Topic 2 P2 [172 marks]

1a.

An elastic climbing rope is tested by fixing one end of the rope to the top of a crane. The otherend of the rope is connected to a block which is initially at position A. The block is released fromrest. The mass of the rope is negligible.

The unextended length of the rope is 60.0 m. From position A to position B, the block falls freely.

At position B the rope starts to extend. Calculate the speed of the block at position B.

Markschemeuse of conservation of energy

OR

v = u + 2as

v = « » = 34.3 «ms »

[2 marks]

2 2

√2 × 60.0 × 9.81 –1

At position C the speed of the block reaches zero. The time taken for the block to fall between B

[2 marks]

1b.

At position C the speed of the block reaches zero. The time taken for the block to fall between Band C is 0.759 s. The mass of the block is 80.0 kg.

Determine the magnitude of the average resultant force acting on the block between Band C.

Markschemeuse of impulse F × Δt = Δp

OR

use of F = ma with average acceleration

OR

F =

3620«N»

Allow ECF from (a).

[2 marks]

ave

80.0×34.30.759

Sketch on the diagram the average resultant force acting on the block between B and

[2 marks]

1c. Sketch on the diagram the average resultant force acting on the block between B andC. The arrow on the diagram represents the weight of the block.

Markschemeupwards

clearly longer than weight

For second marking point allow ECF from (b)(i) providing line is upwards.

[2 marks]

1d. Calculate the magnitude of the average force exerted by the rope on the blockbetween B and C.

[2 marks]

[2 marks]

Markscheme3620 + 80.0 × 9.81

4400 «N»

Allow ECF from (b)(i).

[2 marks]

1e.

For the rope and block, describe the energy changes that take place

between A and B.

Markscheme(loss in) gravitational potential energy (of block) into kinetic energy (of block)

Must see names of energy (gravitational potential energy and kinetic energy) – Allow forreasonable variations of terminology (eg energy of motion for KE).

[1 mark]

1f. between B and C.

[1 mark]

[1 mark]

Markscheme(loss in) gravitational potential and kinetic energy of block into elastic potential energy ofrope

See note for 1(c)(i) for naming convention.

Must see either the block or the rope (or both) mentioned in connection with the appropriateenergies.

[1 mark]

1g. The length reached by the rope at C is 77.4 m. Suggest how energy considerationscould be used to determine the elastic constant of the rope.

Markschemek can be determined using EPE = kx

correct statement or equation showing

GPE at A = EPE at C

OR

(GPE + KE) at B = EPE at C

Candidate must clearly indicate the energy associated with either position A or B for MP2.

[2 marks]

12

2

A small ball of mass m is moving in a horizontal circle on the inside surface of a frictionless

[2 marks]

2a.

A small ball of mass m is moving in a horizontal circle on the inside surface of a frictionlesshemispherical bowl.

The normal reaction force N makes an angle θ to the horizontal.

State the direction of the resultant force on the ball.

Markschemetowards the centre «of the circle» / horizontally to the right

Do not accept towards the centre of the bowl

[1 mark]

2b. On the diagram, construct an arrow of the correct length to represent the weight of theball.

[1 mark]

[2 marks]

Markschemedownward vertical arrow of any length

arrow of correct length

Judge the length of the vertical arrow by eye. The construction lines are not required. Alabel is not required

eg:

[2 marks]

2c. Show that the magnitude of the net force F on the ball is given by thefollowing equation.

F =mg

tan θ

[3 marks]

MarkschemeALTERNATIVE 1

F = N cos θmg = N sin θdividing/substituting to get result

ALTERNATIVE 2

right angle triangle drawn with F, N and W/mg labelled

angle correctly labelled and arrows on forces in correct directions

correct use of trigonometry leading to the required relationship

tan θ = [3 marks]

=OA

mg

F

2d. The radius of the bowl is 8.0 m and θ = 22°. Determine the speed of the ball. [4 marks]

Markscheme = m

r = R cos θ

v =

v = 13.4/13 «ms »

Award [4] for a bald correct answer

Award [3] for an answer of 13.9/14 «ms ». MP2 omitted

[4 marks]

mg

tanθv2

r

√ /√ /√gRcos2θsin θ gRcosθtanθ 9.81×8.0cos22tan22–1

–1

2e. Outline whether this ball can move on a horizontal circular path of radius equal tothe radius of the bowl.

Markschemethere is no force to balance the weight/N is horizontal

so no / it is not possible

Must see correct justification to award MP2

[2 marks]

A second identical ball is placed at the bottom of the bowl and the first ball is

[2 marks]

2f. A second identical ball is placed at the bottom of the bowl and the first ball isdisplaced so that its height from the horizontal is equal to 8.0 m.

The first ball is released and eventually strikes the second ball. The two balls remain in contact.Determine, in m, the maximum height reached by the two balls.

Markschemespeed before collision v = « =» 12.5 «ms »

«from conservation of momentum» common speed after collision is initial speed «v = = 6.25 ms »

h = « » 2.0 «m»

Allow 12.5 from incorrect use of kinematics equations

Award [3] for a bald correct answer

Award [0] for mg(8) = 2mgh leading to h = 4 m if done in one step.

Allow ECF from MP1

Allow ECF from MP2

[3 marks]

√2gR –1

12 c

12.52

–1

=vc2

2g6.252

2×9.81

At position B the rope starts to extend. Calculate the speed of the block at position B.

[3 marks]

3a. At position B the rope starts to extend. Calculate the speed of the block at position B.

Markschemeuse of conservation of energy

OR

v = u + 2as

v = « » = 34.3 «ms »

[2 marks]

2 2

√2 × 60.0 × 9.81 –1

3b. Determine the magnitude of the average resultant force acting on the block between Band C.

[2 marks]

[2 marks]

Markschemeuse of impulse F × Δt = Δp

OR

use of F = ma with average acceleration

OR

F =

3620«N»

Allow ECF from (a).

[2 marks]

ave

80.0×34.30.759

3c. Sketch on the diagram the average resultant force acting on the block between B andC. The arrow on the diagram represents the weight of the block.

Markschemeupwards

clearly longer than weight

For second marking point allow ECF from (b)(i) providing line is upwards.

[2 marks]

Calculate the magnitude of the average force exerted by the rope on the block

[2 marks]

3d. Calculate the magnitude of the average force exerted by the rope on the blockbetween B and C.

Markscheme3620 + 80.0 × 9.81

4400 «N»

Allow ECF from (b)(i).

[2 marks]

3e. between A and B.

Markscheme(loss in) gravitational potential energy (of block) into kinetic energy (of block)

Must see names of energy (gravitational potential energy and kinetic energy) – Allow forreasonable variations of terminology (eg energy of motion for KE).

[1 mark]

3f. between B and C.

[2 marks]

[1 mark]

[1 mark]

Markscheme(loss in) gravitational potential and kinetic energy of block into elastic potential energy ofrope

See note for 1(c)(i) for naming convention.

Must see either the block or the rope (or both) mentioned in connection with the appropriateenergies.

[1 mark]

3g. The length reached by the rope at C is 77.4 m. Suggest how energy considerationscould be used to determine the elastic constant of the rope.

Markschemek can be determined using EPE = kx

correct statement or equation showing

GPE at A = EPE at C

OR

(GPE + KE) at B = EPE at C

Candidate must clearly indicate the energy associated with either position A or B for MP2.

[2 marks]

12

2

An elastic climbing rope is tested by fixing one end of the rope to the top of a crane. The other

[2 marks]

3h.

An elastic climbing rope is tested by fixing one end of the rope to the top of a crane. The otherend of the rope is connected to a block which is initially at position A. The block is released fromrest. The mass of the rope is negligible.

The unextended length of the rope is 60.0 m. From position A to position B, the block falls freely.

In another test, the block hangs in equilibrium at the end of the same elastic rope. The elasticconstant of the rope is 400 Nm . The block is pulled 3.50 m vertically below the equilibriumposition and is then released from rest.

Calculate the time taken for the block to return to the equilibrium position for the firsttime.

MarkschemeT = 2π = 2.81 «s»

time = = 0.702 «s»

Award [0] for kinematic solutions that assume a constant acceleration.

[2 marks]

–1

√ 80.0400T4

Calculate the speed of the block as it passes the equilibrium position.

[2 marks]

3i. Calculate the speed of the block as it passes the equilibrium position.

MarkschemeALTERNATIVE 1

ω = = 2.24 «rad s »v = 2.24 × 3.50 = 7.84 «ms »

ALTERNATIVE 2

kx = mv OR 400 × 3.5 = 80v

v = 7.84 «ms »

Award [0] for kinematic solutions that assume a constant acceleration.

Allow ECF for T from (e)(i).

[2 marks]

2π2.81

–1

–1

12

2 12

2 12

2 12

2

–1

Hydrogen atoms in an ultraviolet (UV) lamp make transitions from the first excited state to the

[2 marks]

4a.

Hydrogen atoms in an ultraviolet (UV) lamp make transitions from the first excited state to theground state. Photons are emitted and are incident on a photoelectric surface as shown.

Show that the energy of photons from the UV lamp is about 10 eV.

MarkschemeE = –13.6 «eV» E = – = –3.4 «eV»

energy of photon is difference E – E = 10.2 «≈ 10 eV»

Must see at least 10.2 eV.

[2 marks]

1 213.6

4

2 1

The photons cause the emission of electrons from the photoelectric surface. The work function

[2 marks]

4b.

The photons cause the emission of electrons from the photoelectric surface. The work functionof the photoelectric surface is 5.1 eV.

Calculate, in J, the maximum kinetic energy of the emitted electrons.

Markscheme10 – 5.1 = 4.9 «eV»

4.9 × 1.6 × 10 = 7.8 × 10 «J»

Allow 5.1 if 10.2 is used to give 8.2×10 «J».

–19 –19

−19

4c. Suggest, with reference to conservation of energy, how the variable voltage sourcecan be used to stop all emitted electrons from reaching the collecting plate.

MarkschemeEPE produced by battery

exceeds maximum KE of electrons / electrons don’t have enough KE

For first mark, accept explanation in terms of electric potential energy difference ofelectrons between surface and plate.

[2 marks]

[2 marks]

[2 marks]

4d. The variable voltage can be adjusted so that no electrons reach the collecting plate.Write down the minimum value of the voltage for which no electrons reach the collectingplate.

Markscheme4.9 «V»

Allow 5.1 if 10.2 is used in (b)(i).

Ignore sign on answer.

[1 mark]

4e.

The electric potential of the photoelectric surface is 0 V. The variable voltage is adjusted so thatthe collecting plate is at –1.2 V.

On the diagram, draw and label the equipotential lines at –0.4 V and –0.8 V.

[1 mark]

[2 marks]

Markschemetwo equally spaced vertical lines (judge by eye) at approximately 1/3 and 2/3

labelled correctly

[2 marks]

4f. An electron is emitted from the photoelectric surface with kinetic energy 2.1 eV.Calculate the speed of the electron at the collecting plate.

Markschemekinetic energy at collecting plate = 0.9 «eV»

speed = «

» = 5.6 × 10 «ms »

Allow ECF from MP1

[2 marks]

√ 2×0.9×1.6×10−199.11×10−31

5 –1

5a. State the direction of the resultant force on the ball.

[2 marks]

[1 mark]

Markschemetowards the centre «of the circle» / horizontally to the right

Do not accept towards the centre of the bowl

[1 mark]

5b. On the diagram, construct an arrow of the correct length to represent the weight of theball.

Markschemedownward vertical arrow of any length

arrow of correct length

Judge the length of the vertical arrow by eye. The construction lines are not required. Alabel is not required

eg:

[2 marks]

Show that the magnitude of the net force F on the ball is given by the

[2 marks]

5c. Show that the magnitude of the net force F on the ball is given by thefollowing equation.

MarkschemeALTERNATIVE 1

F = N cos θmg = N sin θdividing/substituting to get result

ALTERNATIVE 2

right angle triangle drawn with F, N and W/mg labelled

angle correctly labelled and arrows on forces in correct directions

correct use of trigonometry leading to the required relationship

tan θ = [3 marks]

F =mg

tan θ

=OA

mg

F

The radius of the bowl is 8.0 m and θ = 22°. Determine the speed of the ball.

[3 marks]

5d. The radius of the bowl is 8.0 m and θ = 22°. Determine the speed of the ball.

Markscheme = m

r = R cos θ

v =

v = 13.4/13 «ms »

Award [4] for a bald correct answer

Award [3] for an answer of 13.9/14 «ms ». MP2 omitted

[4 marks]

mg

tanθv2

r

√ /√ /√gRcos2θsin θ gRcosθtanθ 9.81×8.0cos22tan22–1

–1

5e. Outline whether this ball can move on a horizontal circular path of radius equal tothe radius of the bowl.

[4 marks]

[2 marks]

Markschemethere is no force to balance the weight/N is horizontal

so no / it is not possible

Must see correct justification to award MP2

[2 marks]

5f. A second identical ball is placed at the bottom of the bowl and the first ball isdisplaced so that its height from the horizontal is equal to 8.0 m.

The first ball is released and eventually strikes the second ball. The two balls remain in contact.Determine, in m, the maximum height reached by the two balls.

[3 marks]

Markschemespeed before collision v = « =» 12.5 «ms »

«from conservation of momentum» common speed after collision is initial speed «v = = 6.25 ms »

h = « » 2.0 «m»

Allow 12.5 from incorrect use of kinematics equations

Award [3] for a bald correct answer

Award [0] for mg(8) = 2mgh leading to h = 4 m if done in one step.

Allow ECF from MP1

Allow ECF from MP2

[3 marks]

√2gR –1

12 c

12.52

–1

=vc2

2g6.252

2×9.81

5g.

The ball is now displaced through a small distance x from the bottom of the bowl and is thenreleased from rest.

The magnitude of the force on the ball towards the equilibrium position is given by

where R is the radius of the bowl.

Outline why the ball will perform simple harmonic oscillations about the equilibriumposition.

mgx

R

[1 mark]

Markschemethe «restoring» force/acceleration is proportional to displacement

Direction is not required

[1 mark]

5h. Show that the period of oscillation of the ball is about 6 s.

Markschemeω = «

» = «= 1.107 s »

T = « =

=» 5.7 «s»

Allow use of or g = 9.8 or 10

Award [0] for a substitution into T = 2π

[2 marks]

√ gR

√ 9.818.0 –1

2πω2π

1.107

√ Ig

The amplitude of oscillation is 0.12 m. On the axes, draw a graph to show the variation

[2 marks]

5i. The amplitude of oscillation is 0.12 m. On the axes, draw a graph to show the variationwith time t of the velocity v of the ball during one period.

Markschemesine graph

correct amplitude «0.13 m s »

correct period and only 1 period shown

Accept ± sine for shape of the graph. Accept 5.7 s or 6.0 s for the correct period.

Amplitude should be correct to ± square for MP2

eg: v /m s

[3 marks]

–1

12

–1

A wheel of mass 0.25 kg consists of a cylinder mounted on a central shaft. The shaft has a

[3 marks]

6a.

A wheel of mass 0.25 kg consists of a cylinder mounted on a central shaft. The shaft has aradius of 1.2 cm and the cylinder has a radius of 4.0 cm. The shaft rests on two rails with thecylinder able to spin freely between the rails.

The stationary wheel is released from rest and rolls down a slope with the shaft rolling on therails without slipping from point A to point B.

The moment of inertia of the wheel is 1.3 × 10 kg m . Outline what is meant by themoment of inertia.

Markschemean object’s resistance to change in rotational motion

OR

equivalent of mass in rotational equations

OWTTE

[1 mark]

–4 2

In moving from point A to point B, the centre of mass of the wheel falls through a

[1 mark]

6b. In moving from point A to point B, the centre of mass of the wheel falls through avertical distance of 0.36 m. Show that the translational speed of the wheel is about 1 ms after its displacement.

MarkschemeΔKE + Δrotational KE = ΔGPE

OR

mv + I = mgh

× 0.250 × v + × 1.3 × 10 × = 0.250 × 9.81 × 0.36

v = 1.2 «m s »

[3 marks]

–1

12

2 12

v2

r2

12

2 12

–4 v2

1.44×10−4

–1

6c. Determine the angular velocity of the wheel at B.

Markschemeω «=

» = 100 «rad s »

[1 mark]

1.20.012

–1

The wheel leaves the rails at point B and travels along the flat track to point C. For a short time

[3 marks]

[1 mark]

6d.

The wheel leaves the rails at point B and travels along the flat track to point C. For a short timethe wheel slips and a frictional force F exists on the edge of the wheel as shown.

Describe the effect of F on the linear speed of the wheel.

Markschemeforce in direction of motion

so linear speed increases

[2 marks]

6e. Describe the effect of F on the angular speed of the wheel.

[2 marks]

[2 marks]

Markschemeforce gives rise to anticlockwise/opposing torque on

wheel ✓ so angular speed decreases ✓

OWTTE

[2 marks]

7a.

A girl on a sledge is moving down a snow slope at a uniform speed.

Draw the free-body diagram for the sledge at the position shown on the snow slope. [2 marks]

Markschemearrow vertically downwards labelled weight «of sledge and/or girl»/ W/mg/gravitationalforce/F /F AND arrow perpendicular to the snow slope labelled reactionforce/R/normal contact force/N/F

friction force/F/f acting up slope «perpendicular to reaction force»

Do not allow G/g/“gravity”.

Do not award MP1 if a “driving force” is included.

Allow components of weight if correctly labelled.

Ignore point of application or shape of object.

Ignore “air resistance”.

Ignore any reference to “push of feet on sledge”.

Do not award MP2 for forces on sledge on horizontal ground

The arrows should contact the object

g gravitational

N

7b. After leaving the snow slope, the girl on the sledge moves over a horizontal regionof snow. Explain, with reference to the physical origin of the forces, why the verticalforces on the girl must be in equilibrium as she moves over the horizontal region.

Markschemegravitational force/weight from the Earth «downwards»

reaction force from the sledge/snow/ground «upwards»

no vertical acceleration/remains in contact with the ground/does not move vertically asthere is no resultant vertical force

Allow naming of forces as in (a)

Allow vertical forces are balanced/equal in magnitude/cancel out

When the sledge is moving on the horizontal region of the snow, the girl jumps off the

[3 marks]

7c. When the sledge is moving on the horizontal region of the snow, the girl jumps off thesledge. The girl has no horizontal velocity after the jump. The velocity of the sledgeimmediately after the girl jumps off is 4.2 m s . The mass of the girl is 55 kg and the mass ofthe sledge is 5.5 kg. Calculate the speed of the sledge immediately before the girl jumps from it.

Markschememention of conservation of momentum

OR

5.5 x 4.2 = (55 + 5.5) «v»

0.38 «m s »

Allow p=p′ or other algebraically equivalent statementAward [0] for answers based on energy

–1

–1

7d. The girl chooses to jump so that she lands on loosely-packed snow rather than frozenice. Outline why she chooses to land on the snow.

[2 marks]

[3 marks]

Markschemesame change in momentum/impulse

the time taken «to stop» would be greater «with the snow»

therefore F is smaller «with the snow»

OR

force is proportional to rate of change of momentum therefore F is smaller «with the snow»

Allow reverse argument for ice

F = ΔpΔt

7e.

The sledge, without the girl on it, now travels up a snow slope that makes an angle of 6.5˚ to thehorizontal. At the start of the slope, the speed of the sledge is 4.2 m s . The coefficient ofdynamic friction of the sledge on the snow is 0.11.

Show that the acceleration of the sledge is about –2 m s .

Markscheme«friction force down slope» = µmg cos(6.5) = «5.9 N»

«component of weight down slope» = mg sin(6.5) «= 6.1 N»

«so a = » acceleration = = 2.2 «m s »

Ignore negative signs

Allow use of g = 10 m s

–1

–2

Fm

125.5

–2

–2

Calculate the distance along the slope at which the sledge stops moving. Assume that

[3 marks]

7f. Calculate the distance along the slope at which the sledge stops moving. Assume thatthe coefficient of dynamic friction is constant.

Markschemecorrect use of kinematics equation

distance = 4.4 or 4.0 «m»

Alternative 2

KE lost=work done against friction + GPE

distance = 4.4 or 4.0 «m»

Allow ECF from (e)(i)

Allow [1 max] for GPE missing leading to 8.2 «m»

7g. The coefficient of static friction between the sledge and the snow is 0.14. Outline,with a calculation, the subsequent motion of the sledge.

Markschemecalculates a maximum value for the frictional force = « µR=» 7.5 «N»

sledge will not move as the maximum static friction force is greater than the component ofweight down the slope

Allow correct conclusion from incorrect MP1

Allow 7.5 > 6.1 so will not move

[2 marks]

[2 marks]

8a.

A non-uniform electric field, with field lines as shown, exists in a region where there is nogravitational field. X is a point in the electric field. The field lines and X lie in the plane of thepaper.

Outline what is meant by electric field strength.

Markschemeforce per unit charge

acting on a small/test positive charge

8b. An electron is placed at X and released from rest. Draw, on the diagram, the direction ofthe force acting on the electron due to the field.

Markschemehorizontally to the left

Arrow does not need to touch X

The electron is replaced by a proton which is also released from rest at X. Compare,

[2 marks]

[1 mark]

8c. The electron is replaced by a proton which is also released from rest at X. Compare,without calculation, the motion of the electron with the motion of the proton afterrelease. You may assume that no frictional forces act on the electron or the proton.

Markschemeproton moves to the right/they move in opposite directions

force on each is initially the same

proton accelerates less than electron initially «because mass is greater»

field is stronger on right than left «as lines closer»

proton acceleration increases «as it is moving into stronger field»

OR

electron acceleration decreases «as it is moving into weaker field»

Allow ECF from (b)

Accept converse argument for electron

The diagram below shows part of a downhill ski course which starts at point A, 50 m above level

[4 marks]

9a.

The diagram below shows part of a downhill ski course which starts at point A, 50 m above levelground. Point B is 20 m above level ground.

A skier of mass 65 kg starts from rest at point A and during the ski course some of thegravitational potential energy transferred to kinetic energy.

From A to B, 24 % of the gravitational potential energy transferred to kinetic energy.Show that the velocity at B is 12 m s .

Markscheme

«m s »

Award GPE lost = 65 × 9.81 × 30 = «19130 J»

Must see the 11.9 value for MP2, not simply 12.

Allow g = 9.8 ms .

–1

v2 = 0.24 gh12v = 11.9 –1

–2

Some of the gravitational potential energy transferred into internal energy of the skis,

[2 marks]

9b. Some of the gravitational potential energy transferred into internal energy of the skis,slightly increasing their temperature. Distinguish between internal energy andtemperature.

Markschemeinternal energy is the total KE «and PE» of the molecules/particles/atoms in an object

temperature is a measure of the average KE of the molecules/particles/atoms

Award [1 max] if there is no mention of molecules/particles/atoms.

9c. The dot on the following diagram represents the skier as she passes point B.Draw and label the vertical forces acting on the skier.

[2 marks]

[2 marks]

Markschemearrow vertically downwards from dot labelled weight/W/mg/gravitationalforce/F /F AND arrow vertically upwards from dot labelled reaction force/R/normalcontact force/N/F

W > R

Do not allow gravity.Do not award MP1 if additional ‘centripetal’ force arrow is added.Arrows must connect to dot.Ignore any horizontal arrow labelled friction.Judge by eye for MP2. Arrows do not have to be correctly labelled or connect to dot forMP2.

g gravitational

N

9d. The hill at point B has a circular shape with a radius of 20 m. Determine whether theskier will lose contact with the ground at point B.

[3 marks]

MarkschemeALTERNATIVE 1recognition that centripetal force is required / seen

= 468 «N»

W/640 N (weight) is larger than the centripetal force required, so the skier does not losecontact with the ground

ALTERNATIVE 2

recognition that centripetal acceleration is required / seen

a = 7.2 «ms »

g is larger than the centripetal acceleration required, so the skier does not lose contact withthe ground

ALTERNATIVE 3

recognition that to lose contact with the ground centripetal force ≥ weight

calculation that v ≥ 14 «ms »

comment that 12 «ms » is less than 14 «ms » so the skier does not lose contact with theground

ALTERNATIVE 4

recognition that centripetal force is required / seen

calculation that reaction force = 172 «N»

reaction force > 0 so the skier does not lose contact with the ground

Do not award a mark for the bald statement that the skier does not lose contact with theground.

mv2

r

v2

r

–2

–1

–1 –1

mv2

r

The skier reaches point C with a speed of 8.2 m s . She stops after a distance of 24–1

9e. The skier reaches point C with a speed of 8.2 m s . She stops after a distance of 24m at point D.

Determine the coefficient of dynamic friction between the base of the skis and the snow.Assume that the frictional force is constant and that air resistance can be neglected.

MarkschemeALTERNATIVE 10 = 8.2 + 2 × a × 24 therefore a = «−»1.40 «m s »

friction force = ma = 65 × 1.4 = 91 «N»

coefficient of friction = = 0.14

ALTERNATIVE 2KE = mv = 0.5 x 65 x 8.2 = 2185 «J»

friction force = KE/distance = 2185/24 = 91 «N»

coefficient of friction = = 0.14

Allow ECF from MP1.

–1

2 −2

9165×9.81

12

2 2

9165×9.81

9f.

At the side of the course flexible safety nets are used. Another skier of mass 76 kg falls normallyinto the safety net with speed 9.6 m s .

Calculate the impulse required from the net to stop the skier and state an appropriateunit for your answer.

–1

[3 marks]

[2 marks]

Markscheme«76 × 9.6»= 730Ns OR kg ms–1

9g. Explain, with reference to change in momentum, why a flexible safety net is less likelyto harm the skier than a rigid barrier.

Markschemesafety net extends stopping time

F = therefore F is smaller «with safety net»

OR

force is proportional to rate of change of momentum therefore F is smaller «with safety net»

Accept reverse argument.

ΔpΔt

A glider is an aircraft with no engine. To be launched, a glider is uniformly accelerated from rest

[2 marks]

10a.

A glider is an aircraft with no engine. To be launched, a glider is uniformly accelerated from restby a cable pulled by a motor that exerts a horizontal force on the glider throughout the launch.

The glider reaches its launch speed of 27.0 m s after accelerating for 11.0s. Assume that the glider moves horizontally until it leaves the ground. Calculatethe total distance travelled by the glider before it leaves the ground.

Markschemecorrect use of kinematic equation/equations

148.5 or 149 or 150 «m»

Substitution(s) must be correct.

–1

10b. The glider and pilot have a total mass of 492 kg. During the acceleration the glider issubject to an average resistive force of 160 N. Determine the average tension in thecable as the glider accelerates.

[2 marks]

[3 marks]

Markschemea = or 2.45 «m s »

F – 160 = 492 × 2.45

1370 «N»

Could be seen in part (a).Award [0] for solution that uses a = 9.81 m s

2711

–2

–2

10c. The cable is pulled by an electric motor. The motor has an overall efficiency of 23 %.Determine the average power input to the motor.

[3 marks]

MarkschemeALTERNATIVE 1

«work done to launch glider» = 1370 x 149 «= 204 kJ»

«work done by motor»

«power input to motor» or 80.4 or 81 k«W»

ALTERNATIVE 2

use of average speed 13.5 m s

«useful power output» = force x average speed «= 1370 x 13.5»

power input = « » 80 or 80.4 or 81 k«W»

ALTERNATIVE 3

work required from motor = KE + work done against friction «» = 204 «kJ»

«energy input»

power input k«W»

Award [2 max] for an answer of 160 k«W».

= 204×10023

= × = 80204×100231

11

–1

1370 × 13.5 × =10023

= 0.5 × 492 × 272 + (160 × 148.5)

= work required from motor×10023

= = 80.388300011

10d. The cable is wound onto a cylinder of diameter 1.2 m. Calculate the angular velocity ofthe cylinder at the instant when the glider has a speed of 27 m s . Includean appropriate unit for your answer.

–1[2 marks]

Markscheme « »

rad s

Do not accept Hz.Award [1 max] if unit is missing.

ω = =vr

= 45270.6–1

10e. After takeoff the cable is released and the unpowered glider moves horizontallyat constant speed. The wings of the glider provide a lift force. The diagram showsthe lift force acting on the glider and the direction of motion of the glider.

Draw the forces acting on the glider to complete the free-body diagram. The dotted lines showthe horizontal and vertical directions.

Markscheme

drag correctly labelled and in correct direction

weight correctly labelled and in correct direction AND no other incorrect force shown

Award [1 max] if forces do not touch the dot, but are otherwise OK.

Explain, using appropriate laws of motion, how the forces acting on the glider

[2 marks]

10f. Explain, using appropriate laws of motion, how the forces acting on the glidermaintain it in level flight.

Markschemename Newton's first law

vertical/all forces are in equilibrium/balanced/add to zeroORvertical component of lift mentioned

as equal to weight

10g. At a particular instant in the flight the glider is losing 1.00 m of vertical height for every6.00 m that it goes forward horizontally. At this instant, the horizontal speed of theglider is 12.5 m s . Calculate the velocity of the glider. Give your answer to an appropriatenumber of significant figures.

Markschemeany speed and any direction quoted together as the answer

quotes their answer(s) to 3 significant figures

speed = 12.7 m s or direction = 9.46 or 0.165 rad «below the horizontal» or gradient of

–1

–1 º

− 16

A student investigates how light can be used to measure the speed of a toy train.

[2 marks]

[3 marks]

11a.

A student investigates how light can be used to measure the speed of a toy train.

Light from a laser is incident on a double slit. The light from the slits is detected by a lightsensor attached to the train.

The graph shows the variation with time of the output voltage from the light sensor as the trainmoves parallel to the slits. The output voltage is proportional to the intensity of light incident onthe sensor.

Explain, with reference to the light passing through the slits, why a series of voltagepeaks occurs.

[3 marks]

Markscheme«light» superposes/interferes

pattern consists of «intensity» maxima and minimaORconsisting of constructive and destructive «interference»

voltage peaks correspond to interference maxima

11b. The slits are separated by 1.5 mm and the laser light has a wavelength of 6.3 x 10 m.The slits are 5.0 m from the train track. Calculate the separation between two adjacentpositions of the train when the output voltage is at a maximum.

Markscheme« » 2.1 x 10 «m»

If no unit assume m.Correct answer only.

–7

s = = =λDd

6.3×10−7×5.01.5×10−3

–3

11c. Estimate the speed of the train.

[1 mark]

[2 marks]

Markschemecorrect read-off from graph of 25 m s

v = « » 8.4 x 10 «m s »

Allow ECF from (b)(i)

= =xt

2.1×10−3

25×10−3–2 –1

11d. In another experiment the student replaces the light sensor with a sound sensor. Thetrain travels away from a loudspeaker that is emitting sound waves ofconstant amplitude and frequency towards a reflecting barrier.

The sound sensor gives a graph of the variation of output voltage with time along the track thatis similar in shape to the graph shown in the resource. Explain how this effect arises.

MarkschemeALTERNATIVE 1

«reflection at barrier» leads to two waves travelling in opposite directions

mention of formation of standing wave

maximum corresponds to antinode/maximum displacement «of air molecules»ORcomplete cancellation at node position

A mass is suspended from the ceiling of a train carriage by a string. The string makes

[2 marks]

12. A mass is suspended from the ceiling of a train carriage by a string. The string makesan angle θ with the vertical when the train is accelerating along a straight horizontaltrack.

What is the acceleration of the train?

A. g sin θB. g cos θ C. g tan θ D.

MarkschemeC

g

tanθ

13a.

Curling is a game played on a horizontal ice surface. A player pushes a large smooth stoneacross the ice for several seconds and then releases it. The stone moves until friction brings it torest. The graph shows the variation of speed of the stone with time.

The total distance travelled by the stone in 17.5 s is 29.8 m.

Determine the coefficient of dynamic friction between the stone and the ice duringthe last 14.0 s of the stone’s motion.

[4 marks]

[3 marks]

MarkschemeALTERNATIVE 1

«deceleration» « »

F = 0.243 × m

ALTERNATIVE 2

distance travelled after release = 23.85 «m»KE lost = 5.81m «J»

Award [3] for a bald correct answer.

Ignore sign in acceleration.

Allow ECF from (a) (note that x candidate answer to (a) ).

Ignore any units in answer.

Condone omission of m in solution.

Allow g = 10 N kg (gives 0.024).

= 3.4114.0 = 0.243m s−2

μ = = 0.0250.243×mm×9.81

μd = = = 0.025KE lostmg×distance5.81m

23.85mg

μ = 0.0073

–1

13b. The diagram shows the stone during its motion after release.

Label the diagram to show the forces acting on the stone. Your answer should include thename, the direction and point of application of each force.

[3 marks]

Markschemenormal force, upwards, ignore point of application

Force must be labeled for its mark to be awarded. Blob at poa not required. Allow OWTTE for normal force. Allow N, R, reaction.The vertical forces must lie within the middle third of the stone

weight/weight force/force of gravity, downwards, ignore point of application

Allow mg, W but not “gravity”.

Penalise gross deviations from vertical/horizontal once only

friction/resistive force, to left, at bottom of stone, point of application must be on theinterface between ice and stone

Allow F, µR. Only allow arrows/lines that lie on the interface. Take the tail of the arrow asthe definitive point of application and expect line to be drawn horizontal.

Award [2 max] if any force arrow does not touch the stone

Do not award MP3 if a “driving force” is shown acting to the right. This need not be labelledto disqualify the mark. Treat arrows labelled “air resistance” as neutral.

N.B: Diagram in MS is drawn with the vertical forces not direction of travel collinear forclarity

A company designs a spring system for loading ice blocks onto a truck. The ice block is placed

14a.

A company designs a spring system for loading ice blocks onto a truck. The ice block is placedin a holder H in front of the spring and an electric motor compresses the spring by pushing H tothe left. When the spring is released the ice block is accelerated towards aramp ABC. When the spring is fully decompressed, the ice block loses contact with the spring atA. The mass of the ice block is 55 kg.

Assume that the surface of the ramp is frictionless and that the masses of the spring and theholder are negligible compared to the mass of the ice block.

(i) The block arrives at C with a speed of 0.90ms . Show that the elastic energystored in the spring is 670J.

(ii) Calculate the speed of the block at A.

Markscheme(i)

OR«E =»E +E

OR669 J «E = 669 ≈ 670J»Award [1 max] for use of g=10Nkg , gives 682 J.

(ii)

If 682J used, answer is 5.0ms .

−1

≪ Eel= ≫ mv2 + mgh12

el P K

≪ Eel= ≫ × 55 × 0.902+55 × 9.8 × 1.212

el

–1

× 55 × v2 = 670J12

v =≪ √ = ≫ 4.9ms−12×67055–1

[4 marks]

14b. Describe the motion of the block

(i) from A to B with reference to Newton's first law.

(ii) from B to C with reference to Newton's second law.

Markscheme(i)no force/friction on the block, hence constant motion/velocity/speed

(ii)force acts on block OR gravity/component of weight pulls down slope

velocity/speed decreases OR it is slowing down OR it decelerates

Do not allow a bald statement of “N2” or “F = ma” for MP1.Treat references to energy as neutral.

14c. On the axes, sketch a graph to show how the displacement of the block varies withtime from A to C. (You do not have to put numbers on the axes.)

[3 marks]

[2 marks]

Markschemestraight line through origin for at least one-third of the total length of time axis covered bycandidate line

followed by curve with decreasing positive gradient

Ignore any attempt to include motion before A.

Gradient of curve must always be less than that of straight line.

14d. The spring decompression takes 0.42s. Determine the average force that the springexerts on the block.

Markscheme

F=642≈640NAllow ECF from (a)(ii).

F ≪= ≫=ΔpΔt

55×4.90.42

14e. The electric motor is connected to a source of potential difference 120V and draws acurrent of 6.8A. The motor takes 1.5s to compress the spring.

Estimate the efficiency of the motor.

[2 marks]

[2 marks]

Printed for Jyvaskylan Lyseon lukio

© International Baccalaureate Organization 2019 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markscheme«energy supplied by motor =» 120 × 6.8 × 1.5 or 1224 J OR«power supplied by motor =» 120 × 6.8 or 816 We = 0.55 or 0.547 or 55% or 54.7%

Allow ECF from earlier results.

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