers Mead Community School 8
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.
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Distance ..................................... m (3)
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(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.
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Braking force ........................................ N (3)
Bucklers Mead Community School 10
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.
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....................................................................................................................................
.................................................................................................................................... (2)
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(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);
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Rate .......................... m/s² (2)
(ii) the braking force, if the mass of the car is 900 kg;
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Braking force ............................... N (2)
(iii) the braking distance.
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Braking distance .............................. m (2)
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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.
Bucklers Mead Community School 13
(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)
= ×
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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
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Speed .................... m/s (3)
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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)
Bucklers Mead Community School 15
(ii) Use the following equation to calculate the kinetic energy, in joules, of the orbiter
while it is in orbit.
kinetic energy = ½ mv2
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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.
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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.
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Force = ............................ newtons (1)
(Total 9 marks)
13. (a) (i) gravity/weight 1
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(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]
Bucklers Mead Community School 17
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)
Bucklers Mead Community School 18
(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
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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
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(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
Bucklers Mead Community School 20
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
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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)
Bucklers Mead Community School 21
(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
Bucklers Mead Community School 22
(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.
Bucklers Mead Community School 23
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 ...................................
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........................................................................................................................... (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 ...................................
...........................................................................................................................
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...........................................................................................................................
........................................................................................................................... (3)
Bucklers Mead Community School 24
(b) Use the graph to help you to calculate the distance travelled by the ball bearing between
points C and D.
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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]
Bucklers Mead Community School 25
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)
Bucklers Mead Community School 26
(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]