ultimate momentum edexel physics past papers questions & mark scheme 2001-2009

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A2 PHYSICS UNIT 4 PHYSICS ON THE MOVE EDEXCEL PAST PAPERS 2001-2009 MOMENTUM NEWTON 2ND & 3RD LAW ELASTIC COLLISION INELASTIC COLLISION EXPLOSION DERYK NG A LEVEL ACADEMY UCSI UNIVERSITY

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UNIT 4 NEW SYLLABUS PHYSICS ON THE MOVE EDEXCEL MOMENTUM

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Page 1: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

A2 PHYSICS UNIT 4 PHYSICS ON THE MOVE

EDEXCEL PAST PAPERS 2001-2009MOMENTUM

NEWTON 2ND & 3RD LAWELASTIC COLLISION

INELASTIC COLLISIONEXPLOSION

DERYK NGA LEVEL ACADEMYUCSI UNIVERSITY

Page 2: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

PHY1 JANUARY 2001

1

1. The diagram shows a trolley moving down a gentle slope.

Add forces to a copy of the diagram below to produce

a free-body force diagram for the trolley. [3]

The trolley is photographed by a multiflash

technique. The result is shown opposite.

What evidence is there that the trolley is moving

with constant velocity? [1]

State the acceleration of the trolley down the slope. [1]

What does the value of the acceleration indicate about the forces acting on the trolley? [1]

2. The 'London Eye' is a large wheel which

rotates at a slow steady speed in a vertical plane

about a fixed horizontal axis. A total of 800

passengers can ride in 32 capsules equally

spaced around the rim. A simplified diagram

is shown opposite.

On the wheel, the passengers travel at a speed

of about 0.20 m s-1 round a circle of radius

60 m. Calculate how long the wheel takes to

make one complete revolution. [2]

What is the change in the passenger's velocity

when he travels from point B to point D? [2]

When one particular passenger ascends from

point A to point C his gravitational potential

energy increases by 80 kJ.

Calculate his mass. [3]

Sketch a graph showing how the passenger's

gravitational potential energy would vary with

time as he ascended from A to C.

Add a scale to each axis. [3]

Discuss whether it is necessary for the motor driving the wheel to supply this gravitational potential energy. [2]

3. Define linear momentum. [1]

The principle of conservation of linear momentum is a consequence of Newton's laws of motion. An examination

candidate is asked to explain this, using a collision between two trolleys as an example. He gives the following answer,

which is correct but incomplete. The lines of his answer are numbered on the left for reference.

1. During the collision the trolleys push each other.

2. These forces are of the same size but in opposite directions.

3. As a result, the momentum of one trolley must increase at the same rate as the momentum of the other decreases.

4. Therefore the total momentum of the two trolleys must remain constant.

In which line of his argument is the candidate using Newton's second law? [1]

In which line is he using Newton's third law? [1]

The student is making one important assumption which he has not stated. State this assumption. Explain at what point it

comes into the argument. [2]

Describe how you could check experimentally that momentum is conserved in a collision between two trolleys. [4]

4. State the number of protons and the number of neutrons in 14

6C. [2]

The mass of one nucleus of 14

6C = 2.34 x 10-26

kg. The nucleus of carbon-14 has a radius of 2.70 x 10-15

m.

Show that the volume of a carbon-14 nucleus is about 8 x 10-44

m3. [2]

Determine the density of this nucleus. [2]

How does your value compare with the densities of everyday materials? [1]

Carbon-14 is a radioisotope with a half-life of 5700 years. What is meant by the term half-life? [2]

Calculate the decay constant of carbon-14 in s-1. [2]

Page 3: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

PHYSICS M&R JUNE 2001

1

1. A car traveling at 30ms -1

collides with a wall. The driver wearing a seatbelt, is brought to rest in 0. 070 s.

The driver has a mass of 50 kg. Calculate the momentum of the driver before the crash. [2]

Calculate the average resultant force exerted on the driver during impact. [3]

Explain why the resultant force is not the same as the force exerted on the driver by the seatbelt. [1]

2. Alpha particle radiation has a short range in matter. With reference to the effect of alpha particles on atoms. Explain

why they only travel a short distance. [2]

A worker in the nuclear industry accidentally swallows some liquid that emits alpha particles.

The plant manager tells him not to worry as the swallowed liquid will be excreted within a day.

However, the health physicist investigating the accident is still anxious to determine the half-life of the radioisotope

involved . Explain the significance of the radioactive half-life for the health of the worker. [2]

3. A child is crouching at rest on the ground.

Opposite are the free-body force diagrams for the child and the Earth.

Copy out and complete the table describing forces A, B and C. [4]

All the forces A, B, C and D are of equal magnitude.

Why are forces A nd B equal in magnitude?

Why must forces B and D be equal in magnitude? [2]

The child now jumps vertically upwards. With reference to the forces shown, explain what he must do to jump, and why

he moves upwards. [3]

4. Two campers have to carry a heavy container of water between them.

One way to make this easier is to pass a pole through the handle as shown.

The container weighs 400 N and the weight of the pole may be neglected.

What force must each person apply? [1]

An alternative method is for each person to hold a rope tied to the handle

as shown in the second diagram.

Draw a free-body diagram for the container when held by the ropes. [2]

The weight of the container is 400 N and the two ropes are at 40o to

the horizontal.

Show that the force each rope applies to the container is about 300 N. [3]

Suggest two reasons why the first method of carrying the container

is easier. [2]

Two campers using the rope method find that the container keeps

bumping on the ground. A bystander suggests that they move further

apart so that the ropes are more nearly horizontal. Explain why this

would not be a sensible solution to the problem. [1]

Page 4: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

PHYSICS M&R JANUARY 2002

1

1. Copy out and complete the equations below by writing the correct physical quantities, in words,

inside the brackets. [4]

VOLUME = MASS AVERAGE VELOCITY = ( ) WEIGHT = ( )

( ) TIME MASS

DECAY CONSTANT = ACTIVITY

( )

2. The diagram opposite shows a

trolley running down a slope.

Copy out and complete the

diagram to show an experimental

arrangement you could use to

determine how the trolley's position varies with

time. [2]

The data is used to produce a velocity-time graph

for the trolley.

Opposite is the graph for the motion from point A

to point B. Time is taken to be zero as the trolley

passes A, and the trolley passes B 0.70 s later.

The motion shown on the graph can be described by the

equation v = u + at. Use information from the graph to

determine values for u and a. [3]

Determine the distance AB. [3]

Sketch a graph to show how the displacement x

of the trolley from point A varies with time t.

Add a scale to each axis. [3]

3. When the jet engines on an aircraft are started,

fuel is burned and the exhaust gases emerge from

the back of the engines at high speed. With

reference to Newton's second and third laws of

motion, explain why the aircraft accelerates forward.

You may be awarded a mark for the clarity of your answer. [4]

4. A disk has a radius of 0.24m. Two 5.0 N forces act on the disk as shown on the first diagram below.

What is the resultant moment about O? Find the magnitude and direction of the resultant force on the disk. [6]

The two 5.0N forces are now positioned in such a way that the resultant force on the disk becomes zero whilst the

resultant moment about 0 remains the same as before. The second diagram below shows one of the forces. Copy this

diagram and sdd an arrow to show the new position and direction of the other force. [2]

Page 5: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

PHYSICS M&R JANUARY 2002

2

5. The diagram below and opposite shows the positions of two ice skaters at intervals of 0.50 s

Up to 0.50 s, skater A and his partner B are

gliding together across the ice. Between

0.50s and l.00s, A pushes B away from him

in the direction of travel. After l.00s, they

continue to glide separately across the ice.

Determine the speed of the skaters before

they separate. [2]

The masses of the skaters A and B are 75 kg

and 55 kg respectively. Use these figures,

together with information from the diagram,

to show that momentum is approximately

conserved when they separate. [5]

Why should momentum be approximately

conserved in this situation? [1]

Without further calculation, explain whether

you would expect the total kinetic energy of

the skaters to increase, decrease or remain the same when they separate. [3]

6. A granite block is suspended at rest just below the surface of

water in a tank (Figure i).The block is now released and falls

0.80m to the bottom (Figure ii).

The volume of the block is 3.0 x 10-3 m3, and the density of

granite is 2700 kgm-3

.

Calculate the gravitational potential energy lost by the block as

it falls. [3]

Although the water level has not changed, the water has gained

gravitational potential energy. Explain why. [1]

The gravitational potential energy gained by the water is less than that lost by the granite block. Explain this. [2]

7. Samples of two different isotopes of iron have been prepared. Compare the compositions of their nuclei.

The samples have the same chemical properties. Suggest a physical property which would differ between them. [3]

Tritium (hydrogen-3) is an emitter of beta particles. Complete the nuclear equation for this decay. [3]

Describe how you would verify experimentally that tritium emits only beta particles. [4]

8. A radioactive source contains barium-140. The initial activity of the source is 6.4 x 108 Bq. Its decay constant is

0.053 day-1. Calculate the half-life in days of barium-140.

Calculate the initial number of barium-140 nuclei present in the source. [4]

The graph below represents

radioactive decay.

Copy this graph and add a suitable

scale to each axis, so that the graph

correctly represents the decay of this

barium source.

A radium-226 source has the same

initial activity as the barium-140 source.

Its half-life is 1600 years. On the same

axes sketch a graph to show how the

activity of the radium source would

vary over the same period. [3]

Page 6: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

What happens to this lost g.p.e.?

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

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

At the bottom of the slope the cyclist turns round and pedals back up at the same steady speed of 8.4 m s–1. Give an estimate of the rate at which the cyclist does work as he climbs the hill.

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

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

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

Rate of working = ......................................... (2)

(Total 6 marks)

6. Define momentum and state its unit.

Definition: ........................................................................................................................

Unit: ........................................................ (2)

A stationary nucleus of thorium-226 decays by alpha particle emission into radium. The equation for the decay is:

Th22690 → + Ra222

88 He42

State the value of the momentum of the thorium nucleus before the decay .......................... (1)

After the decay, both the alpha particle and the radium nucleus are moving.

Which has the greater speed? Justify your answer.

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

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

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

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

The Nobel School 7

Page 7: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

What can be said about the directions of travel of the two particles?

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

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

(Total 6 marks)

7. Indium-115 (symbol In, proton number 49) decays by beta-minus emission to tin (symbol Sn). Write down a nuclear equation representing this decay.

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

Indium-115 has a half-life of 4.4 × 1014 years. Calculate its decay constant.

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

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

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

Decay constant = ................................................ (2)

A radioactive source contains 2.3 × 1021 nuclei of indium-115. Calculate the activity of this source in becquerels.

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

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

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

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

Activity = ...................................................... Bq

State how this activity compares with a normal background count rate.

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

(Total 7 marks)

The Nobel School 8

Page 8: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

Radioactive decay is a random process. Explain what this means.

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

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

In what way is the experiment a model of a random process?

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

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

What is meant by the half-life of a radioisotope?

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

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

Does the model illustrate half-life? Justify your answer.

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

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

(Total 7 marks)

7. A wooden mallet is being used to hammer a tent peg into hard ground.

The Nobel School 7

Page 9: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

The head of the mallet is a cylinder of diameter 0.100 m and length 0.196 m. The

density of the wood is 750 kg m–3. Show that the mass of the head is approximately 1.2 kg.

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

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

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

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

The head strikes the tent peg as shown at a speed of 4.20 m s–1 and rebounds at 0.58 m s–1. Calculate the magnitude of its momentum change in the collision.

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

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

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

Momentum change = ............................... (3)

The head is in contact with the peg for 0.012 s. Estimate the average force exerted on the peg by the head during this period.

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

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

Average force = ....................................... (2)

Give a reason why your value for the force will only be approximate.

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

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

With reference to your calculations above, discuss whether a mallet with a rubber head of the same mass would be more or less effective for hammering in tent pegs.

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

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

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

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

(Total 11 marks)

The Nobel School 8

Page 10: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

Using the principle of moments, determine force X. (You should ignore the vertical forces on the

tree stump.)

……….…………………………………………………………………………………………….

……….…………………………………………………………………………………………….

……….…………………………………………………………………………………………….

X = .................................................................... (2)

Explain why, in this moments calculation, it is reasonable to ignore the vertical forces on the stump.

……….…………………………………………………………………………………………….

……….…………………………………………………………………………………………….

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

Calculate force Y.

……….…………………………………………………………………………………………….

……….…………………………………………………………………………………………….

Y = .................................................................... (2)

If a tree stump is to be removed, it is a mistake to cut it off very close to the ground first. Using the concept of moments, explain why.

……….…………………………………………………………………………………………….

……….…………………………………………………………………………………………….

……….…………………………………………………………………………………………….

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

(Total 8 marks)

5. A model truck A of mass 1.2 kg is travelling due west with a speed of 0.90 m s–1. A second truck B of mass 4.0 kg is travelling due east towards A with a speed of 0.35 m s–1.

Calculate the magnitude of the total momentum of the trucks.

……….…………………………………………………………………………………………….

……….…………………………………………………………………………………………….

Total momentum = .......................................... (2)

The trucks collide and stick together. Determine their velocity after the collision.

The Nobel School 6

Page 11: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

1. Listed below are six physical quantities:

energy force moment momentum power velocity

Select from this list the quantity or quantities fitting each description below. You may use each quantity once, more than once or not at all. A quantity which can be measured in newton second (N s).

………………………………………………………………………………………………….

A quantity which equals the rate of change of another quantity in the list.

…………………………………………………………………………………………………..

A quantity which equals the product of two other quantities in the list.

………………………………………………………………………………………………….

A quantity with base units kg m2 s–3.

………………………………………………………………………………………………….

Two quantities which have the same base units.

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

2 .......................................................................................................................................... (Total 5 marks)

2. A neutron of mass 1.7 × 10–27 kg travelling at 2.96 ×107 m s–1 collides with a stationary nucleus of nitrogen of mass 23.3 × 10–27 kg. Calculate the magnitude of the momentum of the neutron before it collides with the nucleus of nitrogen.

…………………………………………………………………………………………….….

…………………………………………………………………………………………….….

Momentum of neutron = ......................................... (2)

Given that the neutron ‘sticks’ to the original nucleus after the collision, calculate the speed of the new heavier nucleus of nitrogen.

…………………………………………………………………………………………….….

…………………………………………………………………………………………….….

…………………………………………………………………………………………….….

…………………………………………………………………………………………….….

Speed = .................................................................... (3)

The Nobel School 1

Page 12: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

An elastic collision is one where kinetic energy is conserved. Make suitable calculations to

determine whether this collision is elastic.

…………………………………………………………………………………………….….

…………………………………………………………………………………………….….

…………………………………………………………………………………………….….

…………………………………………………………………………………………….….

…………………………………………………………………………………………….….

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

(Total 8 marks)

3. An athlete runs a 100 m race. The idealised graph below shows how the athlete’s velocity v changes with time t for a 100 m sprint.

u/

u

m s–1

max

00

2 4 6 8 10t s/

12

By considering the area under the graph, calculate the maximum velocity vmax of the athlete.

…………………………………………………………………………………………….….

…………………………………………………………………………………………….….

…………………………………………………………………………………………….….

…………………………………………………………………………………………….….

Maximum velocity = ........................................ (3)

The Nobel School 2

Page 13: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

1. Complete the table to show the missing physical quantity for each unit.

Unit Physical quantity

m s–1 Velocity

m s–2

kg m–3

N m

kg m s–1

N m s–1

(Total 5 marks)

2. The diagram shows two magnets, M1 and M2, on a wooden stand. Their faces are magnetised as shown so that the magnets repel each other.

Magnet M

Magnet M

1

1

2

Wooden pole

Bottom face of Mis a North pole

Wooden base

The Nobel School 1

Page 14: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

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

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

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

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

(Total 8 marks)

8. When electrons are fired at nucleons many of the electrons are scattered.

When the electrons have low energy, the scattering is elastic.

However, when the electrons have sufficiently high energy, deep inelastic scattering occurs.

(a) What is meant by inelastic in this situation?

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

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

(b) What is revealed about the structure of the nucleon by deep inelastic scattering?

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

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

(c) What quantity is conserved during both elastic and inelastic scattering?

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

(d) Historically, physicists found that electrons of low energy could not be used to find out information about the nucleus of neutral atoms. Suggest why.

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

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

(Total 4 marks)

The Nobel School 11

Page 15: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

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10

5. (a) A toy truck of mass 80 g is released from a height h and rolls down a slope as shownbelow.

What would the height h have to be for the truck to reach a speed of 4.0 m s–1 at thebottom of the slope? You may assume that any friction at its axles is negligible.

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

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

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

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

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

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

Height = ....................................................................(3)

(b) On reaching the bottom, it joins magnetically to two stationary trucks, identical to thefirst, and the trucks all move off together.

(i) State the law of conservation of linear momentum.

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

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

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

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

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

*N24600A01020*

h

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(ii) Use this law to calculate the speed of the trucks immediately after the collision.

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

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

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

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

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

Speed = .....................................................................(2)

(c) One of the stationary trucks has a total frictional force of 0.12 N at its axles. Howmuch time does it take for the three trucks to stop moving if this is the only frictionalforce acting?

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

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

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

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

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

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

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

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

Time = .......................................................................(3)

Turn over

Q5

(Total 10 marks)

*N24600A01120*

Page 17: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

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2

1. What physical quantity does the gradient of each of the following graphs represent? Giveyour answers in the table below the graphs.

*N22362A0216*

Q1

(Total 4 marks)

(i) (ii)

(iii) (iv)

Displacement

Time

Time

Time

Time

Velocity

Momentum Work done

Graph Physical quantity represented by the gradient

(i)

(ii)

(iii)

(iv)

Page 18: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

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4. (a) State Newton’s second law of motion in terms of momentum.

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

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

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

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

(b) A wind blows steadily against a tree. The area of the tree perpendicular to thedirection of the wind is 10 m2 and the velocity of the wind is 20ms–1.

(i) Show that the mass of air hitting the tree each second is about 250 kg. (Densityof air is 1.23 kg m–3.)

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

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

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

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

(ii) Calculate the momentum of this mass of air when it is moving at 20 m s–1.

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

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

Momentum = .........................................................

(iii) Assuming that all the air is stopped by the tree, state the magnitude of the forceexerted on the tree by the wind.

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

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

*N22362A0616*

Q4

(Total 6 marks)

Page 19: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

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*N26139A01016*

5. During a game of football the ball, mass 0.42 kg, is kicked towards the goal. It hits one of the posts and rebounds directly back into play. The diagram shows the ball as it is just colliding with the post. At impact its speed is 27 m s–1.

(a) Calculate the ball’s momentum at impact.

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

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

Momentum = ...........................................(2)

(b) The ball’s speed at the moment it loses contact with the post is 20 m s–1 in the opposite direction. Calculate its momentum at this instant.

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

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

Momentum = ...........................................(2)

(c) (i) The ball remains in contact with the post for 0.22 s. Determine the average force exerted on the ball due to the collision.

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

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

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

Average force = ......................................(3)

(ii) Show the direction of this force on the diagram.(1)

FootballPost

Direction of motion

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Turn over*N26139A01116*

(d) State one difference and one similarity between this force and the force that acts on the post due to the impact of the ball.

Difference: .....................................................................................................................

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

Similarity: .....................................................................................................................

.......................................................................................................................................(2) Q5

(Total 10 marks)

Page 21: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

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*N26139A01216*

6. An elastic collision is one in which kinetic energy is conserved.

A proton, mass 1.67 × 10–27 kg, travelling with a speed of 2.40 × 106 m s–1, has a head-on elastic collision with a stationary helium nucleus. After the collision the helium nucleus, mass 6.65 × 10–27 kg, moves off with a speed of 9.65 × 105 m s–1.

(a) Show that the kinetic energy of the helium nucleus is approximately 3 × 10–15 J.

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

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

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

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

(b) (i) How much kinetic energy is lost by the proton?

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

(ii) Hence determine the speed of the proton after the collision.

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

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

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

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

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

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

Speed of proton = ...............................................(3)

(c) Name a quantity, other than kinetic energy, that is conserved in this collision.

.......................................................................................................................................(1) Q6

(Total 7 marks)

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*N21066A01016*

Q6

(Total 8 marks)

6. A uranium-238 nucleus, decays to a thorium-234 nucleus according to the nuclear equation

23892U → 234

90Th + particle

(a) Identify the particle.

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

(b) The particle is emitted with a speed of 1.41 × 107 m s–1 and a kinetic energy of 6.58 × 10–13 J. Use this data to show that the momentum of the particle at the instant it is emitted is about 9 × 10–20 kg m s–1.

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

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

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

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

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

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

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

(c) When a decaying uranium-238 nucleus is at rest, the thorium-234 nucleus moves with a speed of 2.4 × 105 m s–1 in the opposite direction to the particle.

Explain with the aid of a calculation how this is consistent with the principle of conservation of momentum.

Mass of thorium-234 nucleus = 3.89 × 10–25 kg

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

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

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

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

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

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

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

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*N30615A0616*

3. A boy throws a ball vertically upwards. It rises a maximum distance of 28.0 m in 2.4 s.

(a) (i) Calculate the distance of the ball above the point of release 3.8 s after it was thrown.

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

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

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

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

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

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

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

Distance = .........................................(4)

(ii) State an assumption you have made.

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

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(b) The ball is caught by the boy.

Explain why moving his hands downwards as he catches the ball will reduce the force that the ball applies to his hand. You may be awarded a mark for the clarity of your answer.

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.......................................................................................................................................(5) Q3

(Total 10 marks)

Page 29: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

PHY1 JANUARY 2001

6

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Page 31: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

1. Density (1) Displacement [not distance] (1)

Acceleration due to gravity/gravitational field strength/acceleration of free fall/acceleration g [not gravity or gravitational force/pull] [not g] [not acceleration by itself] (1) Number of nuclei/atoms [not particles, molecules] [not amount of substance] [not nuclides, decayed nuclei] (1)

[4]

2. Diagram: Shown and labelled Ticker timer at top or Strobe light (1) Tape from trolley through timer or camera [consequent] (1) OR Motion sensor pointing at trolley or video (1) Connection to datalogger/computer or rule [both consequent] (1) OR Three or more light gates (1) Connection to datalogger/computer [consequent] (1) [Two light gates connected to ‘timer’ – max 1] [Rule and stop clock - max 1]

Values for υ and a: 2 0.95 m s–1 [2 s.f.] (1) Use of gradient or formula (1) 0.79 m s–2 [no e.c.f. if u = 0] (1) Distance AB: 3 AB = ‘area’ under graph, or quote appropriate equation of motion (1) Physically correct substitutions (1) 0.86 m [allow 0.9 m] [e.c.f. wrong u or a] (1) Graph: 3 Smooth curve rising from origin, getting steeper (1) Initial gradient non-zero [consequent] (1) (0.70, 0.86) matched (e.c.f. on distance) (1) 3

[11]

3. Engine (or plane) pushes/forces the gases (1) Gases push forwards/opposite direction on engine (or plane) (1) Relating this to Newton’s third law (1) Force accelerates plane/increases its momentum, by Newton’s second law/ F =ma (1) Max 3 Clarity of answer (1) 1

[4]

4. Resultant moment: Use of moment = force × distance (1) 1.2 N m (1) Resultant force:

The Nobel School 1

Page 32: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

Resultant force = ( )

45cosN0.52)N0.5()N0.5( 22

××+

(1)

= 7(.1) N [not √(50)] (1) at 45° [can be on diagram] (1) approx. N.E. [can be on diagram] (1) 6

Diagram: Any arrow vertically downwards [2 arrows 0/2] [ignore values] (1) Any line through O (1) 2

5 N

[8]

5. [Accept use of 0.5 s or 0.6 s intervals in calculations. Values using 0.6 s are shown in square brackets.] (1) 2

Speed:

Use of distance/time (1)

6.4 m s–1 [5.3 m s–1]

Conservation of momentum: Final υA = 4.4 m s–1 (2.2/0.5) [3.7 (2.2/0.6)] (1) OR υB = 8.4 m s–1 (4.2/0.5) [7.0 (4.2/0.6)] (1) Any use of p = mυ

Use of pi = (∑m) ui (1) Use of pf = (∑mυf) (1) pi = 832 N s/kg m s–1 [689], pf = 792 N s/kg m s–1 [662.5] (1) (both ± 5) [consequent on all 4 above] 5

Reason: On ice friction is small / No external forces/friction (1) 1

Explanation: Total kinetic energy increases (1) Skater does work pushing partner away [conditional] (1) 2

[10]

6. Calculation of g.p.e: Use of m = ρV (1) Use of Ep = mgh [m = 8.1 × 10x kg] (1) 64 J (1) 3

Explanation: (Some) water has moved up (1) 1

Why g.p.e. is less: Water has less mass (1)

The Nobel School 2

Page 33: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009
Page 34: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009
Page 35: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

Half-life

Time taken for activity/count rate to drop by half/time taken for half the atoms/nuclei to decay (1) 1

How model illustrates half-life

Yes, if children were told to flip coin at regular time interval OR Yes, because about half of the children flipped a head each time OR No, because time is not part of the experiment (1) 1

[7]

7. Mass of head of mallet Selecting density x volume (1) Correct substitutions (1) Mass = 1.15 (kg) [3 significant figures, minimum] (1) 3

Momentum change

p = mυ used (1) Δp = 1.15 or 1.2 kg (4.20 + 0.58) m s–1 (1) = 5.50 / 5.74 kg m s–1/N s (1) 3

Average force

Their above / 0.012 s (1)

F = 458/478 N [e.c.f. Δp above] (1) 2

Value for force

Handle mass/weight/ head weight/force exerted by user (handle) neglected (1) 1

Effectiveness of mallet with rubber head

Δt goes up/Δp goes up (1)

⇒ less force, less effective/more force, more effective [consequent] (1) 2 [11]

8. Vehicle movement

mgh and ½ mυ 2 [Both required] / mgh and mgh / ½ mυ 2 and ½ mυ 2 (1) 1

Expression for speed

Kinetic energy gained = gravitational potential energy lost / mgh = ½ mυ 2 (1) υ = ( )gh2 (1) 2

The Nobel School 3

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Page 37: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

1. Six physical quantities

Momentum (1)

Power / force (1)

Power (1)

Power (1)

Moment and energy (1) [5]

2. Momentum of neutron

Use of p = mυ (1)

p = 5.03 × 10–20 N s/kg m s–1 (1) 2

Speed of nucleus

Total mass attempted to be found (1) Conservation of momentum used (1) υ = 2.01 × 106 m s–1 [ecf from p above only] (1) 3

Whether collision was elastic

Use of k.e. = ½ mυ2 (1)

ke = 7.45 × 10–13 (J) / 5.06 × 10–14 (J) (ecf) (1)

A correct comment based on their two values of ke. (1) 3 [8]

3. Maximum velocity

Area = 100 m (1)

Attempt to find area of trapezium by correct method (1)

υ = 10 m s–1 (1) 3

Sketch graph

Horizontal line parallel to x axis

Some indication that acceleration becomes 0 m s–2

The initial acceleration labelled to be υmax ÷ 2 [ initial a = 5 (m s–2) (1) (ecf)]

t = 2 (s) where graph shape changes (1) 4 [7]

4. Nuclear equation for decay of isotope

C148

(1) )

/ (1) ) [Either way round] 3

N147

e01−

)(01β

−−

The Nobel School 1

Page 38: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

2

6731 Unit Test PHY1

Unit

Physical Quantity

m s-1 Velocity

m s-2 Acceleration / deceleration

kg m-3 Density

N m Moment / energy / (gravitational)potential energy / kinetic energy/heat/work (done) / torque

kg m s-1 Momentum / impulse

1.

N m s-1 Power

5

Page 39: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

11

8. (a) Inelastic scattering

Kinetic energy is not conserved / (some) kinetic energy is ‘lost’

1

(b) Structure

There are point charges/quarks/smaller particles within the nucleon OR mass not uniform

1

(c) Quantity conserved

(d)

Momentum / energy / charge / mass No information Electron was repelled (by the (outer) electron shell(s)) OR captured to make an ion.

1

1 ___

4

Page 40: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

8

(c) Time for trucks to stop

[Do not penalise candidates for using a total frictional force of 0.36 N. 3/3 possible] Either Correct use of power = f × v and ½ mv2 [Do not penalise power of 10 errors or not dividing by 2 in f × V equation] Use of energy divided by power Answer in range 2.6 s to 2.7 s [ecf their value for u]

P = 0.12 N × 233.1 m s-1 = 0.08 W

KE = ½ × (3) × 80× (10-3) kg × (1.33 m s-1)2 = 0.21 J

powerEnergy =

W08.0J21.0

t = 2.6(5) s [accept 2.6 or 2.7 as rounding] OR Use of F = ma

Use of either v = u + at i.e. or a = tv∆

Answer in range 2.6 s to 2.7 s (–)0.12 N = (3) × 80× (10-3) kg × a (a = (–)0.5 m s-2)

0 = 1.33 m s-1 – 0.5 m s-2 × t or (-)0.5 m s-2 = t

1s m33.1)( −−

t = 2.6(6) s OR Select Ft = ∆p Substitution (-)0.12t = (-3) × 80 × (10-3) kg × 1.33 m s-1 [Allow omission of any bracketed value] Answer in range 2.6 s to 2.7 s

3

10

Page 41: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

7

5. (a) From what height?

Use of mg∆h and ½mv2 [ignore power of 10 errors] mg ∆h = ½ mv2 [shown as formulae without substitution, or as numbers substituted into formulae] Answer [0.8(2) m] [It is possible to get 0.8 m by a wrong method:

• If v2 = u2 + 2as is used, award 0 marks • If you see v2/a then apply bod and up to 2/3 marks - the 2nd and 3rd

marks. Note that v2/g is correct and gains the first 2 marks, with the 3rd mark if 0.8 m is calculated]

80 × (10-3) kg × 9.81 N kg-1 × ∆h = ½ × 80 (× 10-3) kg × (4 m s-1)2

h = 13

213

kgN81.9 kg10 80)ms4( kg10 80 5.0

−−

−−

×××××

= 0.8(2) m

3

(b)(i) Law of conservation of linear momentum Provided no external[other/resultant/outside] force acts The total momentum (of a system) does not change / total momentum before(collision)= total momentum after (collision) [Total seen at least once] [Ignore all references to elastic and inelastic] [Do not credit simple statement that momentum is conserved]

2

(ii) Speed of trucks after collision Any correct calculation of momentum Use of conservation of momentum leading to the answer 1.3(3) m s-1

80 × (10-3) kg × 4 m s-1 = 240 × (10-3) kg × u, giving u = 1.3(3) m s-1 2

Page 42: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

3

6731 Unit Test PHY1 1.

Graph Physical quantity represented by

the gradient (i) (Constant) velocity [Not speed. Not

velocity change.] (ii) (Constant) acceleration (iii) Force (iv) Power

Ignore references to units eg velocity m s-1 or dimensions L T-1

4

Page 43: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

7

4. (a)Newton’s Second Law of Motion

(The) force (acting on a body) is proportional/equal to the rate of change of momentum and acts in the direction of the momentum change [accept symbols if all correctly defined for the first of these marks] [ignore any information that is given that is not contradictory] b(i)Calculate the mass Correct calculation for volume of air reaching tree per second [Do not penalise unit error or omission of unit] Correct value for mass of air to at least 3 sig fig [246 kg. No ue.] [If 1.23 × 10 × 20 = 246 kg is seen give both marks. Any order for the numbers] Example 20 m s−1 × 10 m2 = 200 m3 1.23 kg m−3 × 200 m3 = 246 kg b(ii)Calculate the momentum Answer: [ (246 kg × 20 m s−1 =) 4920 kg m s−1. ] [Accept (250 kg × 20 m s−1 =) 5000 kg m s−1. Accept 4900 kg m s-1. Ecf value for mass. Ignore signs in front of values.] b(iii)Magnitude of the force Answer: [ F = 4920 N or 5000 N or 4900 N.] [ Ecf value from b(ii). Ignore signs in front of values]

2

2

2

6

Page 44: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

5 a Momentum at impact

p = mv seen or used Answer [11 kg m s-1] eg momentum = 0.42 kg × 27 m s-1 = 11.34 kg m s-1

2

b Momentum at release Minus 8.4 kg m s-1

2ci Average force(ecf momenta values)

Use of F = tp∆∆ ie for using a momentum value divided by

0.22 Adding momentum values Answer [88.0 N - 89.8 N]

F = s 22.0

s m kg3.11s m kg4.8 11 −− −−

F = (−) 89.5 N Or Use of F = ma Adding velocities to calculate acceleration Answer [88.0 N – 89.8 N]

Eg acceleration = s22.0

sm27sm20 -1-1 −− ( = − 213.6 m s -2)

Force = 0.42 kg × −213.6 m s-2 = (−)89.7(2) N

3cii Direction of force on diagram

Right to left [Accept arrow drawn anywhere on the diagram. Label not required]

1

d Difference and similarity Difference: opposite direction / acts on different object Similarity: same type of force / same size / acts along same line / act for same time / same size impulse [‘Magnitude’ and ‘size’ on their own is sufficient. ‘They are equal’ is OK. Accept; they are both contact forces; they are both electrostatic forces]

2

10

Page 45: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

6 a EK of helium nucleus

Use of EK = 21 mv2

Answer [ 3.1 × 10-15 J. No ue. Min 2 sig fig required]

eg EK = 21 × 6.65 x 10-27 kg × ( 9.65 × 105 m s-1)2

= 3.096 × 10-15 J

2

bi Loss of EK of proton [ecf their value for EK of helium nucleus] 3 × 10-15 J or 3.1 × 10-15 J

1bii Speed of proton after collision

[ecf their value for loss of EK of proton, but not if they have given it as zero] Calculation of initial Ek of proton Subtraction of 3.1 × 10-15 J [= 1.7 × 10-15 J] Answer [(1.40 – 1.50) × 106 m s-1]

eg EK = 21 1.67 × 10-27 kg x (2.4 × 106 m s-1)2 (= 4.8 × 10-15 J)

EK after collision = 4.8 × 10-15 J - 3.1 × 10-15 J ( =1.7 × 10-15 J)

v = (gk1067.15.0

J107.127

15

××× )0.5 = 1.43 × 106 m s-1

Or Use of the principle of conservation of momentum. Correct expression for the total momentum after the collision Answer [ (1.40 – 1.50) × 106 m s-1] Eg 1.67 × 10-27 kg x 2.4 × 106 m s-1 = 6.65 × 10-27 kg × (9.65 × 105 m s-1) + 1.67 × 10-27 kg × V V = −1.44 × 106 m s-1

[For both these solutions allow the second marking point to candidates who incorrectly write: the mass of the proton as 1.6 × 10−27 kg or 1.7 × 10−27 kg, or the mass of the helium as 6.6 × 10−27 kg or 6.7 × 10−27 kg or the velocity as 9.6 × 105 m s−1 or 9.7 × 105 m s−1]

3

c Other factor conserved Momentum / mass / charge / total energy

1

7

Page 46: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

Question Number

Answer Mark

6 (a) Identify particle Alpha (particle) / Helium nucleus / He / He / α / α / alpha /

alpha /α

42

42

42

42

42

42

(1)

b Momentum of particle Momentum equation [In symbols or with numbers] (1) Either

Correct substitution into 21 mv2 = energy (1)

Use the relationship to determine the mass [6.6 x 10-27 kg] (1) Answer [9.3 x 10-20 (kg m s-1 ) Must be given to 2 sig fig. No unit error] (1) Or

Rearrangement of EK = 21 mv2 to give momentum ie

v2EK (1)

Correct substitution (1) Answer [9.3 x 10-20 kg m s-1. Must be given to 2 sig fig. No unit error] (1)

Eg 21 m( 1.41 x 107 m s-1)2 = 6.58 x 10-13 J

m = 217

13

)1041.1(1058.62

smxJxx

= 6.6 x 10-27 kg

momentum = 6.6 x 10-27 kg x 1.41 x 107 m s-1

= 9.3 x 10-20 (kg m s-1) Or

Momentum = 17

13

sm10x1.41J10x6.58x2

= 9.3 x 10-20 (kg m s-1)

(4) c Consistent with the principle of conservation of momentum

(Since total) momentum before and after (decay) = 0 (1) State or show momentum / velocity are in opposite directions (1) [Values of momentum or velocity shown with opposite signs would get this

mark] Calculation ie 3.89 x 10-25 kg x 2.4 x 105 m s-1 = 9(.3) x 10-20 (kg m s-1) (1) Eg 3.89 x 10-25 kg x 2.4 x 105 m s-1 = 9(.3) x 10-20 kg m s-1

(3)

Total for question (8)

Page 47: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

6731 Unit Test PHY1 June 08 v4 Question Number

Answer Mark

1_(a) Add missing information For four correct responses in the ‘vector or scalar’ column (1) For the ‘base unit’ column :- 4 correct responses (3) 3 correct responses (2) 2 correct responses (1)

Quantity Base unit Vector or scalar m vector kg m2 s-2 scalar kg m2 s-3 scalar kg m s-1 vector

[ Accept answers where the units are not combined eg kg m s-3 m for power] [Do not accept answers written in dimension symbols eg M L2 T-2 for gravitational potential energy]

(4)

Total for question (4)

Page 48: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

Question Number

Answer Mark

3_(a) Principle of conservation of linear momentum Provided no external [other/resultant/outside] force acts (1) [For this mark accept ‘closed system’ or ‘closed environment’] [Do not accept fixed for closed] The total momentum (of a system) does not change[is constant] / total momentum before (collision) = total momentum after (collision) [‘Total’ or ‘sum’ should be seen at least once, do not accept ‘all’] (1) [Accept a formula for this mark eg mava + mbvb = (ma + mb)V, the symbols do not have to be defined if written in a clear form such as this] [Ignore all references to elastic and inelastic. Do not credit simple statement that ‘total momentum is conserved’]

(2)

(b) i Measuring velocity Tickertape Light gate(s)/sensor Motion sensor Video (1) Tickertimer Datalogger/PC/timer

[They do not have to be shown connected to the light gate]

Datalogger/PC [They do not have to be shown connected to the motion sensor]

Metre rule / markings on the track

(1)

[The points above maybe labelled on the diagram] [Do not give these first 2 marks for ruler and stopwatch] Description of distance measured and corresponding time or

v = td

or any mention of a distance against time graph[mention

of gradient not required for this mark] (1) [Candidates who have described a ruler and stopwatch method

can get this final mark]

(3)

(b) ii Further measurements The mass(es) of both A and B / the trolleys (1) [Give this mark even when other unnecessary(but not

conflicting) information is given] [Accept ‘Weigh the mass of the two trolleys’ but not ‘weigh the

two trolleys’ or ‘weigh the mass of the trolley’] [Do not accept bald answers ‘mass’ or ‘masses’]

(1)

(b) iii Explain constant velocity requirement [In place of resultant accept unbalanced or net throughout] Either (For the law to be demonstrated) there must be no external [accept ‘outside’] force /resultant force / friction acting (1) [do not accept closed system] (If the trolley(s) are moving with constant velocity) the external[accept ‘outside’] force / resultant force / (effect of)friction (acting on the system)is zero. (1) [Award mark for converse statement ie ‘(if the trolley(s) are changing speed) the external [accept ‘outside’] force / resultant force / (effect of)friction (acting on the system)is not zero] [Award this second mark for candidates who state friction has been compensated for ]

(2)

Page 49: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

Or There must be no external [accept ‘outside’] force / resultant force / friction acting [do not accept closed system] (1) if acceleration is zero (1) Or The velocity / speed measurements required are the velocities / speeds (at the instant) when the trolleys collide(1) [ Award this mark for statements such as ‘the velocities / speeds measured would not be the speeds they have when they collide’] Measurement of these velocities is impossible / difficult (1) [Award no marks for arguments involving just energy]

Total for question (8)

Page 50: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

Question Number

Answer Mark

6_(a) Show speed is about 2 m s-1 Either Substitution into force x distance (1) Equates work done and kinetic energy (1) Or Substitution into equation for force (1) [Give this mark even if the negative value for force (or acceleration) is omitted] Correct use of v2 = u2 + 2as or two appropriate equations (1) [Do not give this mark for use of +1.53 m s-2 and/or u = 0] Answer [(1.94 – 1.97) (m s-1)] [ At least 3 sig fig. No unit error] (1) Eg Work done = 2.75 N x 1.25 m

21

1.80kg x v2 = 2.75 N x 1.25 m

v = 1.95 (m s-1) Or

a = mF

= kg1.80

N2.75− = - 1.53 m s-2

v2 = u2 + 2as 0 = u2 + 2 x - 1.53 m s-2 x 1.25 m u = 1.95 (m s-1)

(3)

(b) Momentum Momentum equation [In symbols or numbers] (1) Answer [(3.5 - 3.6) kg m s-1 or N s. Ecf candidates value for speed] (1) Eg 1.8 kg x 1.95 m s-1 = 3.51 kg m s-1

(2) (c) Momentary force

Selects F = t∆p

or v = u + at and F = ma [May just write

F =t

u) - (v m ] (1)

[If the formulae are not seen, but are clearly used give this mark] Average value of unbalanced force [(5.0 – 5.2) (N)] (1) Average value of momentary force [(7.7 - 7.9) N](1) [Ecf candidate’s value of momentum from b]

Eg F = t∆p

Or v = u + at ; 2 ms-1 = (0 +) a x 0.7 s

= s0.7

smkg3.51 1−

F = ma; F = 1.8 kg xs0.7s m 2 -1

= 5.0 (N)

= 5.0 (N) Average value of force applied = 5.0 N + 2.75 N = 7.75 N

(3)

Total for question (8)

Page 51: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

• Question • Number

• Answer • Mark

• 3 (a)(i) • The height from the ground • Either • Deducts 2.4 s from 3.8 s / 1.4s seen

(1)

• Selects s = (ut) + 21

at2 or 2 appropriate equations.

(1) • Subtracts value obtained for second mark from 28 m / value for

distance fallen seen [9.6(1) m, 9.8 m if 10 m s-2 is used] (1)

• Answer [18 m] (1)

• • Eg t = 3.8 s -2.4 s = 1.4 s

• s = (ut) + 21

at2

• = 21

x 9.81 m s-2 x (1.4 s)2

• = 9.6(1) m [9.8 m if g = 10 m s-2 used] • height = 28 m – 9.6(1) m = 18 (.39 m) [18(.2) if g = 10 m s-2

used] • • Or • Use of equation to calculate initial velocity at point of release /

23.5 m s-1 seen [allow this mark even if the candidate confuses v and u and uses positive ‘g’ value in their calculation] (1)

• Selects s = (ut) + 21

at2 or 2 appropriate equations.

(1) • Uses minus g and correctly applies values to v and u

throughout •

(1) • Answer [ Allow answers in the range (18 – 19) m ]

(1) • • Eg • V = u + at

• • • • • • • • • • • • • •

• (4)

Page 52: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

• 0 = u – 9.81 m s-2 x 2.4 s • u = 23 .5(4) m s-1 • s = ut + ½at2 • = 23.54 m s-1 x 3.8 s - ½ 9.81 m s-2 x 3.82 s2 • = 18. 4(7) m

• (a)(ii) • Assumption made • • That ball falls with constant acceleration / that ball’s

acceleration is 9.8(1) m s-2 [or 10 m s-2] / that (air) resistance (force) is negligible / time at zero velocity is negligible / the ball is caught close to the Earth(’s surface) [Do not accept ‘force of gravity acts downwards’ or ‘no force’. Accept ‘no friction’ or ‘no resistance’

• Where a mixture of wrong and right answers are given do not award mark eg ‘no resultant force acts, no air resistance’] (1)

• • •

• (1)

• (b) • Why force is reduced • QWOC

(1) • Either • (To catch ball) velocity of ball has to be reduced (to zero) or

change in velocity is the same or the relative velocity between the ball and hand is reduced (1)

• (By moving his hand as described) time to do this is lengthened or acceleration is reduced[ Phrases such as ‘the change is slower’ are fine for this mark] (1)

• [Be generous in allowing these two marks above. If these points are made in addition to others which are inaccurate or contradictory, in general, award the marks eg many candidates are writing that the impulse is reduced, but in addition make other perfectly correct statements]

• Therefore force applied by the hand or the force applied to the ball is reduced (1)

• By Newton’s third law / an equal but opposite (reduced) force is applied by the ball or is applied to the hand (1)

• [It is essential that the link to N3 is made for this mark. An answer which only states ‘the force applied by the ball to the hand is reduced’ simply repeats what is already in the stem of the question]

• Or • (To catch ball) momentum of the ball has to be reduced (to

zero) or impulse is the same or momentum change is the same (1)

• (By moving his hand as described) time to do this is lengthened[ Phrases such as ‘the change is slower’ are fine for this mark] (1)

• [ For ‘rate of change of momentum is reduced’ give both these marks]

• [See advice above] • Therefore force applied by the hand or the force applied to

the ball is reduced (1)

• • • • • • • • • • • • • • • • • • • • • • • •

• (5)

Page 53: Ultimate Momentum Edexel Physics Past Papers Questions & Mark Scheme 2001-2009

• By Newton’s third law / an equal but opposite (reduced) force is applied by the ball or is applied to the hand (1)

• [See advice above] • Or • (To catch ball) kinetic energy has to be reduced (to zero)

(1) • (By moving his hand as described) means that the work

required to do this takes place over a longer distance (1)

• [See advice above] • • Therefore force applied by the hand or the force applied to

the ball is reduced (1) • By Newton’s third law / an equal but opposite (reduced) force is

applied by the ball or is applied to the hand (1)

• [See advice above] • •

• • Total • 10

• Question • Number

• Answer • Mark

• 4(a) • Show weight is ~ 0.3 N • • Use of πr2t to find volume or 3.5(3) (x 10-6 m3) seen

(1) • [Award this mark even when the diameter value is use for the

radius] • • Appropriate values substituted into density equation

(1) • Answer [0.31 N. No ue but must have 2 d.p. Accept values in

range 0.305 N – 0.314N. Because this is a ‘show that’ question this mark must only be given if it is clear that the candidate has used g to get the weight. A bald answer gets no marks here ] (1)

• Eg volume = 4

m) 10 x (30 x 2-3π x 5 x 10-3 m = 3.53 x 10-6 m3

• Mass = 3.53 x 10-6 m3 x 8900 kg m-3 = 3.14 x 10-2 kg • Weight = 3.14 x 10-2 kg x 9.81 N kg-1 = 0.308 N •

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• (3)

• (b) (i) • State Newton’s first law • • A body[allow ‘it’] will remain at rest or will move with uniform

speed in a straight line / uniform velocity / zero acceleration (1)

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