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I GOT HOLD OF THIS PAPER FROM ANOTHER FORUM THANKS TO THAT PERSON BUT I HOLD CREDITS FOR COMBING IT 
I HAVE ALSO GOT HOLD OF THE ANSWER THAT SOME ONE MADE
-
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EDEXCEL UNIT 1 PHYSICS MAY 2013 PAPER
- Thread starter e1f123
- Start date
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Multiple Choice:
1. Which pair of quantities does not contain a vector and a scalar?
A - Acceleration and Time
B - Force and Displacement
C - Mass and Acceleration
D - Velocity and Time
2. A wire of length 80cm has force F applied. The new length of the wire is 84cm.
Strain is given by...
A - 4/84
B - 4/80
C - 80/84
D - 84/80
3. Which of the following is a derived SI quantity?
A - Force
B - Length
C - Second
D - Watt
4. A projectile launched at 45 degrees to the horizontal. Ignoring air resistance which graphs show how the horizontal and vertical components vary with time for the projectile until it lands?
A - Vertical [Upsidedown U shape], Horizontal [Horizontal line]
B - Vertical [Upsidedown U shape], Horizontal [Diagonal line: Origin to Positive]
C - Vertical [Diagonal line: Positive to Negative], Horizontal [Horizontal line]
D - Vertical [Diagonal line: Positive to Negative], Horizontal[Diagonal line: Origin to Positive]
5. The graph shows stress against strain up to the breaking point for two materials X and Y.
Which row in the table correctly identifies the behaviour of each material?
--------------X----------------------Y---------------
A ----------Brittle------------Ductile---------
B ----------Ductile--------------Brittle-------------
C ---------Ductile----------------Hard--------------
D ---------Brittle-----------------Hard--------------
6. A ball of volume Vb and density Pb is released in a volume V1 of liquid with density P1.
The Upthrust on the ball is given by:
A - Vb * Pb * g
B - Vb * P1 * g
C - V1 * Pb * g
D - V1 * P1 * g
7. A hanging basket of weight W is supported by three chains of equal length, each at an angle # to the vertical.
The Tension, T, in each chain is given by:
A - T = 3W/Cos#
B - T = 3W/Sin#
C - T = W/3Cos#
D - T = W/3Sin#
8. Which of the following descriptions of a material implies that it undergoes significant plastic deformation?
A - brittle
B - hard
C - malleable
D - stiff
9. A trolley rolls down a slope from rest. The trolley moves through a vertical heighth while rolling a distance s along the slope.
The maximum possible speed is given by:
A - 2gs
B - 2gh
C - Rt/(2gs)
D - Rt/(2gh)
10. An apple is at rest on the ground.
The diagram shows three forces of equal magnitude.
W: weight of apple
P: push of apple on ground
R : normal contact force of ground on apple
Which row in the table shows Newton's first and third laws being applied correctly?
-------------------Newton's First Law--------------Newton's Third Law----------
A -----------------------P=W-------------------------------------R=P--------------------
B -----------------------R=P--------------------------------------W=R-------------------
C -----------------------W=R-------------------------------------P=W-------------------
D ---------------------W=R---------------------------------R=P------------------
- Messages
- 40
- Reaction score
- 18
- Points
- 8
Written Section:
11 Viscosity is sometimes given units of kg m^-1 s^-1 and sometimes Pa s
Show these are equivalent. [2]
Pa * s
(Force * s)/Area
(Mass * Acceleration * s)/Area
(Kg * m * s^-2 * s)/m^2
You cancel the m with the m^-2 to get m^1. You cancel s^-2 with s to get s^-1.
=> kg*m^-1*s^-1
Essentially you need to show the base units of Pa s can be shown to be kg M^-1 s^-1. As long as you did that with some clarity that should be your two marks in the bag no problem.
12 (a) State what is meant by centre of gravity. [1]
A point from which the weight of a body or system may be considered to act. (or something to that effect, wording does not need to be exact)
(b) The picture shows a snooker cue. It is made from wood of uniform density and takes the form of a rod with decreasing diameter towards one end.
(i) On the picture, mark the position of the centre of gravity of the snooker cue.[1]
Draw a point on the snooker cue between the center and the left side (thickest diameter end). This does not need to be perfectly at a certain point, just acknowledge that the center of gravity is more towards the thicker end.
(ii) State a simple method to test if this is the correct position.[1]
Place the snooker cue on a pivot, if it falls off it is the incorrect position. (or something to the same effect)
13. Queues of cars often form behind cyclists on narrow, rural roads.
Sometimes cars that would norrnally travel at 65 km hour^-1 muy be limited to about 20 km hour^-1 by a cyclist.
(a) Show that 65 km hour^-1 is about 18 m s^-1.[1]
65000m per (60*60) seconds
65000/3600 = 18.06 m s^-1
(b) The graph shows the amount of carbon dioxide emitted per kilometre by a typical car at different speeds.
During a 10 minute journey a cyclist, travelling at 5 m s^-1, has an average of three cars queuing behind him. The cars would otherwise be travelling at 18 m s^-1. The cars emit more carbon dioxide because they are travelling slowly.
(i) Calculate the extra carbon dioxide emitted by the 3 cars due to traveling at this reduced speed.[4]
Distance of journey = (10*60 seconds)*5 m s^-1 = 3000m (3km)
CO2 emissions/kg km^-1 at 18 m s^-1 = 0.18kg ~ 0.185 (from graph)
Emissions of 1 car at 18 m s^-1 = 3*0.18 = 0.54 kg
CO2 emissions/kg km^-1 at 5 m s^-1 = 0.26
Emissions from 1 car at 5 m s^-1 = 3*0.26 = 0.78 kg
Extra emissions from 1 car at the reduced speed = 0.78-0.54 = 0.24 kg.
Total extra emissions for all 3 cars = 3*0.24 = 0.72kg(or similar depending on your reading of emissions at 18 m s^-1)
(ii) If the cyclist had made the same journey in his car at 18 m s^-1, his car would have emitted 0.54 kg of carbon dioxide. Comment on the significance of this.[1]
The significance is that there would have been less total pollution if he had made the journey in his car at 18 m s^-1.
Show these are equivalent. [2]
Pa * s
(Force * s)/Area
(Mass * Acceleration * s)/Area
(Kg * m * s^-2 * s)/m^2
You cancel the m with the m^-2 to get m^1. You cancel s^-2 with s to get s^-1.
=> kg*m^-1*s^-1
Essentially you need to show the base units of Pa s can be shown to be kg M^-1 s^-1. As long as you did that with some clarity that should be your two marks in the bag no problem.
12 (a) State what is meant by centre of gravity. [1]
A point from which the weight of a body or system may be considered to act. (or something to that effect, wording does not need to be exact)
(b) The picture shows a snooker cue. It is made from wood of uniform density and takes the form of a rod with decreasing diameter towards one end.
(i) On the picture, mark the position of the centre of gravity of the snooker cue.[1]
Draw a point on the snooker cue between the center and the left side (thickest diameter end). This does not need to be perfectly at a certain point, just acknowledge that the center of gravity is more towards the thicker end.
(ii) State a simple method to test if this is the correct position.[1]
Place the snooker cue on a pivot, if it falls off it is the incorrect position. (or something to the same effect)
13. Queues of cars often form behind cyclists on narrow, rural roads.
Sometimes cars that would norrnally travel at 65 km hour^-1 muy be limited to about 20 km hour^-1 by a cyclist.
(a) Show that 65 km hour^-1 is about 18 m s^-1.[1]
65000m per (60*60) seconds
65000/3600 = 18.06 m s^-1
(b) The graph shows the amount of carbon dioxide emitted per kilometre by a typical car at different speeds.
During a 10 minute journey a cyclist, travelling at 5 m s^-1, has an average of three cars queuing behind him. The cars would otherwise be travelling at 18 m s^-1. The cars emit more carbon dioxide because they are travelling slowly.
(i) Calculate the extra carbon dioxide emitted by the 3 cars due to traveling at this reduced speed.[4]
Distance of journey = (10*60 seconds)*5 m s^-1 = 3000m (3km)
CO2 emissions/kg km^-1 at 18 m s^-1 = 0.18kg ~ 0.185 (from graph)
Emissions of 1 car at 18 m s^-1 = 3*0.18 = 0.54 kg
CO2 emissions/kg km^-1 at 5 m s^-1 = 0.26
Emissions from 1 car at 5 m s^-1 = 3*0.26 = 0.78 kg
Extra emissions from 1 car at the reduced speed = 0.78-0.54 = 0.24 kg.
Total extra emissions for all 3 cars = 3*0.24 = 0.72kg(or similar depending on your reading of emissions at 18 m s^-1)
(ii) If the cyclist had made the same journey in his car at 18 m s^-1, his car would have emitted 0.54 kg of carbon dioxide. Comment on the significance of this.[1]
The significance is that there would have been less total pollution if he had made the journey in his car at 18 m s^-1.
14 The gravitational field strength on the Moon is about l/6 of the gravitational field strength on the Earth.
(a) On the Moon, an astronaut dropped a golf ball. He later wrote "When I dropped the ball, it took about three seconds to land.'
Show that the astronaut would need to be over 7m tall for the ball to take 3 s to land.[2]
9.81 * (1/6) = 1.64 m s^-2
u = 0, a = 1.64, s = ?, t = 3
s = ut + 1/2at^2
s = 1/2*1.64*3^2
s = 7.38m
(b) The astronaut hit the ball with a golf club. He wrote "The ball, which would have gone thirty to forty yards on the Earth, went over two hundred yards. The ball stayed up in the black sky for almost thirty seconds."
Assume an initial velocity of 18 m s^-1 at 34 degrees to the horizontal.
(i) Show that the astronaut's suggested time of flight of 30 s is over twice the actual
value.[3]
Inital vertical velocity = 18sin34 = 10.07m s^-1
u = 10.1, v = 0, a = -1.64, t = ?
v = u + at
0 = 10.1 + -1.64t
t = 10.1/1.64 = 6.1 seconds (half journey)
Whole journey = 6.1*2 = 12.2 seconds which makes 30 s over twice the actual flight time.
(ii) Show that the value given for the initial velocity leads to a value for the horizontal distance travelled by the ball in agreement with his stated value. [3]
200 yards: 183 m
Horizontal velocity = 18cos34 = 14.9 m s^-1
Speed = Distance/Time (horizontal velocity is constant)
14.9 = distance / 12.2
Distance = 14.9*12.2 = 181.8m (aprox)
*(c) A projectile would have a greater range on the Moon than the Earth because of the lower gravitational field strength and because of the lack of an atmosphere.
Explain how each of these factors would increase the range of the projectile.[3]
Lower gravitational field strength => longer time in air
Lack of atmosphere => constant horizontal velocity
Both of those result in longer range of projectile (You need to have said things to this effect)
15 The photographs show an exercise device and someone using it. The device contains two rubber cords which atre extended when the device is used.
A student investigates the properties of the device by hanging weights on it and measuring the extension.
The student obtains a graph for her results.
(a) The student notices that her graph is a straight line between A and B and concludes that the device obeys Hooke's law.
Comment on this conclusion. [2]
I said it was incorrect as the force is not directly proportional to the extension for the entire graph until the limit of proportionality, this graph only has a section of it proportional. (Not sure about this one)
(b) (i) Describe how the student could use the graph to obtain an estimate of the total work done.[2]
Calculate the energy stored in each square (0.025*10), find the total of the squares under the graph for the total work done (or something to that effect)
(ii) The student sets up a spreadsheet to investigate the work done in stretching the device each time a weight is added.
Explain why this spreadsheet results in an over-estimate for the total work done.[2]
Not too sure about this one, can't really remember what i put INPUT REQUESTED!
(c) The student eats a packet of crisps and then uses the exercise device. The energy
content in a packet of crisps is 540 kJ. During exercise this energy is converted and
25% of it is transferred to mechanical work.
The student extends the device fully 15 times in 1 minute. An accurate value for the work done in fully extending the device is 14.7 J.
Calculate the time it would take the student, working at this rate, to transfer 25% of the energy from the crisps to mechanical work.[3]
540000 * 0.25 = 135000 J
Full extensions to transfer that energy = 135000/14.7 = 9184 extensions
Time = 9184/15 = 612 minuets (or ~36700 seconds)
(NOTE: It doesn't specify time units, either is fine)
(d) Explain whether more or less work would be done applying the same maximum total stretching force to a similar exercise device with rubber cords of twice the cross sectional area [2]
You will pull it a shorter distance with the same force so less work would be done (work = force * distance).
(a) On the Moon, an astronaut dropped a golf ball. He later wrote "When I dropped the ball, it took about three seconds to land.'
Show that the astronaut would need to be over 7m tall for the ball to take 3 s to land.[2]
9.81 * (1/6) = 1.64 m s^-2
u = 0, a = 1.64, s = ?, t = 3
s = ut + 1/2at^2
s = 1/2*1.64*3^2
s = 7.38m
(b) The astronaut hit the ball with a golf club. He wrote "The ball, which would have gone thirty to forty yards on the Earth, went over two hundred yards. The ball stayed up in the black sky for almost thirty seconds."
Assume an initial velocity of 18 m s^-1 at 34 degrees to the horizontal.
(i) Show that the astronaut's suggested time of flight of 30 s is over twice the actual
value.[3]
Inital vertical velocity = 18sin34 = 10.07m s^-1
u = 10.1, v = 0, a = -1.64, t = ?
v = u + at
0 = 10.1 + -1.64t
t = 10.1/1.64 = 6.1 seconds (half journey)
Whole journey = 6.1*2 = 12.2 seconds which makes 30 s over twice the actual flight time.
(ii) Show that the value given for the initial velocity leads to a value for the horizontal distance travelled by the ball in agreement with his stated value. [3]
200 yards: 183 m
Horizontal velocity = 18cos34 = 14.9 m s^-1
Speed = Distance/Time (horizontal velocity is constant)
14.9 = distance / 12.2
Distance = 14.9*12.2 = 181.8m (aprox)
*(c) A projectile would have a greater range on the Moon than the Earth because of the lower gravitational field strength and because of the lack of an atmosphere.
Explain how each of these factors would increase the range of the projectile.[3]
Lower gravitational field strength => longer time in air
Lack of atmosphere => constant horizontal velocity
Both of those result in longer range of projectile (You need to have said things to this effect)
15 The photographs show an exercise device and someone using it. The device contains two rubber cords which atre extended when the device is used.
A student investigates the properties of the device by hanging weights on it and measuring the extension.
The student obtains a graph for her results.
(a) The student notices that her graph is a straight line between A and B and concludes that the device obeys Hooke's law.
Comment on this conclusion. [2]
I said it was incorrect as the force is not directly proportional to the extension for the entire graph until the limit of proportionality, this graph only has a section of it proportional. (Not sure about this one)
(b) (i) Describe how the student could use the graph to obtain an estimate of the total work done.[2]
Calculate the energy stored in each square (0.025*10), find the total of the squares under the graph for the total work done (or something to that effect)
(ii) The student sets up a spreadsheet to investigate the work done in stretching the device each time a weight is added.
Explain why this spreadsheet results in an over-estimate for the total work done.[2]
Not too sure about this one, can't really remember what i put INPUT REQUESTED!
(c) The student eats a packet of crisps and then uses the exercise device. The energy
content in a packet of crisps is 540 kJ. During exercise this energy is converted and
25% of it is transferred to mechanical work.
The student extends the device fully 15 times in 1 minute. An accurate value for the work done in fully extending the device is 14.7 J.
Calculate the time it would take the student, working at this rate, to transfer 25% of the energy from the crisps to mechanical work.[3]
540000 * 0.25 = 135000 J
Full extensions to transfer that energy = 135000/14.7 = 9184 extensions
Time = 9184/15 = 612 minuets (or ~36700 seconds)
(NOTE: It doesn't specify time units, either is fine)
(d) Explain whether more or less work would be done applying the same maximum total stretching force to a similar exercise device with rubber cords of twice the cross sectional area [2]
You will pull it a shorter distance with the same force so less work would be done (work = force * distance).
16 The 'Stealth'roller coaster at the Thorpe Park theme park is advertised as reaching 135 km hour^-1 from rest in 2.3 seconds.
Most roller coasters are driven slowly up to the top of a slope at the start of the ride.
However the carriages on Stealth are initially accelerated horizontally from rest at ground level by a hydraulic launch system, before rising to the top of the first slope.
(a) (i) Calculate the average acceleration of the carriages [2]
135 km hourr : 37.5 m s^-1
37.5/2.3 = 16.3 m s^-2
(ii) Calculate the minimum average power which must be developed by the launch system. [3]
mass of carriages and passengers : 10 000 kg
u = 0 v = 37.5 a = 16.3 t = 2.3 s = ?
s = 1/2 * 16.3 * 2.3^2 = 43.1m
Power = Work Done / Time
Power = (Force * Distance) / Time
Power = ((mass * acceleration) * Distance) / Time
=> ((10000*16.3)*43.1)/2.3 = 3.05x10^6 W
(iii) Suggest why the power in (ii) is a minimum value. [1]
Because it must overcome resistance forces such as air resistance/track friction
*(b) The force required to launch ostealth'is not always the same. The ride is monitoredand the data from preceding launches is used to calculate the required force.
If the mass of the passengers for a particular ride is significantly more than for
preceding launches, this can lead to 'rollback'. This is when the carriages do not quite reach the top of the first slope and return backwards to the start.
Explain why 'rollback'would occur in this situation. [3]
F = ma, with a greater mass and the same force you get a lower acceleration a = F/m. So doesn't reach a high enough velocity, KE = 1/2mv^2, not enough energy to overcome GPE.
(c) Suggest why roller coasters may have agreater acceleration when the lubricating oil between the movingpafts has had time to warm up.[2]
Lubricating oil is higher temperature, means lower viscosity. Therefore there is lower friction due to the moving parts and a greater resultant force.
Most roller coasters are driven slowly up to the top of a slope at the start of the ride.
However the carriages on Stealth are initially accelerated horizontally from rest at ground level by a hydraulic launch system, before rising to the top of the first slope.
(a) (i) Calculate the average acceleration of the carriages [2]
135 km hourr : 37.5 m s^-1
37.5/2.3 = 16.3 m s^-2
(ii) Calculate the minimum average power which must be developed by the launch system. [3]
mass of carriages and passengers : 10 000 kg
u = 0 v = 37.5 a = 16.3 t = 2.3 s = ?
s = 1/2 * 16.3 * 2.3^2 = 43.1m
Power = Work Done / Time
Power = (Force * Distance) / Time
Power = ((mass * acceleration) * Distance) / Time
=> ((10000*16.3)*43.1)/2.3 = 3.05x10^6 W
(iii) Suggest why the power in (ii) is a minimum value. [1]
Because it must overcome resistance forces such as air resistance/track friction
*(b) The force required to launch ostealth'is not always the same. The ride is monitoredand the data from preceding launches is used to calculate the required force.
If the mass of the passengers for a particular ride is significantly more than for
preceding launches, this can lead to 'rollback'. This is when the carriages do not quite reach the top of the first slope and return backwards to the start.
Explain why 'rollback'would occur in this situation. [3]
F = ma, with a greater mass and the same force you get a lower acceleration a = F/m. So doesn't reach a high enough velocity, KE = 1/2mv^2, not enough energy to overcome GPE.
(c) Suggest why roller coasters may have agreater acceleration when the lubricating oil between the movingpafts has had time to warm up.[2]
Lubricating oil is higher temperature, means lower viscosity. Therefore there is lower friction due to the moving parts and a greater resultant force.
- Messages
- 40
- Reaction score
- 18
- Points
- 8
17. Many hand held devices such as smartphones and tablet computers contain
accelerometers. These allow changes in orientation of the device to be tracked.
A student models a simple accelerometer by attaching a small mass on a string to the
roof of a car.
(a) (i)Complete a free body force diagram for the mass when the car starts moving.[2]
Tension to North-East as you look at the page, Weight down, any others?
(ii) Draw a vector diagram, in the space below, to show how the resultant force on
the mass is produced.[2]
NOT A CLUE
(iii) When the string is at 7 degrees to the vertical, show that the acceleration of the car is 1 m s^-1
9.81sin7 = 1.2? NOT SURE
(b) Sketch the positions of the mass and string when the car is moving in the same
direction and is:
(i) moving with constant velocity[1]
Vertical
(ii) undergoing a much greater acceleration than in (a)(iii)[1]
Higher than (a)(iii) on the left side.
(iii)deceleration[1]
Up on the right side
(these explanations are a bit rubbish, hopefully you know what I mean)
(c) Explain why the string would not become horizontal, however great the acceleration.[2]
There is always a vertical component/weight down. (2 marks?...)
(d) Suggest why many devices contain 3 accelerometers, arranged at right angles to
each other[1]
Each accelerometer covers a plane of motion (as if it were X,Y,Z axis on a graph)
18 Two balls, one solid, one hollow blah blah blah
The two balls dropped, shows the Velocity/Time graphs.
(a) State how the graphs show that neither ball reaches terminal velocity.[1]
Terminal velocity is where it has reached maximum velocity and so would be a horizontal line on the graph. There is no horizontal line.
(b) (i) By drawing a tangent to the graph, to the show that the acceleration of the hollow ball
at time t:0.60 s is about 7 m s^-2.[2]
Draw tangent, calculate gradient. Can't do too accurately on computer.
(ii) Show that the resultant force on the hollow ball at t:0.60s is about 0.02 N.
mass of hollow ball: 2.4 g[2]
F=ma (need a from (i))
(iii) Show that the drag force on the hollow ball at t:0.60s is about 0.01 N. You
may neglect upthrust.[2]
Sub into stoke's law equation
(iv) Demonstrate that the Stoke's law force is not sufficient to produce this drag
force.[2]
radius of hollow ball: 2.0 cm
viscosity of air: 1.8 x 10-s Pa s
(c) The diagram shows the air flow around the hollow ball as it falls.
(i) Add labels to show laminar flow and turbulent flow.[1]
Label the flows accurately
(ii) Suggest why the drag is much greater than the Stoke's law force[1]
The turbulence causes additional drag
(d) Without further calculation, use the graph to describe the motion of the solid ball.[3]
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thx
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hey, are you sure that the answer to 3-is D notA
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Thank you.
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The is Defo A
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Thank you.
No Problemo
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Ok si I'm screwed
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You've done so much of hard work..
u could've done it after ur exams...
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so guys wat do u think the grade boundaries will be.....it was indeed a difficult ppr....n many like me wuld hav messed up!!!!!!isnt it?????
i hope the boundary is low
i hope the boundary is low