A.2 Forces and Momentum Questions

1A-1A-03

Elastic and inelastic collisions·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

In the moderator of a nuclear reactor, a neutron of mass m moving at speed v collides elastically and head-on with a stationary carbon-12 nucleus of mass 12m. The neutron rebounds along its original line with speed 11v/13.

What fraction of the initial kinetic energy of the neutron is transferred to the carbon nucleus?

Show mark scheme

Marking point	Mark	Notes
Step 1The collision is elastic, so the total kinetic energy is conserved: the energy lost by the neutron is the energy gained by the carbon nucleus.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The kinetic energy of the neutron after the collision is ½m(11v/13)², i.e. (11/13)² = 121/169 = 0.72 of its initial kinetic energy.	—
Step 3Fraction transferred = 1 − 121/169 = 48/169 = 0.28. (Check with momentum: the carbon nucleus moves off at (v + 11v/13)/12 = 2v/13, and ½(12m)(2v/13)² = (48/169) × ½mv².)	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis is (2/13)², the square of the speed ratio of the carbon nucleus: its kinetic energy has been calculated with the neutron mass m instead of 12m.
BThis is 1 − 11/13 = 2/13, the fraction of the speed lost by the neutron. Kinetic energy depends on the square of the speed.
CCorrect: the neutron keeps (11/13)² = 121/169 of its kinetic energy, so 48/169 = 0.28 is transferred.
DThis is (11/13)² = 0.72, the fraction of its kinetic energy that the neutron keeps, not the fraction transferred.

Syllabus understandingA.2 — the elastic and inelastic collisions of two bodies; energy considerations in elastic collisions, inelastic collisions, and explosions; A.3 — the kinetic energy of translational motion as given by E_k = ½mv²; E.4 — the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant Command term: Determine

2A-1A-11

Impulse and force–time graphs·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A stationary ball of mass 0.25 kg is struck. The force on the ball rises linearly from zero to 400 N and then falls linearly back to zero, the whole contact lasting 0.020 s.

What is the speed of the ball immediately after the strike?

Force F on the ball against time t (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1Impulse equals the area under the force–time graph. The graph is a triangle of base 0.020 s and height 400 N.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Impulse = ½ × 400 × 0.020 = 4.0 N s.	—
Step 3Impulse = change in momentum = mv, so v = 4.0/0.25 = 16 m s⁻¹.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis equates the impulse to a kinetic energy, ½mv² = 4.0, giving v = √(2 × 4.0/0.25) = 5.7 m s⁻¹. Impulse is a change of momentum, mv, not of energy.
BThis uses an average force of 100 N — a quarter of the peak — giving 2.0 N s. The mean of a triangle is half the peak, 200 N.
CCorrect: the area under the graph is ½ × 400 × 0.020 = 4.0 N s, and 4.0/0.25 = 16 m s⁻¹.
DThis uses the rectangle 400 × 0.020 = 8.0 N s. The force is not constant — the area is a triangle.

Syllabus understandingA.2 — that the impulse J applied to a body is given by J = FΔt = Δp; that the area under a force–time graph is the impulse Command term: Determine

3A-1A-19

Circular motion·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A small coin rests on a horizontal turntable 0.50 m from the axis. The coefficient of static friction between coin and turntable is 0.40.

What is the greatest angular speed at which the coin remains in place? (g = 9.81 m s⁻²)

Show mark scheme

Marking point	Mark	Notes
Step 1Friction provides the centripetal force: f ≤ μmg, and the required centripetal force is mω²r.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2At the limit mω²r = μmg, so ω² = μg/r — the mass cancels.	—
Step 3ω² = 0.40 × 9.81/0.50 = 7.85, so ω = 2.8 rad s⁻¹.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis is √(μgr) = √(0.40 × 9.81 × 0.50) = 1.4 — the radius has been multiplied instead of divided; ω² = μg/r.
BCorrect: ω = √(μg/r) = √7.85 = 2.8 rad s⁻¹.
CThis is μg = 3.9 — the radius has been left out and no square root taken; ω² = μg/r, not μg.
DThis is ω² — the square root has not been taken.

Syllabus understandingA.2 — that bodies moving along a circular trajectory at a constant speed experience an acceleration that is directed radially towards the centre of the circle, as given by a = v²/r = ω²r; that circular motion is caused by a centripetal force acting perpendicular to the velocity; surface frictional force on a stationary body as given by F_f ≤ μ_sF_N Command term: Determine

4A-1A-24

Connected bodies and contact forces·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Two blocks of mass 2.0 kg and 3.0 kg are in contact on a frictionless horizontal surface. A horizontal force of 10 N pushes on the 2.0 kg block so that both blocks accelerate together.

What is the magnitude of the force exerted by the 2.0 kg block on the 3.0 kg block?

Show mark scheme

Marking point	Mark	Notes
Step 1Treat the pair as one body: a = 10/5.0 = 2.0 m s⁻².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The only horizontal force on the 3.0 kg block is the contact force from the 2.0 kg block: F = 3.0 × 2.0 = 6.0 N.	—
Step 3Check: on the 2.0 kg block, 10 − 6.0 = 4.0 N = 2.0 × 2.0 ✓ (Newton's third law gives an equal 6.0 N back on the 2.0 kg block).	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis is the net force on the 2.0 kg block, not the contact force.
BThis is half of the applied force — the force is not shared equally; it is shared in proportion to the masses.
CCorrect: the contact force must accelerate the 3.0 kg block alone.
DThe 2.0 kg block needs 4.0 N of the 10 N for its own acceleration, so it cannot pass on the full 10 N.

Syllabus understandingA.2 — Newton’s three laws of motion; that free-body diagrams can be analysed to find the resultant force on a system; the normal force F_N Command term: Determine

5A-1A-25

Inelastic collisions·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark4 steps to full marksDetermine

A wooden block of mass 1.98 kg hangs at rest from light vertical strings. A pellet of mass 20 g, moving horizontally, hits the block and stays embedded in it. The block and pellet swing together until their centre of mass has risen through a vertical height of 0.20 m. Air resistance is negligible.

What was the speed of the pellet just before it hit the block?

Show mark scheme

Marking point	Mark	Notes
Step 1During the swing mechanical energy is conserved: ½(m + M)v² = (m + M)gh, so v = √(2gh) = √(2 × 9.81 × 0.20) = 1.98 m s⁻¹ just after the impact.	—	All 4 steps must be completed — there is no mark for a part-answer.
Step 2The impact is very short and the strings are vertical, so there is no horizontal external force: momentum is conserved in the collision (kinetic energy is not, because the collision is inelastic).	—
Step 3mu = (m + M)v gives u = (2.00/0.020) × 1.98 = 198 m s⁻¹ ≈ 2.0 × 10² m s⁻¹.	—
Step 4Only about 1 % of the pellet's kinetic energy (3.9 J out of 392 J) remains after the collision.	✓ 1	Answer C

Answer: C · 4 stages of work, one mark

Every option, and why

AThis applies conservation of kinetic energy to the whole process, ½mu² = (m + M)gh. Kinetic energy is not conserved in the embedding collision, where about 99 % of it is transferred to internal energy.
BThis uses v = √(gh) = 1.40 m s⁻¹ for the speed after the impact, omitting the factor 2 that comes from ½v² = gh.
CCorrect: v = √(2gh) = 1.98 m s⁻¹ after the impact, and momentum conservation gives u = 100 × 1.98 = 198 m s⁻¹.
DThis uses v = 2gh = 3.92 m s⁻¹, forgetting to take the square root, and then conserves momentum: 100 × 3.92 = 392 m s⁻¹.

Syllabus understandingA.2 — the conservation of linear momentum; the elastic and inelastic collisions of two bodies; A.3 — the principle of the conservation of energy; transfers between kinetic energy and gravitational potential energy, ΔE_p = mgΔh Command term: Determine

6A-1A-28

Friction on an incline·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A block of mass m is at rest on a rough plane inclined at 30° to the horizontal, as shown.

What is the magnitude of the friction force on the block?

Block at rest on a rough incline; the three forces are shown.

Show mark scheme

Marking point	Mark	Notes
Step 1The block is in equilibrium, so the forces along the slope balance: friction (up the slope) equals the component of weight down the slope.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Component of weight along the slope = mg sin 30°.	—
Step 3f = ½mg. Note that static friction takes whatever value is needed (up to its maximum μN), so μmg cos 30° is only an upper limit.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AWithout friction the block would slide down the slope.
BCorrect: f = mg sin 30° = ½mg.
CThis is the normal force mg cos 30°, perpendicular to the slope, not the friction.
DThis is the maximum static friction; the actual friction is only as large as is needed for equilibrium.

Syllabus understandingA.2 — that forces acting on a body can be represented in a free-body diagram; surface frictional force on a stationary body as given by F_f ≤ μ_sF_N; Newton’s first law applied to translational equilibrium Command term: Determine

7A-1A-29

Vertical circular motion·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A ball of mass m on a string moves in a vertical circle of radius r. At the top of the circle its speed is v = √(2gr).

What is the tension in the string at the top of the circle?

Ball at the top of a vertical circle. Tension and weight both act towards the centre.

Show mark scheme

Marking point	Mark	Notes
Step 1At the top both the tension and the weight act downwards, towards the centre: T + mg = mv²/r.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2mv²/r = m(2gr)/r = 2mg.	—
Step 3T = 2mg − mg = mg.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

ATension is zero only at the minimum speed √(gr); here the speed is larger.
BCorrect: T = mv²/r − mg = 2mg − mg.
CThis is the centripetal force itself; the weight provides part of it.
DThis adds the weight instead of subtracting it — the result at the bottom of the circle at this speed.

Syllabus understandingA.2 — that bodies moving along a circular trajectory at a constant speed experience an acceleration that is directed radially towards the centre of the circle, as given by a = v²/r; that circular motion is caused by a centripetal force acting perpendicular to the velocity (guidance: circular motion in a vertical plane, at the top or bottom); tension Command term: Determine

8A-1A-37

Newton's third law·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

A crate stands on the floor of a lift that is accelerating upwards.

Which force forms a Newton's third law pair with the weight of the crate?

Show mark scheme

Marking point	Mark	Notes
Step 1The weight of the crate is the gravitational force exerted by the Earth on the crate.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2A Newton's third law pair consists of two forces of the same type, acting on two different bodies, equal in magnitude and opposite in direction.	—
Step 3So the partner is the gravitational force exerted by the crate on the Earth.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis force acts on the same body (the crate) and is a contact force, not gravitational; here it is also larger than the weight because the crate accelerates upwards.
BThis is the third-law partner of the normal force of the floor on the crate, not of the weight.
CCorrect: same type of force (gravitational), acting on the other body of the interaction (the Earth).
DThe resultant force is not an interaction between two bodies; it is the vector sum of the forces on the crate.

Syllabus understandingA.2 — Newton's three laws of motion; forces as interactions between bodies (identification of Newton's third law force pairs) Command term: Identify

9A-1A-38

Static and dynamic friction·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A crate of mass 12 kg rests on a horizontal floor. The coefficient of static friction is 0.50 and the coefficient of dynamic friction is 0.35. A horizontal pull on the crate is increased slowly from zero. When the crate starts to move, the pull is kept constant at 60 N. g = 9.81 m s⁻².

Which row gives the friction force on the crate when the pull is 40 N, and the acceleration of the crate once it is moving?

Friction when pull is 40 N / NAcceleration / m s⁻²

Show mark scheme

Marking point	Mark	Notes
Step 1Maximum static friction = μ_sF_N = 0.50 × 12 × 9.81 = 59 N. A pull of 40 N is less than this, so the crate stays at rest and the static friction is just 40 N (equilibrium).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Once sliding, F_f = μ_dF_N = 0.35 × 117.7 = 41.2 N.	—
Step 3a = (60 − 41.2)/12 = 1.6 m s⁻².	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis uses the static coefficient for the moving crate: (60 − 58.9)/12 = 0.095 m s⁻².
BCorrect: static friction matches the 40 N pull; dynamic friction 41.2 N gives a = 1.6 m s⁻².
CThis uses μ_dF_N for a stationary crate (41.2 N); a stationary crate cannot have a friction force larger than the pull, or it would accelerate towards the puller.
DThis takes the static friction as always equal to its maximum μ_sF_N (59 N) and also uses it once the crate slides.

Syllabus understandingA.2 — surface frictional force on a stationary body as given by F_f ≤ μ_sF_N or a body in motion as given by F_f = μ_dF_N; Newton's first and second laws Command term: Determine

10A-1A-39

Hooke's law·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksCalculate

The spring in an angler's spring balance has a length of 0.150 m when unloaded. When a fish of mass 0.60 kg hangs at rest from it, the length of the spring is 0.186 m. g = 9.81 m s⁻².

What is the spring constant of the spring?

Show mark scheme

Marking point	Mark	Notes
Step 1At rest the elastic force balances the weight: kx = mg = 0.60 × 9.81 = 5.89 N.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Extension x = 0.186 − 0.150 = 0.036 m.	—
Step 3k = 5.89/0.036 = 1.6 × 10² N m⁻¹.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis uses the extension in centimetres (5.89/3.6 = 1.6) without converting to metres.
BThis uses the mass instead of the weight (0.60/0.036 = 17), forgetting to multiply by g.
CThis divides by the total length of the spring (5.89/0.186 = 32) instead of the extension.
DCorrect: k = mg/x = 5.89/0.036 = 164 N m⁻¹.

Syllabus understandingA.2 — elastic restoring force F_H following Hooke's law as given by F_H = −kx where k is the spring constant; translational equilibrium Command term: Calculate

11A-1A-40

Buoyancy·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A museum conservator hangs a small metal fragment from a newton-meter. The reading is 5.40 N in air and 3.40 N when the fragment is completely submerged in water of density 1000 kg m⁻³.

What is the density of the metal?

Show mark scheme

Marking point	Mark	Notes
Step 1The buoyancy force is the loss in reading: F_b = 5.40 − 3.40 = 2.00 N.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2F_b = ρVg, so the volume displaced (= volume of the fragment) is V = 2.00/(1000 × 9.81) = 2.04 × 10⁻⁴ m³.	—
Step 3Mass = 5.40/9.81 = 0.550 kg, so density = 0.550/2.04 × 10⁻⁴ = 2.7 × 10³ kg m⁻³ (equivalently 1000 × 5.40/2.00).	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis inverts the ratio: 1000 × 2.00/5.40 = 370 kg m⁻³.
BThis treats the reading in water as the buoyancy force: 1000 × 5.40/3.40 = 1.6 × 10³ kg m⁻³.
CThis uses the reading in water as the weight: 1000 × 3.40/2.00 = 1.7 × 10³ kg m⁻³.
DCorrect: density = density of water × (weight/buoyancy force) = 1000 × 5.40/2.00.

Syllabus understandingA.2 — buoyancy F_b acting on a body due to the displacement of the fluid as given by F_b = ρVg where V is the volume of fluid displaced; translational equilibrium Command term: Determine

12A-1A-41

Explosions·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

In deep space a stationary capsule of mass 5.0 kg is split into three parts by a spring mechanism. A 2.0 kg part moves east at 6.0 m s⁻¹ and a 1.0 kg part moves north at 8.0 m s⁻¹, as shown. The third part has mass 2.0 kg.

Which row gives the speed and the direction of motion of the third part?

Top view (not to scale). The third part, of mass 2.0 kg, is not shown.

Speed / m s⁻¹Direction

Show mark scheme

Marking point	Mark	Notes
Step 1The total momentum is zero before and after, so the third part must cancel the vector sum of the other two momenta: 12 kg m s⁻¹ east and 8.0 kg m s⁻¹ north.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Its momentum is √(12² + 8.0²) = 14.4 kg m s⁻¹, directed south-west, so its speed is 14.4/2.0 = 7.2 m s⁻¹.	—
Step 3Its direction makes tan⁻¹(8.0/12) = 34° with west, towards the south: 34° south of west.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis is the direction of the sum of the other two momenta; the third part must move opposite to it.
BCorrect: momentum (−12, −8.0) kg m s⁻¹, speed 7.2 m s⁻¹, 34° south of west.
CThe angle tan⁻¹(12/8.0) = 56° is measured from the wrong axis: it is the angle from south, not from west.
DThis adds the momenta as scalars: (12 + 8.0)/2.0 = 10 m s⁻¹. Momentum is a vector and must be added as one.

Syllabus understandingA.2 — explosions; linear momentum p = mv remains constant unless the system is acted upon by a resultant external force (two-dimensional situations, HL) Command term: Determine

13A-1A-65

Free-body diagrams and translational equilibrium·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A lantern of weight W hangs at rest from the midpoint of a light cable fixed between two posts, as shown. Each half of the cable makes an angle θ with the horizontal.

What is the tension in the cable?

Lantern hanging from the midpoint of the cable (not to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1The point where the lantern hangs is in equilibrium under three forces: the weight W and the tension T in each half of the cable.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Horizontal components T cos θ cancel. Vertically: 2T sin θ = W.	—
Step 3T = W/(2 sin θ). As θ becomes small, the tension becomes very large.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis shares the weight between the two halves of the cable as if they were vertical. Each half supports W/2 only through its vertical component T sin θ.
BThis resolves with the cosine: T cos θ is the horizontal component, which cancels and does not support the weight.
CThis uses the correct vertical component but forgets that two halves of the cable support the lantern: T sin θ = W.
DCorrect: 2T sin θ = W.

14A-1A-66

Circular motion·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A car travels at constant speed v around a flat horizontal bend that is an arc of a circle of radius r. The frictional force from the road on the tyres towards the centre of the bend is F.

The same car later travels at constant speed 2v around a flat bend of radius 2r. What is the frictional force towards the centre now?

Show mark scheme

Marking point	Mark	Notes
Step 1Friction provides the centripetal force: F = mv²/r.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Doubling v multiplies v² by 4; doubling r halves the force.	—
Step 3New force = F × 4/2 = 2F.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis assumes that doubling both quantities cancels, as if F ∝ v/r. The force depends on the square of the speed.
BCorrect: F ∝ v²/r, so the factor is 2²/2 = 2.
CThis includes the effect of the speed but ignores the larger radius, which halves the force needed.
DThis multiplies by the radius instead of dividing (F ∝ v²r). A larger radius means a gentler turn and needs less force.

Syllabus understandingA.2 — that bodies moving along a circular trajectory at a constant speed experience an acceleration directed radially towards the centre of the circle, as given by a = v²/r; that circular motion is caused by a centripetal force acting perpendicular to the velocity; surface frictional force Command term: Deduce

15A-1A-67

Elastic and inelastic collisions·A.2 Forces and momentum

Paper 1AHard1 mark

Multiple choice · 1 mark4 steps to full marksDetermine

Two gliders, P and Q, each of mass m, are on a horizontal frictionless air track. A light spring of spring constant k is fixed to the front of Q, which is at rest. P moves towards Q with speed u and compresses the spring. The spring obeys Hooke's law.

What is the maximum compression of the spring?

Show mark scheme

Marking point	Mark	Notes
Step 1While the spring is being compressed, P is faster than Q. The compression is greatest at the instant the two gliders have the same velocity.	—	All 4 steps must be completed — there is no mark for a part-answer.
Step 2Momentum is conserved: mu = 2mv, so v = u/2.	—
Step 3Kinetic energy at that instant = ½(2m)(u/2)² = ¼mu², so the energy stored in the spring is ½mu² − ¼mu² = ¼mu².	—
Step 4½kx² = ¼mu² gives x = u√(m/(2k)).	✓ 1	Answer C

Answer: C · 4 stages of work, one mark

Every option, and why

AThis stores all the initial kinetic energy in the spring, as if Q were held fixed. Q is free to move, so the gliders keep kinetic energy ¼mu² at maximum compression.
BThis subtracts only the kinetic energy of P at the common speed, ½m(u/2)², from the initial kinetic energy and forgets that Q also moves at u/2: stored energy ⅜mu².
CCorrect: at maximum compression both gliders move at u/2 and the spring stores ¼mu².
DThis takes the stored energy to be the kinetic energy of P alone at the common speed, ½m(u/2)² = ⅛mu².

Syllabus understandingA.2 — the conservation of linear momentum; the elastic and inelastic collisions of two bodies; energy considerations in collisions; A.3 — the elastic potential energy as given by E_H = ½k(Δx)² Command term: Determine

16A-1A-87

Static friction and Newton’s first law·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark2 steps to full marksIdentify

A suitcase rests on a horizontal conveyor belt and moves with the belt without slipping. Air resistance is negligible.

Which row gives the direction of the frictional force exerted by the belt on the suitcase while the belt moves at constant velocity, and while the belt is speeding up?

Belt at constant velocityBelt speeding up

Show mark scheme

Marking point	Mark	Notes
Step 1At constant velocity the suitcase has no acceleration, so by Newton’s first law the resultant force on it is zero. The weight and the normal force balance vertically, and no other horizontal force acts, so the friction must be zero.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2While the belt speeds up, the suitcase accelerates in the direction of motion with the belt. The only horizontal force on it is the static friction from the belt, so the friction acts in the direction of motion (Newton’s second law).	✓ 1	Answer A

Answer: A · 2 stages of work, one mark

Every option, and why

ACorrect: no horizontal force is needed for constant velocity, and the forward acceleration needs a forward static friction force.
BThis assumes that friction always opposes the motion. Static friction opposes relative slipping: without friction the suitcase would slide backwards relative to the accelerating belt, so the friction acts forwards.
CThis assumes that a force is needed to keep a body moving at constant velocity. By Newton’s first law, no resultant force is needed for constant velocity.
DThis treats friction like air resistance, acting against the motion even at constant velocity. A backward force with no forward force to balance it would decelerate the suitcase.

Syllabus understandingA.2 — Newton’s three laws of motion (Newton’s first law applied to translational equilibrium); surface frictional force F_f acting in a direction parallel to the plane of contact between a body and a surface, on a stationary body as given by F_f ≤ μ_sF_N or a body in motion as given by F_f = μ_dF_N Command term: Identify

17A-1A-88

Force as rate of change of momentum·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A trolley moves in a straight line. The graph shows how the momentum p of the trolley varies with time t.

What is the resultant force on the trolley at t = 1.0 s?

Momentum p of the trolley against time t (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1The resultant force is the rate of change of momentum, F = Δp/Δt, which is the gradient of the momentum–time graph.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2From 0 to 3.0 s the graph is a straight line, so the gradient is the same at every instant in this interval, including at 1.0 s.	—
Step 3Gradient = (8.0 − 2.0)/3.0 = 2.0 N.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis takes the change in momentum over the whole 5.0 s: (8.0 − 2.0)/5.0 = 1.2 N. That is the average force over 5.0 s; from 3.0 s to 5.0 s the force is zero.
BCorrect: the gradient of the sloping section is 6.0 N s/3.0 s = 2.0 N.
CThis divides the final momentum by the time, 8.0/3.0 = 2.7 N, ignoring the initial momentum of 2.0 N s. The force depends on the change in momentum.
DThis divides the momentum at 1.0 s by the time: 4.0/1.0 = 4.0 N. The force is the gradient Δp/Δt, not p/t.

Syllabus understandingA.2 — Newton’s second law; that Newton’s second law in the form F = ma assumes mass is constant whereas F = Δp/Δt allows for situations where mass is changing; that a resultant force applied to a system constitutes an impulse J as given by J = FΔt; that the applied external impulse equals the change in momentum of the system Command term: Determine

18A-1A-89

Buoyancy and tension·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A block of wood of volume V and density ρ_b is held completely under water by a light vertical thread tied to the bottom of a tank, as shown. The density of the water is ρ_w, where ρ_w > ρ_b. The block is at rest.

What is the tension in the thread?

The block held under water by the thread (not to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1Three forces act on the block: the buoyancy force ρ_wVg upwards, its weight ρ_bVg downwards and the tension T in the thread, which pulls downwards.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The block is in equilibrium (Newton’s first law): ρ_wVg = ρ_bVg + T.	—
Step 3T = (ρ_w − ρ_b)Vg.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis is the weight of the block. The thread pulls the block down, so it is the buoyancy force that must balance the weight plus the tension.
BThis is the buoyancy force alone; it forgets that the weight of the block also acts downwards and supplies part of the balance.
CThis adds the weight to the buoyancy force, as if the weight acted upwards, or as if the thread had to support the block.
DCorrect: buoyancy force = weight + tension, so T = (ρ_w − ρ_b)Vg.

Syllabus understandingA.2 — that forces acting on a body can be represented in a free-body diagram; that free-body diagrams can be analysed to find the resultant force on a system; buoyancy F_b acting on a body due to the displacement of the fluid as given by F_b = ρVg where V is the volume of fluid displaced; tension Command term: Determine

19A-1A-90

Viscous drag, buoyancy and terminal speed·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A small steel sphere of density ρ_s is released from rest in a tall cylinder of oil of density ρ_o. The drag force on the sphere is given by F_d = 6πηrv. The sphere eventually reaches a terminal speed.

What is the acceleration of the sphere at the instant when its speed is half of its terminal speed?

Show mark scheme

Marking point	Mark	Notes
Step 1The weight W = ρ_sVg acts down; the buoyancy force F_b = ρ_oVg and the drag act up. At the terminal speed the drag is W − F_b.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The drag is proportional to the speed, so at half the terminal speed it is ½(W − F_b), and the resultant force is the other half, ½(W − F_b).	—
Step 3a = ½(ρ_s − ρ_o)Vg/(ρ_sV) = ½g(1 − ρ_o/ρ_s).	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis halves the free-fall acceleration and forgets the buoyancy force, which acts throughout the fall and does not depend on the speed.
BCorrect: at half the terminal speed the drag has half its terminal value, so the resultant force is ½(W − F_b).
CThis takes the drag to be proportional to the square of the speed, so that it is only ¼ of its terminal value. The viscous drag 6πηrv is proportional to v itself.
DThis is the acceleration at the moment of release, when the speed and so the drag are zero.

Syllabus understandingA.2 — viscous drag force F_d acting on a small sphere opposing its motion through a fluid as given by F_d = 6πηrv; buoyancy F_b acting on a body due to the displacement of the fluid as given by F_b = ρVg where V is the volume of fluid displaced; Newton’s three laws of motion; A.1 — the qualitative effect of fluid resistance, including terminal speed Command term: Deduce

20A-1A-91

Impulse in two dimensions·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A tennis ball of mass 0.058 kg moving horizontally at 20 m s⁻¹ strikes a smooth vertical wall at 30° to the wall. It rebounds at the same speed, also at 30° to the wall, as shown in the top view.

What is the magnitude of the impulse exerted by the wall on the ball?

Top view of the ball before and after striking the wall (not to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1The component of the velocity parallel to the wall, 20 cos 30° = 17.3 m s⁻¹, is unchanged, so only the perpendicular component contributes to the change in momentum.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The perpendicular component, 20 sin 30° = 10.0 m s⁻¹, is reversed: its change is 2 × 10.0 = 20.0 m s⁻¹.	—
Step 3Impulse = change in momentum = 0.058 × 20.0 = 1.16 N s ≈ 1.2 N s, directed perpendicular to the wall, away from it.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis uses the change in speed, which is zero. Momentum is a vector: its direction changes, so the change in momentum is not zero.
BCorrect: J = 2mu sin 30° = 2 × 0.058 × 20 × 0.50 = 1.16 N s.
CThis resolves with the cosine, 2mu cos 30° = 2.0 N s, which is twice the momentum component parallel to the wall — the component that does not change.
DThis treats the ball as if it struck the wall head-on and reversed completely: 2mu = 2.3 N s.

Syllabus understandingA.2 — that a resultant force applied to a system constitutes an impulse J as given by J = FΔt; that the applied external impulse equals the change in momentum of the system; that linear momentum as given by p = mv remains constant unless the system is acted upon by a resultant external force (momentum as a vector; two-dimensional situations, HL) Command term: Determine

21A-1A-92

Vertical circular motion·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A car is driven at constant speed over a hump in a road. The top of the hump is an arc of a vertical circle. The driving force on the car balances the air resistance, and the car stays in contact with the road.

Three statements are made about the car at the highest point of the hump.

I. The car is in equilibrium because its speed is constant.
II. The resultant force on the car is directed vertically downwards.
III. The normal force exerted by the road on the car is smaller than the weight of the car.

Which statements are correct?

Show mark scheme

Marking point	Mark	Notes
Step 1At the highest point the car moves on a circle whose centre is below it, so it has a centripetal acceleration v²/r directed vertically downwards, even though its speed is constant.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2By Newton’s second law the resultant force is therefore vertically downwards and non-zero: the car is not in equilibrium (I is wrong, II is correct).	—
Step 3Vertically: mg − F_N = mv²/r, so F_N = mg − mv²/r, which is less than the weight (III is correct).	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis includes I. A constant speed does not mean a constant velocity: the direction of motion changes, so the car accelerates towards the centre of the circle.
BThis includes I and leaves out III. The weight must exceed the normal force to give the downward resultant force.
CCorrect: the resultant force mv²/r is downwards, towards the centre, so F_N is smaller than mg.
DThis includes I. Equilibrium needs zero resultant force, which contradicts statement II.

Syllabus understandingA.2 — that bodies moving along a circular trajectory at a constant speed experience an acceleration that is directed radially towards the centre of the circle — known as a centripetal acceleration — as given by a = v²/r = ω²r = 4π²r/T²; that circular motion is caused by a centripetal force acting perpendicular to the velocity; that a centripetal force causes the body to change direction even if its magnitude of velocity may remain constant (guidance: non-uniform circular motion in a vertical plane, at the top or bottom); the normal force F_N Command term: Deduce

22A-1A-93

Angular velocity and centripetal acceleration·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Point P is on the surface of the Earth at latitude 60°. Point Q is on the equator. Both points move in circles because of the rotation of the Earth about its axis. Treat the Earth as a sphere.

Which row gives the ratio of the angular velocities ω_P/ω_Q and the ratio of the centripetal accelerations a_P/a_Q?

ω_P/ω_Qa_P/a_Q

Show mark scheme

Marking point	Mark	Notes
Step 1Every point on the Earth turns through 2π rad in the same time (one day), so ω is the same for P and Q: ω_P/ω_Q = 1.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2P moves in a circle about the axis of radius R cos 60° = ½R, where R is the radius of the Earth; Q moves in a circle of radius R.	—
Step 3a = ω²r with the same ω, so a_P/a_Q = ½.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis takes the angular velocity to be proportional to the radius of the circle (as the speed is), then uses a = ω²r: ¼ × ½ = ⅛. All points on a rigid rotating body have the same angular velocity.
BThis uses a = v²/r with the speed of P halved but with the radius of the Earth R for both circles: (½)² = ¼. P moves in a smaller circle of radius ½R.
CCorrect: same ω, and a = ω²r is proportional to the radius of the circle, which is halved.
DThis uses a = v²/r as if P and Q had the same speed, so that a ∝ 1/r. The speed v = ωr is halved too.

23A-1A-94

Field forces in a free-body diagram·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A small sphere of mass 2.0 g carries a positive charge. It hangs at rest from a light insulating thread in a uniform horizontal electric field of field strength 5.0 × 10⁴ N C⁻¹. The thread makes an angle of 40° with the vertical, as shown.

What is the charge on the sphere?

The charged sphere at rest in the uniform electric field (not to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1Three forces act on the sphere: weight mg downwards, electric force qE horizontally, and the tension T along the thread.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Equilibrium: T cos 40° = mg and T sin 40° = qE, so qE = mg tan 40°.	—
Step 3q = 2.0 × 10⁻³ × 9.81 × tan 40°/5.0 × 10⁴ = 3.3 × 10⁻⁷ C.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis uses sin 40° in place of tan 40°, i.e. it sets the electric force equal to mg sin 40°: 2.5 × 10⁻⁷ C. The tension, not the weight, is resolved into the two components.
BThis uses cos 40° in place of tan 40°, mixing up which component of the tension balances which force: mg cos 40°/E = 3.0 × 10⁻⁷ C.
CCorrect: q = mg tan 40°/E = 3.3 × 10⁻⁷ C.
DThis inverts the trigonometric ratio, qE = mg/tan 40° (the angle measured from the horizontal instead of the vertical): 4.7 × 10⁻⁷ C.

Syllabus understandingA.2 — that forces acting on a body can be represented in a free-body diagram; that free-body diagrams can be analysed to find the resultant force on a system; tension; the nature and use of the field forces: gravitational force F_g = mg, electric force F_e; D.2 — the electric field strength as given by E = F/q Command term: Determine

24A-1A-95

Explosions and relative velocity·A.2 Forces and momentum

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A child of mass m stands on a trolley of mass M that is at rest on a smooth horizontal floor. The child jumps horizontally off the back of the trolley. Immediately after the jump the child moves with speed u relative to the trolley.

What is the speed of the trolley relative to the floor immediately after the jump?

Show mark scheme

Marking point	Mark	Notes
Step 1No resultant external horizontal force acts on the child and trolley, so their total horizontal momentum stays zero.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Let the trolley move forwards at V relative to the floor. The child moves backwards at u relative to the trolley, so the child’s velocity relative to the floor is V − u.	—
Step 3MV + m(V − u) = 0, so V = mu/(M + m).	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: momentum conservation with the child’s ground velocity V − u gives V = mu/(M + m).
BThis takes u to be the child’s speed relative to the floor: MV = mu. The speed given is relative to the moving trolley.
CThis is the speed of the child relative to the floor, u − V = Mu/(M + m), not the speed of the trolley.
DThis assumes that the child and the trolley have equal kinetic energies, ½MV² = ½mu². It is the momenta that are equal and opposite; the kinetic energies are shared in the inverse ratio of the masses.

25A-1A-96

Hooke’s law·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

The graph shows how the force F needed to stretch each of two light springs, P and Q, varies with the extension x. The top of P is fixed to a support, the top of Q is hooked to the bottom of P, and a load of weight 6.0 N hangs at rest from the bottom of Q.

What is the total extension of the two springs?

Force F against extension x for springs P and Q (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1From the gradients: k_P = 12/0.30 = 40 N m⁻¹ and k_Q = 12/0.20 = 60 N m⁻¹.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The springs are light, so each spring is in equilibrium with a tension of 6.0 N: the full load acts on both P and Q.	—
Step 3Total extension = 6.0/40 + 6.0/60 = 0.15 + 0.10 = 0.25 m (or read the extension of each spring at 6.0 N from the graph).	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis adds the spring constants as if the springs were side by side sharing the load: 6.0/(40 + 60) = 0.060 m.
BThis assumes that only Q, which carries the load directly, is stretched: 6.0/60 = 0.10 m. P must support Q and the load, so it also has a tension of 6.0 N.
CThis shares the load equally between the springs, 3.0 N each: 3.0/40 + 3.0/60 = 0.125 m. Each light spring carries the whole 6.0 N.
DCorrect: each spring has a tension of 6.0 N, so the extensions 0.15 m and 0.10 m add to 0.25 m.

Syllabus understandingA.2 — elastic restoring force F_H following Hooke’s law as given by F_H = −kx where k is the spring constant; tension; that forces acting on a body can be represented in a free-body diagram; that free-body diagrams can be analysed to find the resultant force on a system; Newton’s third law Command term: Determine

26A-1A-97

Newton’s third law and momentum·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Two trolleys, X of mass 0.20 kg and Y of mass 0.60 kg, carry magnets that repel each other. They are held at rest on a horizontal frictionless track and then released, so that they move apart.

What is the ratio (speed of X)/(speed of Y) at any instant after the release?

Show mark scheme

Marking point	Mark	Notes
Step 1By Newton’s third law the magnetic forces on X and Y are equal in magnitude and opposite in direction at every instant, and they act for the same time.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2So the trolleys receive equal and opposite impulses: the total momentum stays zero, and 0.20v_X = 0.60v_Y.	—
Step 3v_X/v_Y = 0.60/0.20 = 3.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis inverts the ratio of the masses: the lighter trolley moves faster, not slower.
BThis assumes that equal forces give equal speeds. Equal forces give equal changes in momentum, so the lighter trolley gains more speed.
CThis assumes that the trolleys have equal kinetic energies, ½mv², giving √(0.60/0.20). It is the momenta that are equal in magnitude.
DCorrect: equal and opposite impulses give equal magnitudes of momentum, so the speeds are in the inverse ratio of the masses, 3.

Syllabus understandingA.2 — Newton’s three laws of motion (Newton’s third law); forces as interactions between bodies; the nature and use of the field force: magnetic force F_m; that linear momentum as given by p = mv remains constant unless the system is acted upon by a resultant external force Command term: Deduce

27A-1A-98

Non-uniform circular motion·A.2 Forces and momentum

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A car travels at 20 m s⁻¹ round a flat horizontal bend that is an arc of a circle of radius 80 m. The coefficient of static friction between the tyres and the road is 0.80. The driver brakes while following the bend. Air resistance is negligible.

What is the greatest rate at which the car can lose speed, at the instant its speed is 20 m s⁻¹, without the tyres slipping?

Show mark scheme

Marking point	Mark	Notes
Step 1The friction force from the road is the only horizontal force. Its maximum magnitude is μ_smg, so the greatest horizontal acceleration is μ_sg = 0.80 × 9.81 = 7.85 m s⁻².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Part of this acceleration must be the centripetal acceleration v²/r = 20²/80 = 5.0 m s⁻², perpendicular to the velocity; the rest is the tangential deceleration, along the velocity.	—
Step 3The two components are perpendicular: a_t = √(7.85² − 5.0²) = 6.0 m s⁻².	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis subtracts the centripetal acceleration from μ_sg as if the two acted along the same line: 7.85 − 5.0 = 2.8 m s⁻². They are perpendicular and combine by Pythagoras.
BCorrect: the friction provides both components, so a_t = √((μ_sg)² − (v²/r)²) = 6.0 m s⁻².
CThis ignores the bend and uses all the available friction for braking: μ_sg = 7.8 m s⁻². The friction must also provide the centripetal force.
DThis adds the two components by Pythagoras, √(7.85² + 5.0²) = 9.3 m s⁻². The total acceleration cannot exceed μ_sg, the limit set by the friction.

Syllabus understandingA.2 — that circular motion is caused by a centripetal force acting perpendicular to the velocity; that a centripetal force causes the body to change direction even if its magnitude of velocity may remain constant; surface frictional force F_f acting in a direction parallel to the plane of contact between a body and a surface, on a stationary body as given by F_f ≤ μ_sF_N or a body in motion as given by F_f = μ_dF_N (guidance: uniform and non-uniform circular motion in a horizontal plane) Command term: Determine

28A-1B-03

Newton's third law·A.2 Forces and momentum

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

A permanent magnet rests on an electronic top-pan balance that reads to ±0.01 g. A stiff horizontal copper wire, held by a clamp that does not touch the magnet or the balance, passes between the poles of the magnet at right angles to its uniform field. The length of wire in the field is 4.0 cm. When there is a current I in the wire, the magnetic force on the wire is vertically upwards.

The balance is not set to zero before the readings are taken. The table and graph show the balance reading for each current.

I / A	0.00	0.50	1.00	1.50	2.00	2.50	3.00
balance reading / g	152.36	152.48	152.61	152.72	152.85	152.97	153.09

Graph drawn to scale

(a)

(i)

State and explain the direction of the force that the wire exerts on the magnet.

(1)

(b)

(i)

Determine the force on the wire when I = 3.00 A.

(2)

(ii)

Determine the magnetic flux density B of the field, using the gradient of the graph.

(2)

(c)

(i)

Compare the percentage uncertainty in Δm at 0.50 A with that at 3.00 A, and suggest how, using the same balance and wire, the percentage uncertainty at small currents could be reduced.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Vertically downwards: by Newton's third law the wire pushes on the magnet with a force equal in magnitude and opposite in direction to the force of the magnet on the wire (so the balance reading increases)	✓ 1	Direction and a reference to Newton's third law both needed.
Part (b)(i)
Subtract the reading at zero current: Δm = 153.09 − 152.36 = 0.73 g	✓ 1	Using 153.09 g alone scores 0.
F = 0.73 × 10⁻³ × 9.81 = 7.2 × 10⁻³ N	✓ 1	Accept 7.1–7.2 × 10⁻³ N.
Part (b)(ii)
Gradient, e.g. (153.09 − 152.36)/3.00 = 0.243 g A⁻¹, i.e. 0.243 × 10⁻³ × 9.81 = 2.39 × 10⁻³ N A⁻¹	✓ 1	Accept 0.237–0.251 g A⁻¹. The gradient is the force per ampere found in (b)(i) — Allow ECF from (b)(i) for a consistent value.
F = BIL, so B = 2.39 × 10⁻³/0.040 = 0.060 T	✓ 1	Accept 0.058–0.062 T. Length must be in m. A single-point value from (b)(i) (7.2 × 10⁻³/(3.00 × 0.040) = 0.060 T) scores [1 max].
Part (c)(i)
Each Δm is a difference of two readings, ±0.02 g: at 0.50 A, 0.02/0.12 = 17 %; at 3.00 A, 0.02/0.73 = 2.7 %	✓ 1	Allow ECF from (b)(i). Accept 8 % and 1.4 % (one reading) for this mark only if the comparison is correct.
Reverse the current at each value so that the reading falls instead of rises: the difference between the two readings is 2Δm with the same ±0.02 g, halving the percentage uncertainty	✓ 1	Accept: take several readings at each current with the current switched on and off and average the differences. Do not accept "use a more precise balance".

Answers: (b)(i) 7.2 × 10⁻³ N · (b)(ii) 0.060 T · (c)(i) 17 % and 2.7 % (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — Newton's third law of motion; forces as interactions between bodies; D.3 — the magnetic force on a current-carrying conductor in a magnetic field as given by F = BIL sin θ; Tools 3 — zero (systematic) error, uncertainty of a difference, gradient of a graph; improving a procedure Command term: Determine

29A-1B-04

Circular motion and orbits·A.2 Forces and momentum

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

Astronomers track five large moons of Uranus over several weeks. For each moon the orbital radius r is found from its greatest angular distance from the planet and the known distance of Uranus from the Earth, and the orbital period T is found by timing repeated positions. The orbits are assumed to be circular, with the centripetal force provided by the gravitational force of Uranus, mass M.

The graph shows T² against r³.

moon	r / 10⁸ m	T / days	r³ / 10²⁴ m³	T² / 10¹⁰ s²
Miranda	1.299	1.415	2.19	1.49
Ariel	1.909	2.521	6.96	4.74
Umbriel	2.660	4.146	18.8	12.8
Titania	4.363	8.710	83.1	56.6
Oberon	5.835	13.471	199	135

Graph drawn to scale

(a)

(i)

Show that T² = (4π²/GM)r³.

(2)

(b)

(i)

Determine the gradient of the graph. Give its unit.

(2)

(ii)

Hence determine M.

(2)

(iii)

A smaller moon of Uranus has an orbital period of 0.762 days. Predict its orbital radius.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Gravitational force provides the centripetal force: GMm/r² = mω²r (or mv²/r)	✓ 1
With ω = 2π/T (or v = 2πr/T): GM/r³ = 4π²/T², which rearranges to the result	✓ 1
Part (b)(i)
Gradient from the line using the full range, e.g. 135 × 10¹⁰/199 × 10²⁴ = 6.82 × 10⁻¹⁵	✓ 1	Accept 6.6–7.0 × 10⁻¹⁵. Both powers of ten must be used.
Unit s² m⁻³	✓ 1
Part (b)(ii)
Gradient = 4π²/GM, so M = 4π²/(6.67 × 10⁻¹¹ × 6.82 × 10⁻¹⁵)	✓ 1	Allow ECF from (b)(i).
M = 8.7 × 10²⁵ kg	✓ 1	Accept 8.4–9.0 × 10²⁵ kg.
Part (b)(iii)
r³ = T²/gradient = (0.762 × 86 400)²/6.82 × 10⁻¹⁵, so r = 8.6 × 10⁷ m	✓ 1	Allow ECF from (b)(i). Accept 8.5–8.8 × 10⁷ m.

Answers: (b)(i) 6.82 × 10⁻¹⁵ s² m⁻³ · (b)(ii) 8.7 × 10²⁵ kg · (b)(iii) 8.6 × 10⁷ m (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — circular motion: the centripetal force and acceleration a = v²/r = ω²r; D.1 — Newton's universal law of gravitation F = Gm₁m₂/r²; Kepler's third law; Tools 3 — linearising a relationship, gradient with units Command term: Determine

30A-1B-06

Terminal speed with a speed-dependent resistive force·A.2 Forces and momentum

Paper 1BHard7 marks

Data-based question7 steps to full marksDetermine

A small cylindrical magnet falls down the centre of a long vertical copper tube. Brass (non-magnetic) discs are fixed to the magnet to change the total mass m. Two thin coils wound round the tube 0.400 m apart, well below the top, are connected to a data logger; each coil gives a voltage pulse as the magnet passes. The time t between the pulses has an uncertainty of ±0.05 s. The magnet falls at its terminal speed v between the coils.

The resistive force on the magnet is modelled as bv, where b is a constant.

m / g	10.0	15.0	20.0	25.0	30.0
t / s	3.27	2.18	1.63	1.29	1.09

Graph drawn to scale. The point for 30.0 g has not been plotted.

(a)

(i)

Calculate v for m = 30.0 g together with its absolute uncertainty.

(2)

(ii)

The lines of maximum and minimum gradient pass through the ends of the error bars of the points for 10.0 g and 30.0 g. Determine the gradient of the line of best fit and its absolute uncertainty.

(2)

(iii)

Hence determine b, with its absolute uncertainty.

(2)

(b)

(i)

The copper tube is replaced by an identical tube with a narrow slit cut along its whole length. Explain why the terminal speed of the magnet increases.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
v = 0.400/1.09 = 0.367 m s⁻¹	✓ 1
Fractional uncertainty = 0.05/1.09 = 4.6 %, so Δv = ±0.02 m s⁻¹	✓ 1	Absolute uncertainty to 1 s.f.; accept ±0.017.
Part (a)(ii)
Best-fit gradient = 12.3 × 10⁻³ m s⁻¹ g⁻¹ (12.3 m s⁻¹ kg⁻¹)	✓ 1	Accept 0.0120–0.0126 m s⁻¹ g⁻¹.
Max: (0.384 − 0.120)/20.0 = 13.2 × 10⁻³; min: (0.350 − 0.124)/20.0 = 11.3 × 10⁻³ m s⁻¹ g⁻¹; uncertainty = ½(max − min) = ±0.9 m s⁻¹ kg⁻¹	✓ 1	Allow ECF from the error bar of (a)(i). Accept ±0.8 to ±1.0 m s⁻¹ kg⁻¹ (±7 % to ±8 %).
Part (a)(iii)
At terminal speed mg = bv, so v = (g/b)m and b = g/gradient = 9.81/12.3 = 0.80 kg s⁻¹	✓ 1	Allow ECF from (a)(ii). The mass must be in kg: 8.0 × 10² scores 0 for this mark.
Same percentage uncertainty (≈ 7 %): b = 0.80 ± 0.06 kg s⁻¹	✓ 1	Allow ECF from (a)(ii). Accept ±0.05 to ±0.07.
Part (b)(i)
The changing magnetic flux induces emfs, and so eddy currents, that circulate round the tube; the slit breaks these current paths, so the induced currents are much smaller. By Lenz's law the induced currents oppose the motion, so the resistive force (b) is smaller and v must be larger before bv = mg	✓ 1	Must link the slit to smaller induced currents and a smaller opposing force.

Answers: (a)(i) 0.37 ± 0.02 m s⁻¹ · (a)(ii) (12.3 ± 0.9) m s⁻¹ kg⁻¹ · (a)(iii) b = 0.80 ± 0.06 kg s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingA.1 — the qualitative effect of fluid resistance, including terminal speed; A.2 — Newton’s first law applied to translational equilibrium; D.4 — the direction of induced emf is determined by Lenz’s law; Tool 3 — error bars, lines of maximum and minimum gradient, uncertainty in a gradient Command term: Determine

31A-1B-08

Impulse from a force–time graph·A.2 Forces and momentum

Paper 1BMedium6 marks

Data-based question6 steps to full marksDetermine

A trolley of mass 0.500 kg runs along a level track and collides head-on with a spring buffer mounted on a force sensor fixed to the end of the track. A motion sensor shows that the trolley moves at 0.80 m s⁻¹ towards the buffer before the collision and at 0.52 m s⁻¹ away from it afterwards; each speed is ±0.01 m s⁻¹. The force sensor was not set to zero before the run.

The graph shows the reading of the force sensor against time t. Friction is negligible.

Graph drawn to scale. Each small square represents 0.0125 s × 0.5 N.

(a)

(i)

State the zero offset of the force sensor.

(1)

(ii)

Estimate the impulse exerted on the trolley by the buffer.

(2)

(iii)

The uncertainty in your answer to (a)(ii) is about ±5 %. Calculate the change in momentum of the trolley and comment on whether it is consistent with your answer to (a)(ii).

(2)

(iv)

Determine the average force exerted on the trolley during the contact.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
1.5 N (the reading when no force acts, before and after the collision)	✓ 1	Accept 1.4–1.6 N.
Part (a)(ii)
Area between the curve and the 1.5 N line from 0.100 s to 0.220 s, by counting squares (about 100 small squares of 0.00625 N s)	✓ 1	Allow ECF from (a)(i). Award [0] for the whole area down to the time axis (≈ 0.82 N s).
Impulse ≈ 0.64 N s	✓ 1	Accept 0.60–0.68 N s. A triangle of base 0.12 s and height 8.4 N (0.50 N s) underestimates the area and scores [1 max].
Part (a)(iii)
Δp = 0.500 × (0.80 − (−0.52)) = 0.66 N s	✓ 1	Award [0] for 0.500 × (0.80 − 0.52) = 0.14 N s (direction ignored).
0.64 ± 0.03 N s includes 0.66 N s, so the impulse and the change in momentum are consistent (impulse = Δp)	✓ 1	Allow ECF from (a)(ii): the comment must agree with the candidate's own range; a candidate who included the offset (0.82 N s) should find them inconsistent.
Part (a)(iv)
F_av = impulse/contact time = 0.64/0.120 = 5.3 N	✓ 1	Allow ECF from (a)(ii). Accept 5.0–5.7 N. Contact time must be read from the graph (0.120 s).

Answers: (a)(i) 1.5 N · (a)(ii) 0.64 N s · (a)(iii) 0.66 N s; consistent · (a)(iv) 5.3 N (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — impulse J = FΔt as the area under a force–time graph and equal to the change of momentum; momentum as a vector; Tools 3 — zero (systematic) error read from a graph, area under a graph, comparing two values with their uncertainties Command term: Determine

32A-1B-11

Viscous drag, buoyancy and terminal speed·A.2 Forces and momentum

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

Steel ball bearings of radius r are released at the surface of a tall cylinder of glycerine. Their terminal speeds v are found by timing each ball between two marks low on the cylinder. The density of steel is 7800 kg m⁻³ and that of glycerine is 1260 kg m⁻³.

Balancing weight, buoyancy F_b = ρ_fVg and viscous drag F_d = 6πηrv gives v = 2r²(ρ_s − ρ_f)g/(9η), where η is the viscosity of glycerine. The graph shows v against r².

r / mm	0.50	1.00	1.50	2.00	2.50
r² / mm²	0.25	1.00	2.25	4.00	6.25
v / cm s⁻¹	0.26	1.01	2.30	4.05	6.40

Graph drawn to scale

(a)

(i)

Determine the gradient of the graph in SI units. Give its unit.

(2)

(ii)

Hence determine η.

(2)

(iii)

Another student analyses the same data but forgets the buoyancy force. Calculate the value of η that this student obtains.

(2)

(iv)

A ball of radius 4.00 mm is found to fall at 12.5 cm s⁻¹. Compare this with the value predicted by the graph and suggest a reason for the difference.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Gradient = 6.40/6.25 = 1.02 cm s⁻¹ mm⁻² = 1.02 × 10⁻²/10⁻⁶ = 1.02 × 10⁴	✓ 1	Accept 1.00–1.04 × 10⁴. Both conversions (cm → m and mm² → m²) needed.
Unit m⁻¹ s⁻¹	✓ 1
Part (a)(ii)
Gradient = 2(ρ_s − ρ_f)g/(9η), so η = 2 × 6540 × 9.81/(9 × 1.02 × 10⁴)	✓ 1	Allow ECF from (a)(i).
η = 1.4 Pa s	✓ 1	Accept 1.35–1.45 Pa s (or kg m⁻¹ s⁻¹).
Part (a)(iii)
Without buoyancy, (ρ_s − ρ_f) is replaced by ρ_s: η is too large by 7800/6540 = 1.19	✓ 1
η = 1.19 × 1.40 = 1.7 Pa s	✓ 1	Allow ECF from (a)(ii). Accept 1.6–1.7 Pa s.
Part (a)(iv)
Predicted v = 1.02 × 16.0 = 16 cm s⁻¹, so the measured speed is about 25 % lower; the drag is larger than 6πηrv because 6πηrv applies only to slow, steady flow and underestimates the drag at this larger speed and radius, or because the large ball is close to the walls of the cylinder (wall effect)	✓ 1	Allow ECF from (a)(i). Prediction and one physically justified reason both needed.

Answers: (a)(i) 1.02 × 10⁴ m⁻¹ s⁻¹ · (a)(ii) 1.4 Pa s · (a)(iii) 1.7 Pa s · (a)(iv) predicted 16 cm s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — viscous drag F_d = 6πηrv; buoyancy F_b = ρVg; terminal speed and translational equilibrium; Tools 3 — linearising a relationship, gradient with units, extrapolating beyond the data range Command term: Determine

33A-1B-12

Hooke's law·A.2 Forces and momentum

Paper 1BEasy7 marks

Data-based question7 steps to full marksDetermine

A student investigates a spring for a toy. Masses m are hung from the spring and its length L is measured with a metre rule; each length reading is ±1 mm. The graph shows the force F = mg against the extension x.

The spring is then used as a mass–spring oscillator. With a 0.300 kg mass attached, the student uses a stopwatch to time 10 vertical oscillations and obtains 5.0 s. The uncertainty in each stopwatch timing, mainly due to reaction time, is ±0.2 s.

m / g	0	100	200	300	400	500	600	700
L / mm	82	102	122	142	161	182	209	232

The student's results: force F = mg against extension x.

(a)

(i)

The student suggests that the force is proportional to the extension. Test this suggestion using at least three of the data points.

(2)

(ii)

Determine the spring constant k.

(2)

(b)

(i)

Deduce whether the timing is consistent with the period predicted for a mass–spring system.

(2)

(ii)

Suggest how, with the same stopwatch, the uncertainty in the period could be reduced, and state the new uncertainty.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Calculates F/x for at least three loads, e.g. 0.981/0.020 = 49, 2.94/0.060 = 49, 4.91/0.100 = 49 N m⁻¹	✓ 1	Two points only scores [1 max].
Constant (≈ 49 N m⁻¹) up to 500 g, so proportional there; for 600 g and 700 g the ratio falls to 46.3 and 45.8 N m⁻¹, so the limit of proportionality has been exceeded	✓ 1	Must identify the range over which the suggestion holds.
Part (a)(ii)
Gradient of the straight part only (0 to 500 g), e.g. 4.91 N/0.100 m	✓ 1	Allow ECF from (a)(i) for the range used. Including the 600 g and 700 g points scores 0 for this mark.
k = 49 N m⁻¹	✓ 1	Accept 47–51 N m⁻¹.
Part (b)(i)
Predicted T = 2π√(m/k) = 2π√(0.300/49) = 0.49 s	✓ 1	Allow ECF from (a)(ii).
Measured T = 5.0/10 = 0.50 ± 0.02 s, which includes 0.49 s: consistent	✓ 1	Allow ECF. The uncertainty must be divided by 10 as well.
Part (b)(ii)
Time a larger number of oscillations, e.g. 40 oscillations: the ±0.2 s is then shared over 40 periods, ΔT = ±0.005 s	✓ 1	Allow ECF from (b)(i). Accept any number of oscillations > 10 with the matching uncertainty. "Repeat the measurement" alone is not enough.

Answers: (a)(i) proportional up to 500 g · (a)(ii) 49 N m⁻¹ · (b)(i) predicted 0.49 s; measured 0.50 ± 0.02 s; consistent · (b)(ii) ±0.005 s for 40 oscillations (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — the elastic restoring force F_H = −kx (Hooke's law); C.1 — the time period of a mass–spring system T = 2π√(m/k); Tools 3 — testing a proportional relationship with several data points, reducing the uncertainty of a timing by measuring many cycles Command term: Determine

34A-1B-13

Static and dynamic friction·A.2 Forces and momentum

Paper 1BEasy7 marks

Data-based question7 steps to full marksDetermine

A wooden block, loaded with different masses, rests on a horizontal board and is pulled horizontally by a hand-held force sensor. For each normal force F_N the student records the peak reading just before the block starts to slip and the steady reading while it slides at constant speed. Both sets of results are plotted with lines of best fit, which meet the force axis at about 0.2 N instead of at the origin.

F_N / N	2.0	4.0	6.0	8.0	10.0	12.0
Peak force / N	1.10	2.02	2.88	3.75	4.67	5.54
Sliding force / N	0.80	1.47	1.98	2.57	3.22	3.82

Peak force before slipping (gold) and steady force while sliding (navy) against the normal force, with the lines of best fit (drawn to scale).

(a)

(i)

Determine the coefficient of static friction μ_s and the coefficient of dynamic friction μ_d.

(2)

(ii)

Another student calculates μ_s from the first row of data alone as 1.10/2.0 = 0.55. Explain, with reference to the intercept, why this value differs from your answer to (a)(i).

(2)

(b)

(i)

The board, with the unloaded block on it, is tilted slowly. Determine the angle to the horizontal at which the block starts to slide.

(2)

(ii)

State and explain whether the block accelerates once it starts to slide at the angle found in (b)(i).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
μ_s = gradient of the peak-force line, e.g. (5.54 − 1.10)/(12.0 − 2.0) = 0.44	✓ 1	Accept 0.42–0.46.
μ_d = gradient of the sliding-force line, e.g. (3.82 − 0.80)/(12.0 − 2.0) = 0.30	✓ 1	Accept 0.28–0.32. Single ratios such as 1.10/2.0 score 0.
Part (a)(ii)
The intercept shows a zero offset (systematic error): the sensor reads ≈ 0.2 N too high at every force	✓ 1
Correcting gives (1.10 − 0.2)/2.0 = 0.45, in agreement with the gradient (0.44); a gradient is unaffected by the offset, a single ratio is not	✓ 1	Allow ECF from (a)(i).
Part (b)(i)
On the point of slipping mg sin θ = μ_smg cos θ, so tan θ = μ_s	✓ 1
θ = tan⁻¹ 0.44 = 24°	✓ 1	Allow ECF from (a)(i). Accept 23–25°.
Part (b)(ii)
Yes: the component of weight down the slope (mg sin θ = 0.44mg cos θ) exceeds the dynamic friction 0.30mg cos θ, so there is a resultant force down the slope	✓ 1	Allow ECF from (a)(i) and (b)(i): any μ_d < μ_s.

Answers: (a)(i) μ_s = 0.44; μ_d = 0.30 · (b)(i) 24° (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — the surface frictional force F_f ≤ μ_sF_N when the body is stationary and F_f = μ_dF_N when it slides; free-body diagrams on a slope; Tools 3 — gradient of a graph; zero (systematic) error from an intercept Command term: Determine

35A-1B-14

Momentum and variable mass·A.2 Forces and momentum

Paper 1BEasy7 marks

Data-based question7 steps to full marksDetermine

A horizontal jet of water leaves a circular nozzle at speed v and strikes a flat vertical plate mounted on a force sensor. The water spreads out sideways over the plate, losing all of its horizontal velocity. The speed is changed with a valve and found from the flow rate; each speed is ±0.1 m s⁻¹. The density of water ρ is 1000 kg m⁻³.

The rate of change of momentum of the water gives a force on the plate F = ρAv², where A is the cross-sectional area of the jet.

v / m s⁻¹	2.0	3.0	4.0	5.0	6.0	7.0
F / N	0.118	0.241	0.456	0.699	1.009	1.379

Graph drawn to scale. The point for v = 7.0 m s⁻¹ has not been plotted.

(a)

(i)

Calculate v² for v = 7.0 m s⁻¹ together with its absolute uncertainty.

(2)

(ii)

Determine the gradient of the graph of F against v², using the full range of the data. Give its unit.

(2)

(iii)

Hence determine the diameter of the nozzle.

(2)

(iv)

The plate is replaced by a cup that sends the water back towards the nozzle with almost the same speed. Estimate the force on the cup when v = 7.0 m s⁻¹.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
v² = 49 m² s⁻²	✓ 1
Fractional uncertainty = 2 × 0.1/7.0 = 2.9 %, so Δ(v²) = ±1 m² s⁻²	✓ 1	Absolute uncertainty to 1 s.f.; accept ±1.4. ±0.7 (not doubled) scores 0 for this mark.
Part (a)(ii)
Line through the origin and the point (49, 1.379) from (a)(i): gradient = 1.379/49 = 0.028	✓ 1	Allow ECF from (a)(i). Accept 0.027–0.029.
Unit kg m⁻¹ (or N s² m⁻²)	✓ 1
Part (a)(iii)
A = gradient/ρ = 0.028/1000 = 2.8 × 10⁻⁵ m²	✓ 1	Allow ECF from (a)(ii).
d = √(4A/π) = 6.0 × 10⁻³ m	✓ 1	Accept 5.9–6.1 mm. Using A = πd² (radius confused with diameter) scores 0 for this mark.
Part (a)(iv)
The velocity change of each kilogram of water is about 2v instead of v, so F ≈ 2 × 0.028 × 49 ≈ 2.8 N	✓ 1	Allow ECF from (a)(ii). Accept 2 × 1.379 = 2.8 N.

Answers: (a)(i) 49 ± 1 m² s⁻² · (a)(ii) 0.028 kg m⁻¹ · (a)(iii) 6.0 mm · (a)(iv) 2.8 N (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — Newton's second law in the form F = Δp/Δt for situations where mass is changing; Tools 3 — propagation of uncertainty for a power, linearising a relationship, gradient with units using the full data range Command term: Determine

36A-1B-25

Circular motion·A.2 Forces and momentum

Paper 1BMedium7 marks

Data-based question6 steps to full marksDetermine

A rubber bung of mass m is tied to a string that passes through a smooth vertical tube held in the hand. A mass M = 0.200 kg hangs from the lower end of the string, so the tension in the string is Mg. The bung is whirled in a horizontal circle of radius r, kept constant using a marker on the string. The time for 20 revolutions is measured with a stopwatch; each timing has an uncertainty of ±0.3 s.

If the tension provides the centripetal force, T² = (4π²m/Mg)r, where T is the period. The graph shows T² against r.

r / m	time for 20 revolutions / s	T / s	T² / s²
0.300	8.0	0.400	0.160
0.400	9.2	0.460	0.212
0.500	10.1
0.600	11.1	0.555	0.308
0.700	12.0	0.600	0.360
0.800	12.8	0.640	0.410

Graph drawn to scale, with error bars for T² and the line of best fit.

(a)

(i)

Calculate T for r = 0.500 m, together with its absolute uncertainty.

(2)

(ii)

A student suggests that T² is proportional to r. Test this suggestion using at least three of the data points.

(2)

(b)

(i)

Determine m, using the gradient of the graph.

(2)

(ii)

The marker was fixed to the string so that the true radius of the circle was 0.015 m larger than each value of r recorded. State and explain the effect of this on the value of m found in (b)(i).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
T = 10.1/20 = 0.505 s	✓ 1
ΔT = 0.3/20 = ±0.015 s	✓ 1	Accept ±0.02 s. ±0.3 s (uncertainty not divided by 20) scores 0 for this mark.
Part (a)(ii)
Calculates T²/r for at least three points, e.g. 0.533, 0.510, 0.514 and 0.512 s² m⁻¹	✓ 1	Allow ECF from (a)(i) for the 0.500 m point (T² = 0.255 s²). A test using two points only scores [1 max] in total.
The values agree to within about ±2 %, less than the 5–7.5 % uncertainty in T² (twice the percentage uncertainty in T), so the data support the suggestion	✓ 1	The conclusion must refer to the uncertainty. Accept the uncertainty estimated at one point, e.g. 2 × 0.015/0.505 = 6 %.
Part (b)(i)
Gradient using the full range of the line, e.g. (0.410 − 0.160)/(0.800 − 0.300) = 0.50 s² m⁻¹	✓ 1	Accept 0.48–0.52 s² m⁻¹.
m = gradient × Mg/4π² = 0.50 × 0.200 × 9.81/(4π²) = 0.0248 kg	✓ 1	Accept 0.024–0.026 kg. Allow ECF from the candidate's gradient. Using M instead of Mg scores [1 max].
Part (b)(ii)
No effect: every point moves by the same 0.015 m along the r axis, so the line is shifted sideways but its gradient, from which m was found, is unchanged	✓ 1	Allow ECF from (b)(i) (a candidate who used a single point should state that m would be too large). Do not accept "no effect" without a reason.

Answers: (a)(i) 0.505 ± 0.015 s · (a)(ii) ratios constant within the uncertainty; supported · (b)(i) 0.0248 kg · (b)(ii) no effect on m (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — that bodies moving along a circular trajectory at a constant speed experience a centripetal acceleration as given by a = 4π²r/T²; that circular motion is caused by a centripetal force acting perpendicular to the velocity; tension; Tools 3 — uncertainty of a timing divided over many cycles; testing a proportional relationship with several data points; gradient; systematic error Command term: Determine

37A-1B-30

Buoyancy·A.2 Forces and momentum

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

A solid metal cylinder of diameter 3.60 cm hangs from a force sensor with its axis vertical. The cylinder is lowered in steps into a beaker of salt solution. At each step the depth of immersion d of the cylinder is read from a scale marked on its side, and the force-sensor reading F is recorded. The cylinder never touches the beaker.

The table and the graph show the results.

d / cm	0.0	1.0	2.0	3.0	4.0	5.0	6.0	7.0	8.0	9.0
F / N	1.62	1.50	1.39	1.27	1.16	1.05	0.93	0.93	0.92	0.93

Force-sensor reading F against depth of immersion d (drawn to scale).

(a)

(i)

Determine, using the sloping part of the graph, the gradient of the graph. Give its unit in SI units.

(2)

(ii)

Hence determine the density of the salt solution.

(2)

(iii)

The diameter of the cylinder was measured as 3.60 ± 0.02 cm. Calculate the percentage uncertainty in the density found in (a)(ii) that results from this measurement alone.

(1)

(b)

(i)

Use the graph to determine the height of the cylinder.

(1)

(ii)

Determine the density of the metal.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Gradient from the straight sloping section using a wide range, e.g. (1.05 − 1.62)/0.050 m or from a line of best fit through the points from 0 to 5–6 cm = −11.4 N m⁻¹	✓ 1	Accept −11.0 to −12.0 N m⁻¹; accept the magnitude. The depth must be converted to m for this mark; −0.115 N cm⁻¹ scores [1 max] if the unit is given correctly as N cm⁻¹.
Unit N m⁻¹ (or kg s⁻²)	✓ 1
Part (a)(ii)
Buoyancy force = ρAgd, so \|gradient\| = ρAg with A = π(0.0180)² = 1.02 × 10⁻³ m²	✓ 1	Allow ECF from (a)(i). Using the diameter as the radius (A = 4.07 × 10⁻³ m²) scores 0 for this mark.
ρ = 11.4/(1.02 × 10⁻³ × 9.81) = 1.14 × 10³ kg m⁻³	✓ 1	Accept 1.10–1.20 × 10³ kg m⁻³. Allow ECF from (a)(i).
Part (a)(iii)
Percentage uncertainty in A = 2 × (0.02/3.60) × 100 % = 1.1 %, and the density is inversely proportional to A, so ±1.1 %	✓ 1	Accept 1 %. 0.56 % (uncertainty not doubled for the square) scores 0.
Part (b)(i)
The reading stops changing when the cylinder is completely immersed: the sloping line and the horizontal line meet at d = 6.0 cm, so the height is 6.0 cm	✓ 1	Accept 5.8–6.2 cm.
Part (b)(ii)
Weight = 1.62 N; buoyancy force when fully immersed = 1.62 − 0.93 = 0.69 N = ρ_liquidVg, so ρ_metal = ρ_liquid × 1.62/0.69 = 2.7 × 10³ kg m⁻³	✓ 1	Allow ECF from (a)(ii) and (b)(i). ALT: ρ_metal = W/(gAh) = 1.62/(9.81 × 1.02 × 10⁻³ × 0.060) = 2.7 × 10³ kg m⁻³. Accept 2.6–2.8 × 10³ kg m⁻³.

Answers: (a)(i) −11.4 N m⁻¹ · (a)(ii) 1.14 × 10³ kg m⁻³ · (a)(iii) ±1.1 % · (b)(i) 6.0 cm · (b)(ii) 2.7 × 10³ kg m⁻³ (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — buoyancy F_b acting on a body due to the displacement of the fluid as given by F_b = ρVg where V is the volume of fluid displaced; that forces acting on a body can be represented in a free-body diagram; that free-body diagrams can be analysed to find the resultant force on a system (translational equilibrium); B.1 — density ρ = m/V; Tools 3 — gradient with units, intersection of two lines, propagation of uncertainty for a power Command term: Determine

38A-1B-31

Explosions and conservation of momentum·A.2 Forces and momentum

Paper 1BEasy7 marks

Data-based question6 steps to full marksDetermine

Glider A, of mass 0.250 kg, and glider B rest in contact on a level air track, with a compressed spring between them held by a thread. When the thread is burned, the gliders are pushed apart. Each glider carries a card of length 10.0 ± 0.1 cm, and each card passes through a light gate after the gliders separate: gate A is 0.20 m from the starting position and gate B is 0.40 m from it. Each gate time has an uncertainty of ±0.001 s.

The mass of B is changed by adding loads, and the same spring compression is used each time. The table shows the results.

mass of B / kg	time at gate A / s	time at gate B / s
0.250	0.144	0.146
0.350	0.134	0.189
0.450	0.127	0.232
0.550	0.123	0.274
0.650	0.120	0.316

(a)

(i)

Calculate the speed of B when its mass is 0.450 kg, together with its absolute uncertainty.

(2)

(ii)

It is suggested that the total momentum of the gliders after the release is zero. Test this suggestion, using at least three rows of the data.

(2)

(b)

(i)

Determine the energy released by the spring in the run with B of mass 0.450 kg.

(1)

(ii)

In every row the momentum calculated for B is slightly smaller than that calculated for A. Suggest a reason for this and an improvement to the experiment.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
v = 0.100/0.232 = 0.431 m s⁻¹	✓ 1
Percentage uncertainty = 1.0 % + (0.001/0.232) × 100 % = 1.4 %, so Δv = ±0.0062 ≈ ±0.006 m s⁻¹	✓ 1	Absolute uncertainty to 1 s.f.: accept ±0.006 m s⁻¹. Using only the timing uncertainty (±0.001 m s⁻¹) scores 0 for this mark.
Part (a)(ii)
Calculates the momentum of A and of B for at least three rows, e.g. 0.174 and 0.171 kg m s⁻¹; 0.197 and 0.194 kg m s⁻¹; 0.208 and 0.206 kg m s⁻¹	✓ 1	Two rows only scores [1 max] in total. Allow ECF from (a)(i) for the 0.450 kg row.
The momenta are opposite in direction and their magnitudes differ by at most about 1.5 %, which is less than the combined uncertainty of the two momenta (about ±3 %, each being about ±1.3–1.8 %), so the data support the suggestion	✓ 1	The conclusion must refer to the uncertainty and to the opposite directions (total zero).
Part (b)(i)
E_k = ½ × 0.250 × 0.787² + ½ × 0.450 × 0.431² = 0.119 J	✓ 1	Allow ECF from (a)(i). Accept 0.11–0.12 J.
Part (b)(ii)
Gate B is further from the start, so friction/air resistance acts on B for longer before its speed is measured, reducing its measured momentum	✓ 1	Accept: the track is tilted slightly so that B moves uphill. Do not accept "human error" or "the gates are inaccurate".
Place the two gates at equal (and small) distances from the starting position / check that the track is level	✓ 1	The improvement must address the reason given. Do not accept "repeat the readings".

Answers: (a)(i) 0.431 ± 0.006 m s⁻¹ · (a)(ii) supported: equal and opposite momenta within the uncertainty · (b)(i) 0.119 J (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — explosions; that linear momentum as given by p = mv remains constant unless the system is acted upon by a resultant external force; energy considerations in explosions; A.3 — kinetic energy E_k = ½mv²; Tools 3 — propagating uncertainties in a quotient, testing a hypothesis with several data points, systematic error and improvement Command term: Determine

39A-1B-32

Impulse in a bounce·A.2 Forces and momentum

Paper 1BHard7 marks

Data-based question7 steps to full marksDetermine

A rubber ball of mass 0.0580 kg is dropped from rest from a height h onto a force plate, which records the force on the ball during the impact. Software calculates the impulse J exerted on the ball by the plate from the area under the force–time graph; each value of J has an uncertainty of ±0.008 N s. Air resistance is negligible, and the contact is so short that the impulse of the weight during the contact can be ignored.

The ball leaves the plate with a speed k times the speed with which it arrives, where k is a constant. The table and the graph show J against √h; the line of best fit is drawn.

h / m	0.20	0.40	0.60	0.80	1.00	1.20
√h / √m	0.447	0.632	0.775	0.894	1.000	1.095
J / N s ± 0.008	0.200	0.285	0.350	0.399	0.452	0.493

Impulse J against √h, with error bars and the line of best fit (drawn to scale).

(a)

(i)

Show that J = m(1 + k)√(2g)√h, where m is the mass of the ball.

(2)

(b)

(i)

Determine the gradient of the graph and its absolute uncertainty, using lines of maximum and minimum gradient that pass through the error bars of the first and last points.

(2)

(ii)

Hence determine k, together with its absolute uncertainty.

(2)

(c)

(i)

Calculate the fraction of the kinetic energy of the ball that is transferred to other forms in each bounce.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Speed on arrival from energy conservation: ½mv² = mgh, so v = √(2gh)	✓ 1	Symbolic working required.
The velocity reverses, so the change in momentum is mkv − (−mv) = m(1 + k)v = J, giving the result	✓ 1	A change of momentum m(1 − k)v (direction ignored) scores 0 for this mark.
Part (b)(i)
Best-fit gradient = 0.452 N s m^−½	✓ 1	Accept 0.440–0.460 N s m^−½. Unit not required.
Max: (0.501 − 0.192)/(1.095 − 0.447) = 0.477; min: (0.485 − 0.208)/0.648 = 0.427; uncertainty = ½(max − min) = ±0.025	✓ 1	Accept ±0.020 to ±0.030. Award this mark for max and min gradients read from the candidate’s own lines.
Part (b)(ii)
k = gradient/(m√(2g)) − 1 = 0.452/(0.0580 × 4.429) − 1 = 0.76	✓ 1	Allow ECF from (a)(i) and (b)(i). Accept 0.71–0.79.
Δk = Δ(gradient)/(m√(2g)) = 0.025/0.2569 = ±0.10	✓ 1	Allow ECF from (b)(i). Accept ±0.08 to ±0.12. The hidden step: the 1 is exact, so the absolute (not the percentage) uncertainty carries over. Applying the percentage uncertainty of the gradient to k (±0.04) scores 0 for this mark.
Part (c)(i)
Fraction = 1 − k² = 1 − 0.76² = 0.42	✓ 1	Allow ECF from (b)(ii). Accept 0.38–0.50. 1 − k (≈ 0.24) scores 0.

Answers: (b)(i) 0.452 ± 0.025 N s m^−½ · (b)(ii) k = 0.76 ± 0.10 · (c)(i) 0.42 (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — that a resultant force applied to a system constitutes an impulse J as given by J = FΔt; that the applied external impulse equals the change in momentum of the system; that the area under a force–time graph is the impulse; A.3 — conservation of energy, E_k = ½mv² and ΔE_p = mgΔh; Tools 3 — linearising a relationship, gradient, error bars, lines of maximum and minimum gradient, propagation of an absolute uncertainty Command term: Determine

40A-2-01

Collisions in two dimensions·A.2 Forces and momentum

Paper 2Hard14 marks

Short answer & extended response8 steps to full marksDetermine

Puck A, of mass 0.20 kg, slides at 4.0 m s⁻¹ on a horizontal air table and collides with puck B, of mass 0.30 kg, which is at rest. After the collision A moves at 2.0 m s⁻¹ in a direction 60° to its original direction of motion. Friction is negligible on the air table.

(a)

(i)

Determine the speed of B immediately after the collision.

(3)

(ii)

Determine the direction of the velocity of B.

(1)

(b)

(i)

Deduce whether the collision is elastic.

(2)

(ii)

In a second experiment A collides elastically with a puck of equal mass at rest, and the two pucks move off in different directions. By considering momentum and kinetic energy, show that the two pucks then move at 90° to each other.

(2)

(c)

(i)

The pucks are in contact for 5.0 ms. Calculate the average force exerted by A on B.

(2)

(ii)

State the magnitude and direction of the average force exerted by B on A during the contact.

(1)

(d)

(i)

Puck B leaves the air table and slides onto a rough horizontal surface where the coefficient of dynamic friction is 0.15. Calculate the distance B slides before stopping.

(1)

(ii)

Puck B is made of aluminium of specific heat capacity 900 J kg⁻¹ K⁻¹. Half of the thermal energy generated by the friction is retained by the puck. Determine the rise in temperature of the puck.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Components of the momentum of B: parallel to the original motion 0.80 − 0.20 × 2.0 cos 60° = 0.60 kg m s⁻¹	✓ 1
perpendicular: 0.20 × 2.0 sin 60° = 0.35 kg m s⁻¹ (on the opposite side to A)	✓ 1
v_B = √(0.60² + 0.35²)/0.30 = 2.3 m s⁻¹	✓ 1	Accept 2.3 m s⁻¹. A cosine-rule solution scores full marks.
Part (a)(ii)
tan φ = 0.35/0.60, φ = 30° to the original direction of A, on the opposite side of that line from A	✓ 1	Allow ECF from (a)(i) components. Both the angle and the side are needed.
Part (b)(i)
Kinetic energy before = ½ × 0.20 × 4.0² = 1.6 J; after = ½ × 0.20 × 2.0² + ½ × 0.30 × 2.31² = 1.2 J	✓ 1	Allow ECF from (a)(i).
Kinetic energy is not conserved, so the collision is inelastic	✓ 1	The conclusion must follow from the candidate's own values.
Part (b)(ii)
Momentum (equal masses): u = v_A + v_B as vectors, so the three velocities form a closed triangle	✓ 1	No numbers are required.
Kinetic energy: u² = v_A² + v_B², which is Pythagoras' theorem for that triangle, so the angle between v_A and v_B is 90°	✓ 1	Award the second mark only with a reason linking Pythagoras (or a zero cross-term 2v_A·v_B = 0) to the right angle.
Part (c)(i)
F = Δp/Δt = 0.69/(5.0 × 10⁻³)	✓ 1	Allow ECF from (a)(i): Δp = 0.30v_B.
F = 1.4 × 10² N	✓ 1	Accept 1.4 × 10² N.
Part (c)(ii)
1.4 × 10² N, directed opposite to the velocity of B after the collision (Newton's third law)	✓ 1	Allow ECF from (c)(i). Magnitude and direction both needed.
Part (d)(i)
s = v_B²/(2μg) = 2.31²/(2 × 0.15 × 9.81) = 1.8 m	✓ 1	Allow ECF from (a)(i). Accept 1.7–1.9 m.
Part (d)(ii)
Thermal energy generated = kinetic energy of B = ½ × 0.30 × 2.31² = 0.80 J; puck receives 0.40 J	✓ 1	Allow ECF from (a)(i) or (b)(i).
ΔT = 0.40/(0.30 × 900) = 1.5 × 10⁻³ K	✓ 1	Accept 1.5 × 10⁻³ K.

Answers: (a)(i) 2.3 m s⁻¹ · (a)(ii) 30° on the other side · (b)(i) 1.2 J after; inelastic · (c)(i) 1.4 × 10² N · (c)(ii) 1.4 × 10² N, opposite · (d)(i) 1.8 m · (d)(ii) 1.5 × 10⁻³ K (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — linear momentum remains constant unless the system is acted upon by a resultant external force; elastic and inelastic collisions of two bodies; collisions in two dimensions; Newton's second law in the form F = Δp/Δt; Newton's third law; dynamic friction F_f = μ_dF_N; B.1 — specific heat capacity, Q = mcΔT Command term: Determine

41A-2-07

Momentum and variable mass·A.2 Forces and momentum

Paper 2Hard12 marks

Short answer & extended response8 steps to full marksDetermine

An ion thruster on a spacecraft accelerates singly charged xenon ions from rest through a potential difference of 1.20 kV and ejects them as a beam. Xenon is used at a rate of 5.0 mg s⁻¹. The mass of a xenon ion is 131 u. The total mass of the spacecraft is 1200 kg.

(a)

(i)

Show that the speed of the ejected ions is about 4 × 10⁴ m s⁻¹.

(2)

(ii)

Calculate the thrust on the spacecraft.

(1)

(b)

(i)

Determine the current carried by the ion beam.

(2)

(ii)

The beam power P is the kinetic energy given to the ions per second. Show that the ratio of thrust to beam power is F/P = 2/v, where v is the ion speed.

(2)

(iii)

Hence explain why raising the accelerating potential difference, at the same beam power, reduces the thrust.

(1)

(c)

(i)

The thruster runs continuously for 90 days. Determine the change in speed of the spacecraft.

(3)

(ii)

Justify the assumption, made in (c)(i), that the mass of the spacecraft is constant.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
qV = ½mv², with m = 131 × 1.661 × 10⁻²⁷ = 2.18 × 10⁻²⁵ kg	✓ 1
v = √(2 × 1.60 × 10⁻¹⁹ × 1200/2.18 × 10⁻²⁵) = 4.20 × 10⁴ m s⁻¹	✓ 1	Must see 4.20 × 10⁴ or full substitution.
Part (a)(ii)
F = (Δm/Δt)v = 5.0 × 10⁻⁶ × 4.20 × 10⁴ = 0.21 N	✓ 1	Allow ECF from (a)(i). Accept 0.20–0.21 N.
Part (b)(i)
Ions per second = 5.0 × 10⁻⁶/2.18 × 10⁻²⁵ = 2.30 × 10¹⁹ s⁻¹	✓ 1
I = 2.30 × 10¹⁹ × 1.60 × 10⁻¹⁹ = 3.7 A	✓ 1	Accept 3.6–3.7 A.
Part (b)(ii)
F = (Δm/Δt)v and P = ½(Δm/Δt)v²	✓ 1	Symbolic working required.
Dividing: F/P = v/(½v²) = 2/v	✓ 1
Part (b)(iii)
A larger p.d. gives a larger ion speed (v ∝ √V), and F = 2P/v, so the thrust falls (as 1/√V)	✓ 1	Allow ECF from (b)(ii) and (a)(i) relation.
Part (c)(i)
a = F/m = 0.210/1200 = 1.75 × 10⁻⁴ m s⁻²	✓ 1	Allow ECF from (a)(ii).
Time = 90 × 86 400 = 7.78 × 10⁶ s	✓ 1
Δv = 1.75 × 10⁻⁴ × 7.78 × 10⁶ = 1.4 × 10³ m s⁻¹	✓ 1	Accept 1.3–1.4 × 10³ m s⁻¹.
Part (c)(ii)
Xenon used = 5.0 × 10⁻⁶ × 7.78 × 10⁶ = 39 kg, only about 3 % of 1200 kg	✓ 1	Allow ECF from (c)(i) time.

Answers: (a)(ii) 0.21 N · (b)(i) 3.7 A · (c)(i) 1.4 × 10³ m s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — Newton's second law in the form F = Δp/Δt for a changing mass; the conservation of linear momentum; impulse; A.3 — kinetic energy and power; D.2 — the work done on a charge moving through a potential difference, W = qΔV; B.5 — electric current as the rate of flow of charge Command term: Determine

42A-2-09

Elastic and inelastic collisions·A.2 Forces and momentum

Paper 2Medium14 marks

Short answer & extended response8 steps to full marksDetermine

In a railway yard, wagon A of mass 1.2 × 10⁴ kg moving at 1.50 m s⁻¹ runs into wagon B of mass 8.0 × 10³ kg, which is at rest. The buffers of the wagons are springs with hydraulic dampers. The graph shows how the force exerted by A on B varies with time during the contact. Friction from the track is negligible.

Force F exerted by wagon A on wagon B against time t (drawn to scale).

(a)

(i)

Use the graph to estimate the impulse given to wagon B.

(2)

(ii)

Determine the velocities of A and B immediately after the contact, and deduce whether the wagons couple together.

(3)

(b)

(i)

Calculate the kinetic energy transferred to other forms in the collision.

(2)

(ii)

Show that, when a moving wagon of mass m_A couples to a stationary wagon of mass m_B, the fraction of the kinetic energy transferred to other forms is m_B/(m_A + m_B). Confirm that your answer to (b)(i) agrees.

(2)

(iii)

80 % of this energy is transferred to the 4.0 kg of oil in the hydraulic dampers, which has specific heat capacity 1.9 × 10³ J kg⁻¹ K⁻¹. Determine the rise in temperature of the oil.

(2)

(c)

(i)

Determine the maximum deceleration of wagon A during the contact.

(2)

(ii)

Suggest why the buffers are designed to make the contact last as long as possible.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Impulse = area under the F–t graph, found by counting squares or by an equivalent shape	✓ 1
Impulse ≈ 7.2 × 10³ N s	✓ 1	Accept 6.8–7.6 × 10³ N s. A triangle of base 0.40 s and height 28 kN (5.7 × 10³ N s) scores [1 max].
Part (a)(ii)
v_B = 7.2 × 10³/8.0 × 10³ = 0.90 m s⁻¹	✓ 1	Allow ECF from (a)(i).
A receives an equal and opposite impulse: v_A = (1.2 × 10⁴ × 1.50 − 7.2 × 10³)/1.2 × 10⁴ = 0.90 m s⁻¹	✓ 1	Newton's third law or momentum conservation must be used.
Both move at the same velocity after the contact, so they move off together (coupled): a perfectly inelastic collision	✓ 1	Allow ECF from (a)(i): the deduction must follow from the candidate's two velocities.
Part (b)(i)
Before: ½ × 1.2 × 10⁴ × 1.50² = 13500 J; after: ½ × 2.0 × 10⁴ × 0.90² = 8100 J	✓ 1	Allow ECF from (a)(ii).
Transferred = 5400 J	✓ 1	Accept 5.4 × 10³ J.
Part (b)(ii)
Common velocity m_Au/(m_A + m_B); final E_k = m_A²u²/2(m_A + m_B), so final/initial = m_A/(m_A + m_B) and the fraction lost is m_B/(m_A + m_B)	✓ 1	Symbolic working required.
8.0/20 = 0.40 and 5400/13500 = 0.40 ✓	✓ 1	Allow ECF from (b)(i).
Part (b)(iii)
Energy to oil = 0.80 × 5400 = 4320 J	✓ 1	Allow ECF from (b)(i).
ΔT = 4320/(4.0 × 1.9 × 10³) = 0.57 K	✓ 1	Accept 0.56–0.58 K.
Part (c)(i)
Maximum force read from the graph ≈ 28 kN, and the force on A is equal and opposite	✓ 1	Accept 27–29 kN.
a = 28 × 10³/1.2 × 10⁴ = 2.3 m s⁻²	✓ 1	Accept 2.2–2.5 m s⁻².
Part (c)(ii)
For the same impulse (change of momentum), a longer contact time gives a smaller (peak) force on the wagons and their loads	✓ 1	Reference to the same impulse/momentum change needed.

Answers: (a)(i) 7.2 × 10³ N s · (a)(ii) 0.90 m s⁻¹ each; coupled · (b)(i) 5400 J · (b)(iii) 0.57 K · (c)(i) 2.3 m s⁻² (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — the impulse J = FΔt and its representation as the area under a force–time graph; the conservation of linear momentum; elastic and inelastic collisions of two bodies; energy considerations in collisions; Newton's third law; B.1 — specific heat capacity Command term: Determine

43A-2-10

Circular motion·A.2 Forces and momentum

Paper 2Hard13 marks

Short answer & extended response9 steps to full marksDetermine

An astronaut-training centrifuge has a horizontal arm that rotates about a vertical axis. The astronaut sits in a gondola whose centre is 8.0 m from the axis. The gondola swings freely on a pivot, so that its floor is always perpendicular to the total force exerted on the astronaut by the seat. In a test, the magnitude of this force must be 6.0 times the astronaut's weight.

(a)

(i)

Show that the angular speed of the arm needed is about 2.7 rad s⁻¹.

(2)

(ii)

Calculate the rotation rate of the arm in revolutions per minute.

(1)

(iii)

Determine the angle between the floor of the gondola and the vertical.

(2)

(b)

(i)

The astronaut lies with her head 1.8 m closer to the axis than her feet. Determine the difference between the centripetal accelerations of her feet and her head.

(2)

(ii)

Astronauts orbiting the Earth in a space station feel weightless, although the gravitational field strength there is about 90 % of its surface value. Explain this.

(2)

(c)

(i)

The moment of inertia of the arm, gondola and astronaut about the axis is 7.2 × 10⁴ kg m². The arm is brought from rest to the angular speed in (a)(i) in 30 s with uniform angular acceleration. Calculate the torque required.

(1)

(ii)

Calculate the rotational kinetic energy of the arm, gondola and astronaut at the end of the 30 s, and hence the average power delivered to them during the spin-up.

(2)

(iii)

Calculate the number of revolutions made by the arm during the 30 s.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The seat force has a vertical component mg and a horizontal component mω²r, so (ω²r)² + g² = (6.0g)²: ω²r = √35 g = 58.0 m s⁻²	✓ 1	Using ω²r = 6.0g (vertical support ignored) gives 2.71 rad s⁻¹: [0] for this mark.
ω = √(58.0/8.0) = 2.693 rad s⁻¹	✓ 1	Must see 2.69 or better.
Part (a)(ii)
T = 2π/2.693 = 2.33 s, so rate = 60/2.33 = 25.7 rev min⁻¹	✓ 1	Allow ECF from (a)(i). Unit rev min⁻¹ required.
Part (a)(iii)
The total seat force points towards the axis at an angle φ above the horizontal, with tan φ = g/ω²r = 9.81/58.0	✓ 1	Allow ECF from (a)(i).
The floor is perpendicular to this force, so it makes φ = 9.6° with the vertical	✓ 1	Accept 9–10°.
Part (b)(i)
Δa = ω²Δr = 2.693² × 1.8	✓ 1	Allow ECF from (a)(i).
Δa = 13.1 m s⁻² (about 1.3g)	✓ 1	Accept 12.9–13.3 m s⁻².
Part (b)(ii)
The gravitational force on the astronaut and the station provides the centripetal force: both are in free fall with the same acceleration (equal to the local g)	✓ 1	Do not accept "there is no gravity in space".
So the floor exerts no (normal) contact force on the astronaut; the sensation of weight is the contact force, which is zero	✓ 1
Part (c)(i)
τ = Iω/t = 7.2 × 10⁴ × 2.693/30 = 6.5 × 10³ N m	✓ 1	Allow ECF from (a)(i). Accept 6.4–6.5 × 10³ N m. Friction is ignored.
Part (c)(ii)
E_k = ½Iω² = ½ × 7.2 × 10⁴ × 2.693² = 2.61 × 10⁵ J	✓ 1	Allow ECF from (a)(i).
Average power = 2.61 × 10⁵/30 = 8.7 × 10³ W	✓ 1	Allow ECF from the energy. Friction is ignored.
Part (c)(iii)
Δθ = ½(ω_i + ω_f)t = ½ × 2.693 × 30 = 40.4 rad, i.e. 40.4/2π = 6.4 revolutions	✓ 1	Allow ECF from (a)(i). Accept 6.4–6.5 revolutions.

Answers: (a)(ii) 25.7 rev min⁻¹ · (a)(iii) 9.6° · (b)(i) 13.1 m s⁻² · (c)(i) 6.5 × 10³ N m · (c)(ii) 2.61 × 10⁵ J; 8.7 × 10³ W · (c)(iii) 6.4 revolutions (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — that bodies moving along a circular trajectory at a constant speed experience a centripetal acceleration as given by a = ω²r; that free-body diagrams can be analysed to find the resultant force; A.4 — Newton’s second law for rotation τ = Iα; the equations of motion for uniform angular acceleration; the kinetic energy of rotational motion; A.3 — power as the rate of energy transfer; D.1 — Newton’s universal law of gravitation (the gravitational force as the centripetal force in orbit) Command term: Determine

44A-2-20

Static and dynamic friction·A.2 Forces and momentum

Paper 2Hard14 marks

Short answer & extended response9 steps to full marksDetermine

A skier of total mass 72 kg is pulled up a straight snow slope inclined at 12° to the horizontal by a tow rope that makes an angle of 20° with the slope, as shown. The skier moves up the slope at a constant speed of 2.5 m s⁻¹. The coefficient of dynamic friction between the skis and the snow is 0.080 and the coefficient of static friction is 0.10. Air resistance is negligible.

Not to scale. The rope makes 20° with the surface of the slope; the dashed line is parallel to the slope.

(a)

(i)

Draw a labelled free-body diagram for the skier.

(2)

(ii)

Determine the tension in the rope.

(3)

(iii)

Half of the thermal energy generated by friction melts snow at 0 °C under the skis. The specific latent heat of fusion of ice is 3.34 × 10⁵ J kg⁻¹. Determine the mass of snow melted per second.

(2)

(b)

(i)

The skier lets go of the rope while moving at 3.0 m s⁻¹. Show that the deceleration of the skier is about 2.8 m s⁻².

(2)

(ii)

Calculate the distance the skier travels up the slope before stopping.

(1)

(c)

(i)

Deduce whether the skier remains at rest after stopping.

(2)

(ii)

Determine the speed of the skier after sliding 5.0 m back down the slope.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Weight vertically downwards and normal force perpendicular to the slope	✓ 1
Tension along the rope at 20° above the slope, and friction parallel to the slope pointing down it	✓ 1	All four forces correct for 2 marks; three correct for 1.
Part (a)(ii)
Perpendicular to the slope: F_N = mg cos 12° − T sin 20°	✓ 1	The hidden step: the rope lifts the skier and reduces F_N.
Parallel: T cos 20° = mg sin 12° + 0.080F_N	✓ 1
T = mg(sin 12° + 0.080 cos 12°)/(cos 20° + 0.080 sin 20°) = 209 N	✓ 1	Accept 2.1 × 10² N. 202 N (rope taken parallel to the slope) scores [2 max].
Part (a)(iii)
F_N = 72 × 9.81 × cos 12° − 209 × sin 20° = 619 N; friction = 0.080 × 619 = 49.6 N	✓ 1	Allow ECF from (a)(ii).
Rate of heating = 49.6 × 2.5 = 124 W; mass melted per second = 0.5 × 124/3.34 × 10⁵ = 1.9 × 10⁻⁴ kg s⁻¹	✓ 1	Accept 1.8–2.0 × 10⁻⁴ kg s⁻¹.
Part (b)(i)
Now F_N = mg cos 12°, and both the weight component and dynamic friction act down the slope	✓ 1
a = 9.81 × (sin 12° + 0.080 cos 12°) = 2.81 m s⁻²	✓ 1	Must see 2.81 or full substitution.
Part (b)(ii)
s = 3.0²/(2 × 2.81) = 1.6 m	✓ 1	Allow ECF from (b)(i).
Part (c)(i)
To remain at rest, static friction must balance mg sin 12°, needing μ ≥ tan 12° = 0.21	✓ 1
0.10 < 0.21, so the skier slides back down the slope	✓ 1	Or compares 147 N needed with 69 N available.
Part (c)(ii)
Friction now acts up the slope: a = 9.81 × (sin 12° − 0.080 cos 12°) = 1.27 m s⁻²	✓ 1	Using the same acceleration as in (b)(i) scores 0.
v = √(2 × 1.27 × 5.0) = 3.6 m s⁻¹	✓ 1	Accept 3.5–3.6 m s⁻¹.

Answers: (a)(ii) 209 N · (a)(iii) 1.9 × 10⁻⁴ kg s⁻¹ · (b)(ii) 1.6 m · (c)(i) slides back · (c)(ii) 3.6 m s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — free-body diagrams; resultant force in two dimensions; normal force, tension and surface frictional force F_f ≤ μ_sF_N and F_f = μ_dF_N; A.1 — the equations of motion for uniform acceleration; A.3 — power P = Fv; B.1 — specific latent heat Command term: Determine

45A-2-21

Vertical circular motion·A.2 Forces and momentum

Paper 2Medium14 marks

Short answer & extended response9 steps to full marksDetermine

A rope swing hangs from a branch above a river. The rope is 8.0 m long. A person of mass 60 kg holds the end of the rope and steps off a bank, starting from rest with the rope taut at 40° to the vertical. Treat the person as a point mass; air resistance and the mass of the rope are negligible.

(a)

(i)

Show that the speed of the person at the lowest point of the swing is about 6.1 m s⁻¹.

(2)

(ii)

Calculate the angular velocity of the person at the lowest point.

(1)

(iii)

The speed is momentarily constant at the lowest point. Explain why the tension in the rope is nevertheless greater than the weight of the person.

(2)

(iv)

Calculate the tension in the rope at the lowest point.

(2)

(b)

(i)

The rope can safely support a tension of 900 N. Determine the largest angle to the vertical from which the person can safely start.

(3)

(c)

(i)

Starting from 40°, the person lets go at the lowest point, which is 1.8 m above the water. Calculate the horizontal distance travelled before reaching the water.

(2)

(d)

(i)

Calculate the period of the swing if the person instead started from a very small angle.

(1)

(ii)

Explain why the swing starting at 40° is not simple harmonic motion.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Height lost = 8.0 × (1 − cos 40°) = 1.87 m	✓ 1
v = √(2 × 9.81 × 1.87) = 6.06 m s⁻¹	✓ 1	Must see 6.06 or full substitution.
Part (a)(ii)
ω = v/r = 6.06/8.0 = 0.76 rad s⁻¹	✓ 1	Allow ECF from (a)(i).
Part (a)(iii)
The person moves on a circle, so there is a centripetal acceleration v²/r directed upwards towards the branch	✓ 1
This needs a resultant upward force, so T − mg = mv²/r > 0	✓ 1	Do not accept "centrifugal force".
Part (a)(iv)
T = 60 × 9.81 + 60 × 6.06²/8.0	✓ 1	Allow ECF from (a)(i).
T = 864 N	✓ 1	Accept 860–870 N.
Part (b)(i)
From a starting angle θ: v² = 2gL(1 − cos θ), so T = mg(3 − 2 cos θ)	✓ 1	Or: 900 − 589 = 311 N = mv²/r, so v² = 41.5 m² s⁻².
900 = 589 × (3 − 2 cos θ), cos θ = 0.735	✓ 1
θ = 43°	✓ 1	Accept 42–43°.
Part (c)(i)
Fall time = √(2 × 1.8/9.81) = 0.61 s (initial velocity horizontal)	✓ 1
Distance = 6.06 × 0.606 = 3.7 m	✓ 1	Allow ECF from (a)(i).
Part (d)(i)
T = 2π√(L/g) = 2π√(8.0/9.81) = 5.67 s	✓ 1
Part (d)(ii)
The restoring force mg sin θ is not proportional to the angular displacement θ at large angles (sin θ ≈ θ only for small θ)	✓ 1	Reference to the defining condition of SHM needed.

Answers: (a)(ii) 0.76 rad s⁻¹ · (a)(iv) 864 N · (b)(i) 43° · (c)(i) 3.7 m · (d)(i) 5.67 s (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — circular motion and centripetal acceleration a = v²/r = ω²r; non-uniform circular motion in a vertical plane; A.3 — conservation of energy; A.1 — horizontal projectile motion; C.1 — the conditions for simple harmonic motion; the period of a simple pendulum Command term: Determine

46A-2-22

Momentum and variable mass·A.2 Forces and momentum

Paper 2Hard13 marks

Short answer & extended response8 steps to full marksExplain

At a grain terminal, grain falls vertically from a hopper into an open railway wagon at a steady rate of 40 kg s⁻¹. While it is being filled, the wagon is pulled along a straight, horizontal track at a constant speed of 1.5 m s⁻¹ by a cable from an electric winch. Friction on the wagon is negligible.

(a)

(i)

Explain why a horizontal force is needed to keep the wagon moving at constant speed.

(2)

(ii)

Calculate the magnitude of this force.

(1)

(iii)

Explain why this force cannot be found using F = ma.

(1)

(b)

(i)

Show that the power delivered by the cable is twice the rate at which the kinetic energy of the wagon and its contents increases, whatever the values of the speed and the filling rate.

(2)

(ii)

Account for the difference between the two rates in (b)(i).

(2)

(iii)

The grain falls through a vertical height of 2.5 m from the hopper to the floor of the wagon. Determine the total rate at which energy is dissipated as the grain lands in the wagon.

(2)

(c)

(i)

The empty wagon has a mass of 2000 kg. When it contains 1200 kg of grain the cable is released, but grain continues to fall into the wagon for a further 10 s. Determine the speed of the wagon at the end of this time.

(2)

(ii)

Explain why the horizontal momentum is conserved in (c)(i), even though grain is still being added.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Grain arrives with no horizontal velocity and is given a horizontal velocity of 1.5 m s⁻¹, so the horizontal momentum of the wagon and contents increases	✓ 1
A force equal to the rate of change of momentum, F = Δp/Δt, is needed; without it the wagon would slow down	✓ 1
Part (a)(ii)
F = (Δm/Δt)v = 40 × 1.5 = 60 N	✓ 1
Part (a)(iii)
The acceleration is zero yet a force is needed: the mass is changing, so F = Δp/Δt must be used	✓ 1
Part (b)(i)
P = Fv = (Δm/Δt)v²	✓ 1	Symbolic working required.
Rate of increase of E_k = ½(Δm/Δt)v² (speed constant, mass increasing), so P is twice this	✓ 1
Part (b)(ii)
The other half (here 45 W of the 90 W) is dissipated: each portion of grain lands on a moving floor and slides relative to it before moving with it, an inelastic collision	✓ 1	Allow ECF from (a)(ii) for the values.
That energy becomes internal (thermal) energy of the grain and wagon, and sound	✓ 1
Part (b)(iii)
The vertical motion of the grain is stopped on landing: kinetic energy arriving each second = (Δm/Δt)gh = 40 × 9.81 × 2.5 = 981 W, all dissipated	✓ 1	Air resistance on the falling grain is neglected.
Total = 981 + 45 = 1.0 × 10³ W	✓ 1	Allow ECF from (b)(ii) for the 45 W. 981 W alone scores [1 max].
Part (c)(i)
Horizontal momentum conserved: 3200 × 1.5 = (3200 + 40 × 10)v	✓ 1
v = 4800/3600 = 1.33 m s⁻¹	✓ 1	Accept 1.3 m s⁻¹.
Part (c)(ii)
All the external forces (weights, normal force from the track) are vertical and the falling grain brings no horizontal momentum, so there is no resultant external horizontal force	✓ 1	Allow ECF from (c)(i) reasoning.

Answers: (a)(ii) 60 N · (b)(iii) 1.0 × 10³ W · (c)(i) 1.33 m s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — Newton's second law in the form F = ma assumes mass is constant, whereas F = Δp/Δt allows for situations where mass is changing; the conservation of linear momentum; energy considerations in inelastic collisions; A.3 — power P = Fv; the principle of the conservation of energy; gravitational potential energy Command term: Explain

47A-2-23

Explosions·A.2 Forces and momentum

Paper 2Medium12 marks

Short answer & extended response8 steps to full marksDetermine

Two gliders, A of mass 0.40 kg and B of mass 0.60 kg, rest on a horizontal air track with a light spring of spring constant 800 N m⁻¹ compressed by 2.5 cm between them. A thread holds them together. When the thread is burned, the spring pushes the gliders apart and falls away. Friction is negligible.

(a)

(i)

Calculate the elastic potential energy stored in the spring.

(1)

(b)

(i)

State the total momentum of the gliders after the release and explain your answer.

(2)

(ii)

Show that the ratio of the kinetic energies of the gliders is E_k_A/E_k_B = m_B/m_A.

(2)

(iii)

Determine the speed of each glider, assuming that all the stored energy becomes kinetic energy of the gliders.

(2)

(c)

(i)

An americium-241 nucleus at rest decays by alpha emission to a neptunium-237 nucleus. The kinetic energy of the alpha particle is 5.49 MeV. Determine the kinetic energy of the neptunium nucleus. Give your answer in keV.

(2)

(ii)

The atomic masses are: americium-241, 241.056829 u; neptunium-237, 237.048173 u; helium-4, 4.002603 u. Show that the energy released in the decay is about 5.64 MeV.

(2)

(iii)

Suggest why the total kinetic energy of the alpha particle and the neptunium nucleus is less than the energy released.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
E_p = ½ × 800 × 0.025² = 0.25 J	✓ 1
Part (b)(i)
Zero	✓ 1
The total momentum was zero before release and there is no resultant external horizontal force on the system, so the total momentum is conserved	✓ 1	Both ideas needed.
Part (b)(ii)
The momenta are equal in magnitude, p, and E_k = p²/2m	✓ 1	Symbolic working required.
E_k_A/E_k_B = (p²/2m_A)/(p²/2m_B) = m_B/m_A	✓ 1	Allow ECF from (b)(i) (equal and opposite momenta).
Part (b)(iii)
E_k_A = 0.25 × 0.60/1.00 = 0.15 J and E_k_B = 0.10 J	✓ 1	Allow ECF from (a)(i) and (b)(ii).
v_A = √(2 × 0.15/0.40) = 0.87 m s⁻¹; v_B = √(2 × 0.10/0.60) = 0.58 m s⁻¹, in opposite directions	✓ 1
Part (c)(i)
The decay is an "explosion" from rest, so by (b)(ii): E_k_Np/E_k_α = m_α/m_Np ≈ 4/237	✓ 1	Allow ECF from (b)(ii).
E_k_Np = 5.49 × 4/237 = 0.0927 MeV = 93 keV	✓ 1	Answer in keV required for this mark. Accept 92–93 keV.
Part (c)(ii)
Mass defect = 241.056829 − 237.048173 − 4.002603 = 0.006053 u	✓ 1
Energy released = 0.006053 × 931.5 = 5.638 MeV	✓ 1	Must see at least 3 s.f. or the full substitution.
Part (c)(iii)
5.49 + 0.093 = 5.58 MeV, about 0.06 MeV less than 5.64 MeV: the neptunium nucleus is left in an excited state and emits a gamma-ray photon of about 0.06 MeV (60 keV) as it falls to its ground state	✓ 1	Allow ECF from (c)(i) and (c)(ii). Accept a reference to discrete nuclear energy levels; "energy lost as heat" scores [0].

Answers: (a)(i) 0.25 J · (b)(i) 0 · (b)(iii) v_A = 0.87 m s⁻¹, v_B = 0.58 m s⁻¹ · (c)(i) 93 keV (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — explosions; that linear momentum as given by p = mv remains constant unless the system is acted upon by a resultant external force; energy considerations in elastic collisions, inelastic collisions, and explosions; A.3 — the elastic potential energy as given by E_H = ½k(Δx)²; E_k = p²/2m; E.3 — the mass-energy equivalence as given by E = mc² in nuclear reactions; that the spectrum of alpha and gamma radiations provides evidence for discrete nuclear energy levels Command term: Determine

48A-2-34

Newton's second law·A.2 Forces and momentum

Paper 2Easy11 marks

Short answer & extended response7 steps to full marksCalculate

A goods lift and its load have a total mass of 1200 kg. The graph shows how the upward velocity v of the lift varies with time t during one journey.

Upward velocity of the lift against time.

(a)

(i)

Calculate the acceleration of the lift during the first 2.5 s.

(1)

(ii)

Determine the total distance travelled by the lift.

(2)

(b)

(i)

Calculate the tension in the lift cable during the first 2.5 s.

(2)

(ii)

A crate of mass 50 kg rests on the floor of the lift. Calculate the normal force on the crate during the last 2.0 s of the journey.

(2)

(c)

(i)

The cable is wound in by an electric motor. Determine the maximum rate at which the cable does work on the lift during the journey.

(2)

(ii)

During the whole journey the motor takes 3.2 × 10⁵ J from the supply. Determine the efficiency of the lift system.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
a = 2.0/2.5 = 0.80 m s⁻²	✓ 1
Part (a)(ii)
Area under the graph: ½ × 2.5 × 2.0 + 8.0 × 2.0 + ½ × 2.0 × 2.0	✓ 1
= 20.5 m	✓ 1	Accept 20–21 m.
Part (b)(i)
T − mg = ma, so T = 1200 × (9.81 + 0.80)	✓ 1	Allow ECF from (a)(i).
T = 1.27 × 10⁴ N	✓ 1
Part (b)(ii)
Acceleration = −2.0/2.0 = −1.0 m s⁻² (upwards positive)	✓ 1
F_N = m(g + a) = 50 × (9.81 − 1.0) = 4.4 × 10² N	✓ 1	490 N (acceleration ignored) scores 0.
Part (c)(i)
The maximum is at t = 2.5 s, the end of the acceleration, where both the tension (1.27 × 10⁴ N) and the speed (2.0 m s⁻¹) have their largest values	✓ 1	Allow ECF from (b)(i).
P = Fv = 1.27 × 10⁴ × 2.0 = 2.5 × 10⁴ W	✓ 1	2.4 × 10⁴ W (constant-speed value, tension = weight) scores [1 max].
Part (c)(ii)
Useful energy = gain in gravitational potential energy = 1200 × 9.81 × 20.5 = 2.41 × 10⁵ J (the lift starts and ends at rest)	✓ 1	Allow ECF from (a)(ii).
Efficiency = 2.41 × 10⁵/3.2 × 10⁵ = 0.75 (75 %)	✓ 1	Accept 74–76 %.

Answers: (a)(i) 0.80 m s⁻² · (a)(ii) 20.5 m · (b)(i) 1.27 × 10⁴ N · (b)(ii) 4.4 × 10² N · (c)(i) 2.5 × 10⁴ W · (c)(ii) 75 % (the remaining parts are explanations — see the table above)

Syllabus understandingA.1 — velocity–time graphs: the gradient as acceleration and the area as displacement; A.2 — Newton's second law F = ma; the normal force; A.3 — power P = Fv; efficiency Command term: Calculate

49A-2-35

Newton's second law·A.2 Forces and momentum

Paper 2Easy14 marks

Short answer & extended response9 steps to full marksDetermine

A delivery drone of mass 4.0 kg carries a parcel of mass 1.5 kg that hangs below it on a light vertical cable. The drone takes off vertically from rest. Air resistance is negligible throughout.

(a)

(i)

At take-off the total upward thrust on the drone from its rotors is 66.0 N. Calculate the initial acceleration of the drone and parcel.

(2)

(ii)

Determine the tension in the cable during this acceleration.

(2)

(b)

(i)

Shortly afterwards, the drone and parcel rise at a constant speed of 3.0 m s⁻¹. State the thrust on the drone now, with a reason.

(1)

(ii)

Calculate the rate at which the gravitational potential energy of the drone and parcel increases.

(1)

(iii)

When the parcel is 20.0 m above the ground and still rising at 3.0 m s⁻¹, the cable is released and the parcel falls freely. Determine the time taken for the parcel to reach the ground.

(2)

(iv)

The thrust on the drone does not change when the parcel is released. Determine the acceleration of the drone immediately after the release.

(2)

(c)

(i)

The battery has an emf of 22.2 V and negligible internal resistance, and is rated at 10.0 A h. Show that the energy stored in the fully charged battery is about 8 × 10⁵ J.

(2)

(ii)

The energy density of the battery is 0.65 MJ kg⁻¹. Calculate the mass of the battery.

(1)

(iii)

When the drone hovers with the parcel, the battery delivers a power of 520 W. To protect the battery, only 80 % of its stored energy may be used. Determine the longest time for which the drone can hover.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Resultant force = 66.0 − 5.5 × 9.81 = 12.0 N	✓ 1	Award [0] for using the mass of the drone alone.
a = 12.0/5.5 = 2.2 m s⁻²	✓ 1	Accept 2.19 m s⁻².
Part (a)(ii)
Newton's second law for the parcel: T − mg = ma	✓ 1	Allow ECF from (a)(i).
T = 1.5 × (9.81 + 2.19) = 18 N	✓ 1	Accept 18.0 N. 14.7 N (weight of the parcel only) scores [0].
Part (b)(i)
54 N (= 5.5 × 9.81 N): the velocity is constant, so the resultant force is zero and the thrust equals the total weight (Newton's first law)	✓ 1	Accept 53.96 N. The value and the reason are both needed.
Part (b)(ii)
Rate = Fv = mgv = 54.0 × 3.0 = 1.6 × 10² W	✓ 1	Allow ECF from (b)(i). Accept 162 W.
Part (b)(iii)
The parcel starts with a velocity of 3.0 m s⁻¹ upwards: taking upwards as positive, −20.0 = 3.0t − ½ × 9.81 × t²	✓ 1	Award this mark for a correct equation (or for finding the extra rise 0.46 m and then the fall from 20.46 m from rest). An answer of 2.0 s (initial velocity ignored) scores [0].
t = 2.3 s	✓ 1	Accept 2.35 s.
Part (b)(iv)
Resultant force on the drone = 54.0 − 4.0 × 9.81 = 14.7 N upwards	✓ 1	Allow ECF from (b)(i).
a = 14.7/4.0 = 3.7 m s⁻² (upwards)	✓ 1	Accept 3.68 m s⁻². Using the take-off thrust of 66.0 N (6.7 m s⁻²) or the total mass 5.5 kg scores [1 max].
Part (c)(i)
Charge = 10.0 × 3600 = 3.6 × 10⁴ C	✓ 1
Energy = QV = 3.6 × 10⁴ × 22.2 = 7.99 × 10⁵ J	✓ 1	The answer must be given to at least 2 s.f., or the full substitution shown.
Part (c)(ii)
Mass = 7.99 × 10⁵/0.65 × 10⁶ = 1.2 kg	✓ 1	Allow ECF from (c)(i). Accept 1.23 kg.
Part (c)(iii)
Time = 0.80 × 7.99 × 10⁵/520 = 1.2 × 10³ s (about 20 minutes)	✓ 1	Allow ECF from (c)(i). Accept 1230 s. Omitting the 80 % gives 1.5 × 10³ s and scores [0].

Answers: (a)(i) 2.2 m s⁻² · (a)(ii) 18 N · (b)(i) 54 N · (b)(ii) 1.6 × 10² W · (b)(iii) 2.3 s · (b)(iv) 3.7 m s⁻² · (c)(i) 7.99 × 10⁵ J · (c)(ii) 1.2 kg · (c)(iii) 1.2 × 10³ s (the remaining parts are explanations — see the table above)

Syllabus understandingA.1 — the equations of motion for solving problems with uniformly accelerated motion; motion of a body falling freely with an initial velocity; A.2 — Newton's three laws of motion; tension; A.3 — that power developed is the rate of work done, P = Fv; energy density of the fuel sources; B.5 — electric current I = Δq/Δt Command term: Determine

50A-2-36

Circular motion·A.2 Forces and momentum

Paper 2Medium14 marks

Short answer & extended response9 steps to full marksDetermine

In a fairground swing ride, each seat hangs from a light chain of length 4.0 m attached to the rotating top of the ride 3.0 m from its vertical axis. When the top rotates at a steady rate, each chain makes an angle of 40° with the vertical and each seat moves in a horizontal circle, as shown. A seat and its rider have a total mass of 80 kg and may be treated as a point mass at the lower end of the chain.

Swing ride at steady rotation (not to scale).

(a)

(i)

Outline why a resultant force acts on the rider although the rider moves at constant speed.

(1)

(ii)

Show that the radius of the circle in which the rider moves is about 5.6 m.

(1)

(iii)

Determine the tension in the chain.

(2)

(iv)

Determine the speed of the rider.

(3)

(v)

Calculate the period of rotation of the ride.

(1)

(b)

(i)

Show that, for a chain of length L attached a distance d from the axis and rotating at angular speed ω, the angle θ satisfies g tan θ = ω²(d + L sin θ). Hence explain why the chains of empty seats hang at the same angle as those of occupied seats.

(2)

(ii)

Before the ride starts, the seats hang vertically at rest. Determine the energy that must be transferred to one seat and its rider to bring it to the steady motion described.

(2)

(c)

(i)

The rotation rate is increased so that the period of rotation becomes 4.5 s. Use the relation in (b)(i) to show that the chains then make an angle of about 50° with the vertical.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The direction of the velocity changes continuously, so the rider accelerates (towards the centre of the circle); by Newton's second law a resultant force is needed	✓ 1	Both the changing direction and the link to a resultant force are needed.
Part (a)(ii)
r = 3.0 + 4.0 sin 40° = 3.0 + 2.57 = 5.57 m	✓ 1	The substitution or an answer to at least 3 s.f. is needed.
Part (a)(iii)
Vertical equilibrium: T cos 40° = mg	✓ 1	Award [0] for T = mg or for using sin 40°.
T = 80 × 9.81/cos 40° = 1.0 × 10³ N	✓ 1	Accept 1020–1030 N.
Part (a)(iv)
The horizontal component of the tension provides the centripetal force: T sin 40° = mv²/r	✓ 1	Allow ECF from (a)(ii) and (a)(iii). Accept tan 40° = v²/rg.
v² = 5.57 × 9.81 × tan 40° = 45.9 m² s⁻²	✓ 1	Using r = 4.0 sin 40° (the chain alone) scores [1 max] overall.
v = 6.8 m s⁻¹	✓ 1	Accept 6.7–6.8 m s⁻¹.
Part (a)(v)
T = 2πr/v = 2π × 5.57/6.77 = 5.2 s	✓ 1	Allow ECF from (a)(iv). Accept 5.1–5.2 s.
Part (b)(i)
T cos θ = mg and T sin θ = mω²r with r = d + L sin θ; dividing gives g tan θ = ω²(d + L sin θ)	✓ 1	Both component equations and the radius are needed.
The mass cancels, so θ depends only on ω, d and L, which are the same for every seat	✓ 1	Award this mark only for a reason based on the mass cancelling.
Part (b)(ii)
Gain in height = 4.0(1 − cos 40°) = 0.94 m, so gain in gravitational potential energy = 80 × 9.81 × 0.94 = 7.3 × 10² J	✓ 1
Kinetic energy = ½ × 80 × 6.77² = 1.83 × 10³ J; total = 2.6 × 10³ J	✓ 1	Allow ECF from (a)(iv). Accept 2.5–2.6 kJ.
Part (c)(i)
ω = 2π/4.5 = 1.40 rad s⁻¹	✓ 1
With θ = 50°: g tan θ = 9.81 × tan 50° = 11.7 m s⁻² and ω²(d + L sin θ) = 1.95 × (3.0 + 4.0 sin 50°) = 1.95 × 6.06 = 11.8 m s⁻²; the two sides agree to about 1 %, so θ ≈ 50°	✓ 1	Accept a numerical or graphical solution giving 50.4°. Both sides must be evaluated (or the solution shown); a bare statement scores [0] for this mark.

Answers: (a)(ii) 5.57 m · (a)(iii) 1.0 × 10³ N · (a)(iv) 6.8 m s⁻¹ · (a)(v) 5.2 s · (b)(ii) 2.6 kJ (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — that bodies moving along a circular trajectory at a constant speed experience a centripetal acceleration as given by a = v²/r = ω²r; that circular motion is caused by a centripetal force acting perpendicular to the velocity; free-body diagrams; tension; A.3 — gravitational potential energy and kinetic energy Command term: Determine

51A-2-37

Terminal speed with a speed-dependent resistive force·A.2 Forces and momentum

Paper 2Hard14 marks

Short answer & extended response9 steps to full marksDetermine

A skydiver and equipment, of total mass 80 kg, fall vertically from rest. The air resistance on them is modelled as kv², where v is the speed and k is a constant. At t = 25 s the parachute opens, which greatly increases k. The graph shows how the velocity of the skydiver varies with time; it was calculated for air of constant density, equal to that at ground level.

Velocity of the skydiver against time (drawn to scale).

(a)

(i)

State the acceleration of the skydiver at t = 0, and explain your answer.

(1)

(ii)

Show that k is about 0.27 kg m⁻¹ before the parachute opens.

(1)

(iii)

Use the graph and your answer to (a)(ii) to determine the acceleration of the skydiver at t = 5.0 s.

(2)

(b)

(i)

Estimate the distance fallen by the skydiver in the first 25 s.

(2)

(ii)

Determine the energy transferred by air resistance to the surroundings during the first 25 s.

(3)

(c)

(i)

After the parachute has opened, the skydiver falls at a constant speed of 5.0 m s⁻¹. Deduce the ratio (value of k with the parachute open)/(value of k before it opens).

(1)

(ii)

Explain why the skydiver decelerates after the parachute opens, although the velocity is still downwards.

(2)

(d)

(i)

In fact the skydiver jumped from a height of 4.0 km, where the air pressure is 61.6 kPa and the temperature is 262 K. At ground level the pressure is 101 kPa and the temperature is 288 K. Assume that k is proportional to the density of the air, which behaves as an ideal gas. Determine the terminal speed of the skydiver at a height of 4.0 km, before the parachute opens.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
9.81 m s⁻² (downwards): the speed is zero, so there is no air resistance and the weight is the only force	✓ 1	Value and reason both needed.
Part (a)(ii)
At the terminal speed of 54 m s⁻¹ (from the graph) kv² = mg, so k = 80 × 9.81/54² = 0.269 kg m⁻¹	✓ 1	The read value of the terminal speed and an answer to at least 3 s.f. (or the full substitution) are needed.
Part (a)(iii)
Speed read from the graph at 5.0 s ≈ 39 m s⁻¹; resultant force = mg − kv² = 784.8 − 0.269 × 39² = 376 N	✓ 1	Accept 38–40 m s⁻¹. Allow ECF from (a)(ii).
a = 376/80 = 4.7 m s⁻²	✓ 1	Accept 4.4–5.0 m s⁻². A tangent drawn at 5.0 s giving a value in this range is also accepted. Equivalent: a = g(1 − v²/v_t²).
Part (b)(i)
Distance = area under the graph from 0 to 25 s, found by counting squares or by splitting into simple shapes	✓ 1
Distance = 1.1 × 10³ m	✓ 1	Accept 1080–1200 m. 54 × 25 = 1350 m (constant speed assumed) scores [1 max].
Part (b)(ii)
Loss of gravitational potential energy = 80 × 9.81 × 1140 = 8.9 × 10⁵ J	✓ 1	Allow ECF from (b)(i).
Kinetic energy at 25 s = ½ × 80 × 54² = 1.2 × 10⁵ J	✓ 1
Energy transferred = 8.9 × 10⁵ − 1.2 × 10⁵ = 7.8 × 10⁵ J	✓ 1	Accept 7.3–8.3 × 10⁵ J. Allow ECF from the candidate's values.
Part (c)(i)
In both cases kv² = mg, so the ratio is (54/5.0)² = 117 ≈ 1.2 × 10²	✓ 1	Allow ECF from (a)(ii) for the terminal speed. Accept 110–125. 54/5.0 = 11 scores [0].
Part (c)(ii)
When the parachute opens, k increases greatly while v is still about 54 m s⁻¹, so the air resistance becomes much greater than the weight	✓ 1	Vague answers such as "the parachute slows him down" score 0.
The resultant force (and so the acceleration) is upwards, opposite to the velocity; the speed falls until air resistance again equals the weight, at a lower terminal speed	✓ 1	Award this mark for a resultant force opposite to the velocity, linked to deceleration.
Part (d)(i)
For an ideal gas ρ = pM/RT ∝ p/T, so ρ_{4 km}/ρ_ground = (61.6/262)/(101/288) = 0.67	✓ 1	Award [0] for using temperatures in °C. Accept ρ ∝ n/V = p/RT.
v ∝ 1/√k, so v = 54/√0.67 = 66 m s⁻¹	✓ 1	Allow ECF from (a)(ii) for the terminal speed. Accept 65–67 m s⁻¹. 54/0.67 = 81 m s⁻¹ scores [1 max].

Answers: (a)(i) 9.81 m s⁻² · (a)(iii) 4.7 m s⁻² · (b)(i) 1.1 × 10³ m · (b)(ii) 7.8 × 10⁵ J · (c)(i) ≈ 1.2 × 10² · (d)(i) 66 m s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingA.1 — motion with uniform and non-uniform acceleration; the qualitative effect of fluid resistance, including terminal speed; A.2 — Newton’s three laws of motion; A.3 — the change in the total mechanical energy of a system interpreted as the work done on the system by a non-conservative force (guidance); B.3 — the equations governing the behaviour of ideal gases, PV = nRT Command term: Determine

52A-2-45

Connected bodies with friction·A.2 Forces and momentum

Paper 2Easy13 marks

Short answer & extended response9 steps to full marksDetermine

Block A, of mass 2.0 kg, rests on a horizontal table, 1.50 m from a pulley at the edge of the table. It is joined by a light inextensible string that passes over the light, frictionless pulley to block B, of mass 1.5 kg, which hangs 0.60 m above the floor, as shown. The coefficient of dynamic friction between A and the table is 0.25. The blocks are released from rest. Air resistance is negligible.

Blocks A and B joined by a string over the pulley.

(a)

(i)

Draw a labelled free-body diagram for block A while it is moving.

(2)

(ii)

Determine the acceleration of the blocks.

(3)

(iii)

Calculate the tension in the string.

(2)

(iv)

Explain why the tension in the string is less than the weight of B.

(1)

(b)

(i)

Calculate the speed of B just before it reaches the floor.

(1)

(ii)

B stops when it hits the floor and the string becomes slack. Determine the further distance that A slides along the table.

(2)

(iii)

Determine the total energy transferred to thermal energy by the friction on A, from the release until A stops.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Weight (mg or F_g) vertically down and normal force vertically up, drawn with equal lengths	✓ 1	Do not accept "gravity" for the weight.
Tension horizontally towards the pulley and friction horizontally away from the pulley, the tension arrow longer than the friction arrow	✓ 1	All four forces correct and labelled for [2]; three correct for [1]. The extra mark depends on the relative lengths of the horizontal arrows.
Part (a)(ii)
Friction on A = 0.25 × 2.0 × 9.81 = 4.91 N	✓ 1
Treating A and B together, the resultant force is the weight of B minus the friction: 1.5 × 9.81 − 4.91 = 9.81 N	✓ 1	Or write T − 4.91 = 2.0a and 14.7 − T = 1.5a and solve.
a = 9.81/3.5 = 2.80 m s⁻²	✓ 1	Using the mass of B alone (6.5 m s⁻²) or ignoring friction (4.2 m s⁻²) scores [1 max].
Part (a)(iii)
For B: mg − T = ma, so T = 1.5 × (9.81 − 2.80)	✓ 1	Allow ECF from (a)(ii). ALT: for A, T = 2.0 × 2.80 + 4.91.
T = 10.5 N	✓ 1	Accept 10.5–10.6 N.
Part (a)(iv)
B accelerates downwards, so the resultant force on it must be downwards: its weight is greater than the upward tension (Newton’s second law)	✓ 1	A reference to the downward acceleration (or resultant force) of B is needed.
Part (b)(i)
v = √(2 × 2.80 × 0.60) = 1.83 m s⁻¹	✓ 1	Allow ECF from (a)(ii). Accept 1.8 m s⁻¹.
Part (b)(ii)
Only friction now acts horizontally on A: deceleration = μg = 0.25 × 9.81 = 2.45 m s⁻²	✓ 1	Using the acceleration of (a)(ii) scores 0 for this mark.
s = 1.83²/(2 × 2.45) = 0.69 m	✓ 1	Allow ECF from (b)(i). Accept 0.68–0.69 m.
Part (b)(iii)
Work done against friction = friction × total distance = 4.91 × (0.60 + 0.69)	✓ 1	Allow ECF from (b)(ii). ALT: loss of gravitational potential energy of B minus the kinetic energy of B lost at the floor = 1.5 × 9.81 × 0.60 − ½ × 1.5 × 1.83² = 8.83 − 2.52.
= 6.3 J	✓ 1	Accept 6.3 J. Using only the first 0.60 m (2.9 J) scores [1 max].

Answers: (a)(ii) 2.80 m s⁻² · (a)(iii) 10.5 N · (b)(i) 1.83 m s⁻¹ · (b)(ii) 0.69 m · (b)(iii) 6.3 J (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — Newton’s three laws of motion; that forces acting on a body can be represented in a free-body diagram; that free-body diagrams can be analysed to find the resultant force on a system; tension; surface frictional force F_f acting in a direction parallel to the plane of contact between a body and a surface, on a stationary body as given by F_f ≤ μ_sF_N or a body in motion as given by F_f = μ_dF_N; A.1 — the equations of motion for uniformly accelerated motion; A.3 — the work done by a force; the change in total mechanical energy interpreted as work done by a non-conservative force Command term: Determine

53A-2-46

Buoyancy and drag on a balloon·A.2 Forces and momentum

Paper 2Medium12 marks

Short answer & extended response9 steps to full marksDetermine

The envelope of a hot-air balloon has a volume of 2800 m³. The outside air has a density of 1.20 kg m⁻³ and a temperature of 288 K. The air inside the envelope is at the same pressure as the outside air. The mass of the envelope, basket, burner, passengers and 40 kg of sandbags, not including the air inside the envelope, is 650 kg. Treat the air as an ideal gas.

(a)

(i)

The air inside the envelope is heated to 373 K. Show that the density of this air is about 0.93 kg m⁻³.

(2)

(ii)

Calculate the buoyancy force on the balloon.

(1)

(iii)

The balloon is held at rest on the ground with the air inside at 373 K, and is then released. Determine its initial acceleration.

(3)

(b)

(i)

Later, with the air inside kept at a lower, constant temperature, the balloon descends at a constant speed of 2.0 m s⁻¹. The pilot drops the 40 kg of sandbags. The balloon then rises at a constant speed of 2.0 m s⁻¹. Show that the air resistance on the balloon at 2.0 m s⁻¹ is about 200 N.

(2)

(ii)

The air resistance is proportional to the square of the speed. The pilot now drops a further 20 kg of equipment, with the temperature of the air inside unchanged. Determine the new constant speed at which the balloon rises.

(2)

(iii)

Explain, in terms of the motion of the molecules, why the hot air inside the envelope has a smaller density than the outside air, although its pressure is the same.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
At constant pressure, V ∝ T for a fixed mass of gas, so ρ ∝ 1/T (or ρ = pM/RT)	✓ 1	Use of Celsius temperatures scores 0.
ρ = 1.20 × 288/373 = 0.927 kg m⁻³	✓ 1	Must see 0.927 or the full substitution.
Part (a)(ii)
F_b = ρ_airVg = 1.20 × 2800 × 9.81 = 3.30 × 10⁴ N	✓ 1	Using the density of the hot air scores 0.
Part (a)(iii)
Mass of the hot air = 0.927 × 2800 = 2594 kg, so the total mass = 650 + 2594 = 3244 kg	✓ 1	Allow ECF from (a)(i). The hidden step: the hot air inside is part of the accelerating body.
Resultant upward force = 3.30 × 10⁴ − 3244 × 9.81 = 1135 N	✓ 1	Allow ECF from (a)(ii).
a = 1135/3244 = 0.35 m s⁻² (upwards)	✓ 1	Accept 0.32–0.36 m s⁻² (0.32 m s⁻² follows from 0.93 kg m⁻³). Dividing by 650 kg only (1.7 m s⁻²) scores [2 max]; leaving out the weight of the hot air scores [1 max].
Part (b)(i)
Descending: W = F_b + F_d; rising: W − m_sg + F_d = F_b, with the same F_b and the same magnitude of F_d but the drag reversed	✓ 1	Both equilibrium equations, with the drag reversing direction, are needed. Symbolic working.
Subtracting: 2F_d = m_sg, so F_d = ½ × 40 × 9.81 = 196 N	✓ 1	Must see 196 N or the full substitution.
Part (b)(ii)
Before the drop, the resultant of the buoyancy force and the weight is 196 N upwards (it balances the drag), so after dropping 20 kg it is 196 + 20 × 9.81 = 392 N	✓ 1	Allow ECF from (b)(i).
The drag must equal 392 N: v = 2.0 × √(392/196) = 2.8 m s⁻¹	✓ 1	Allow ECF from (b)(i). Accept 2.8 m s⁻¹. 4.0 m s⁻¹ (drag taken as proportional to the speed) scores [1 max].
Part (b)(iii)
The molecules of the hotter air have a greater average kinetic energy (speed), so each collision with a surface gives a greater change in momentum, and the collisions are more frequent	✓ 1	Do not accept "hot air rises" or "the molecules expand".
So fewer molecules per unit volume are needed to exert the same pressure: the number density, and so the density, is smaller	✓ 1	The link to fewer molecules per unit volume is needed.

Answers: (a)(ii) 3.30 × 10⁴ N · (a)(iii) 0.35 m s⁻² · (b)(ii) 2.8 m s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — buoyancy F_b acting on a body due to the displacement of the fluid as given by F_b = ρVg where V is the volume of fluid displaced; that forces acting on a body can be represented in a free-body diagram; that free-body diagrams can be analysed to find the resultant force on a system; Newton’s first and second laws; B.3 — the equations governing the behaviour of ideal gases (PV/T = constant); the pressure of an ideal gas in terms of the momentum change of its molecules; A.1 — the qualitative effect of fluid resistance Command term: Determine

54A-2-47

Impulse in two dimensions·A.2 Forces and momentum

Paper 2Medium11 marks

Short answer & extended response7 steps to full marksDetermine

A football of mass 0.43 kg is moving horizontally towards a player at 12.0 m s⁻¹. The player heads the ball so that it leaves her head at 15.0 m s⁻¹, back the way it came, at 60° above the horizontal. All the motion is in one vertical plane. The graph shows how the force exerted by the head on the ball varies with time t during the contact. The weight of the ball is negligible compared with this force.

Force F exerted on the ball by the head against time t (drawn to scale).

(a)

(i)

Determine the magnitude of the change in momentum of the ball.

(3)

(ii)

Determine the direction of the impulse exerted on the ball.

(2)

(b)

(i)

Use the graph to determine the contact time, and hence calculate the average force exerted on the ball.

(2)

(ii)

State the magnitude of the average force exerted by the ball on the head, and suggest why players are taught to tense their neck muscles when heading a ball.

(2)

(c)

(i)

Calculate the change in the kinetic energy of the ball, and explain how this change is possible.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Taking the original direction of motion as positive: horizontal change = 0.43 × (−15.0 cos 60° − 12.0) = −8.39 kg m s⁻¹	✓ 1	Using +15.0 cos 60° (direction ignored) scores 0 for this mark.
Vertical change = 0.43 × 15.0 sin 60° = 5.59 kg m s⁻¹	✓ 1
\|Δp\| = √(8.39² + 5.59²) = 10.1 kg m s⁻¹	✓ 1	ALT: cosine rule with the angle of 120° between the velocity vectors, Δv = √(15.0² + 12.0² + 2 × 15.0 × 12.0 cos 60°) = 23.4 m s⁻¹. Accept 10 N s. 1.3 N s (difference of the magnitudes) scores 0.
Part (a)(ii)
tan φ = 5.59/8.39, so φ = 34°	✓ 1	Allow ECF from (a)(i) components.
Above the horizontal, directed away from the player (opposite to the original velocity)	✓ 1	Both the angle and the sense are needed for this mark. Accept 146° to the original direction of motion.
Part (b)(i)
Contact time from the graph = 12 ms	✓ 1	Accept 11.5–12.5 ms.
F_av = Δp/Δt = 10.1/0.012 = 8.4 × 10² N	✓ 1	Allow ECF from (a)(i) and the contact time. Accept 8.1–8.8 × 10² N.
Part (b)(ii)
8.4 × 10² N (equal and opposite, Newton’s third law)	✓ 1	Allow ECF from (b)(i).
Tensing the neck makes the head move with the body, so the force acts on a much larger mass and the acceleration of the head (a = F/m) is much smaller	✓ 1	Do not accept "the force is smaller" or "the head absorbs the force".
Part (c)(i)
ΔE_k = ½ × 0.43 × (15.0² − 12.0²) = 17.4 J (an increase)	✓ 1
The head moves towards the ball during the contact, so the force on the ball acts through a displacement in its direction and does positive work; the energy comes from the player’s muscles	✓ 1	A stationary head could not increase the kinetic energy; the idea of work done by the moving head is needed.

Answers: (a)(i) 10.1 kg m s⁻¹ · (a)(ii) 34° above the horizontal, away from the player · (b)(i) 8.4 × 10² N · (b)(ii) 8.4 × 10² N · (c)(i) 17.4 J increase (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — that a resultant force applied to a system constitutes an impulse J as given by J = FΔt; that the applied external impulse equals the change in momentum of the system; that the area under a force–time graph is the impulse; Newton’s third law; that linear momentum as given by p = mv remains constant unless the system is acted upon by a resultant external force (momentum as a vector; two-dimensional situations, HL); A.3 — the work done by a force; E_k = ½mv² Command term: Determine

55A-2-48

Momentum and variable mass·A.2 Forces and momentum

Paper 2Hard14 marks

Short answer & extended response10 steps to full marksDetermine

A helicopter of mass 1.40 × 10³ kg hovers at rest above level ground. In a simple model, the rotor takes air that is at rest above it and gives it a uniform downward speed v as it passes through the circle swept out by the blades. The blades are 5.0 m long, and the density of the air is 1.20 kg m⁻³.

(a)

(i)

Explain, with reference to Newton’s laws of motion, how the rotor produces an upward force on the helicopter.

(2)

(ii)

Show that the upward force on the helicopter is ρAv², where A is the area swept out by the blades and ρ is the density of the air.

(2)

(iii)

Calculate v for the hovering helicopter.

(1)

(b)

(i)

Explain why the engine must supply energy continuously although the helicopter does not move.

(1)

(ii)

Show that the least power that the rotor must deliver is P = √((mg)³/(4ρA)).

(2)

(iii)

Calculate this power.

(1)

(c)

(i)

The helicopter now hovers at a high altitude where the air pressure is 75.0 kPa and the temperature is 270 K. The density of 1.20 kg m⁻³ applies at 101 kPa and 288 K. Determine the least power needed to hover at this altitude.

(3)

(ii)

Back at the original altitude, the helicopter picks up a load that increases its total mass by 30 %. Calculate the factor by which the least power needed to hover increases.

(1)

(iii)

The actual power delivered by the engine is much greater than the value calculated in (b)(iii). Suggest one reason, other than friction in the engine and gearbox, for the difference.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The rotor gives downward momentum to the air, so (Newton’s second law, F = Δp/Δt) it exerts a downward force on the air	✓ 1	Do not accept "the blades push down on the air" without a reference to the change in momentum of the air.
By Newton’s third law the air exerts an equal upward force on the rotor (and so on the helicopter)	✓ 1	Both laws must be named or clearly applied.
Part (a)(ii)
In a time Δt a column of air of length vΔt passes through the area, so the mass of air moved per second is Δm/Δt = ρAv	✓ 1	Symbolic working required.
Each kilogram gains momentum v (from rest), so F = (Δm/Δt)v = ρAv²	✓ 1	F = ma for the helicopter alone scores 0.
Part (a)(iii)
A = π × 5.0² = 78.5 m²; ρAv² = mg gives v = √(1.40 × 10³ × 9.81/(1.20 × 78.5)) = 12.1 m s⁻¹	✓ 1	Allow ECF from (a)(ii). Accept 12 m s⁻¹. Using the diameter as the radius gives 6.0 m s⁻¹: [0].
Part (b)(i)
No work is done on the helicopter, but kinetic energy is given to the air that is pushed down each second (and this air carries the energy away)	✓ 1	Do not accept "to overcome gravity" or "energy is needed to stay up".
Part (b)(ii)
P = ½(Δm/Δt)v² = ½ρAv³ (= ½mgv)	✓ 1	Symbolic working required. Allow ECF from (a)(ii).
Substituting v = √(mg/(ρA)): P = ½mg√(mg/(ρA)) = √((mg)³/(4ρA))	✓ 1
Part (b)(iii)
P = ½ × 1.40 × 10³ × 9.81 × 12.1 = 8.3 × 10⁴ W	✓ 1	Allow ECF from (a)(iii) or (b)(ii). Accept 8.2–8.4 × 10⁴ W.
Part (c)(i)
For an ideal gas ρ ∝ p/T (ρ = pM/RT)	✓ 1	Use of Celsius temperatures scores 0 for the next two marks.
ρ = 1.20 × (75.0/101) × (288/270) = 0.950 kg m⁻³	✓ 1
P ∝ 1/√ρ, so P = 8.3 × 10⁴ × √(1.20/0.950) = 9.3 × 10⁴ W	✓ 1	Allow ECF from (b)(ii) and (b)(iii). Accept 9.2–9.4 × 10⁴ W. P ∝ 1/ρ (1.0 × 10⁵ W) scores [2 max].
Part (c)(ii)
P ∝ m^3/2, so the factor is 1.30^1.5 = 1.48	✓ 1	Allow ECF from (b)(ii). 1.30 or 1.69 scores 0.
Part (c)(iii)
Any one of: the air speed is not uniform across the rotor area (for the same momentum flux, a non-uniform speed carries more kinetic energy); the air is also set rotating (swirl); air resistance (drag) on the moving blades; air is also pushed out sideways (turbulence)	✓ 1	Accept other valid physics. "Energy is lost as heat" alone scores 0.

Answers: (a)(iii) 12.1 m s⁻¹ · (b)(iii) 8.3 × 10⁴ W · (c)(i) 9.3 × 10⁴ W · (c)(ii) 1.48 (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — that Newton’s second law in the form F = ma assumes mass is constant whereas F = Δp/Δt allows for situations where mass is changing; Newton’s three laws of motion; forces as interactions between bodies; A.3 — power as the rate of energy transfer; E_k = ½mv²; B.3 — the equations governing the behaviour of ideal gases, PV = nRT Command term: Determine

56A-2-49

Circular motion·A.2 Forces and momentum

Paper 2Medium11 marks

Short answer & extended response8 steps to full marksDetermine

The drum of a washing machine has a radius of 0.25 m and rotates about a horizontal axis. Its curved wall has many small holes. The graph shows how the rotation frequency f of the drum varies with time t at the start of a spin cycle. A wet sock of mass 0.12 kg is pressed against the wall of the drum and moves with it.

Rotation frequency f of the drum against time t (drawn to scale).

(a)

(i)

Calculate the angular velocity of the drum when it rotates at its greatest rate.

(1)

(ii)

Determine the centripetal acceleration of the sock at this rate, as a multiple of g.

(2)

(iii)

Calculate the magnitude of the centripetal force on the sock at this rate.

(1)

(iv)

Determine the difference between the normal force exerted by the wall on the sock when the sock is at the bottom of the drum and when it is at the top, and comment on its size.

(2)

(b)

(i)

Explain why water leaves the clothes through the holes in the wall of the drum.

(2)

(c)

(i)

Use the graph to determine the angular acceleration of the drum between t = 30 s and t = 50 s, and the number of revolutions made by the drum in the first 50 s.

(2)

(ii)

Explain why, between t = 30 s and t = 50 s, the resultant force on the sock is not directed towards the centre of the drum.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
ω = 2πf = 2π × 20 = 126 rad s⁻¹	✓ 1	Accept 126 rad s⁻¹. 20 rad s⁻¹ scores 0.
Part (a)(ii)
a = ω²r = 125.7² × 0.25 = 3.9 × 10³ m s⁻²	✓ 1	Allow ECF from (a)(i).
= 3.9 × 10³/9.81 ≈ 402 × g	✓ 1	Allow ECF from (a)(i). Accept 400g.
Part (a)(iii)
F = mω²r = 0.12 × 3.9 × 10³ = 474 N	✓ 1	Allow ECF from (a)(ii). Accept 470–475 N.
Part (a)(iv)
At the bottom F_N − mg = mω²r; at the top F_N + mg = mω²r, so the difference is 2mg = 2 × 0.12 × 9.81 = 2.35 N	✓ 1	Both equations, with the weight towards the centre at the top and away from it at the bottom, are needed.
This is only about 0.5 % of the centripetal force, so the weight of the sock is negligible at this rate	✓ 1	Allow ECF from (a)(iii). The comment must refer to the size compared with the centripetal force.
Part (b)(i)
A centripetal force is needed to keep anything moving in a circle; the clothes and water are pushed towards the centre by the wall	✓ 1	Do not accept "centrifugal force throws the water out".
At a hole there is no wall to exert a force on the water, so the water continues in a straight line along the tangent (Newton’s first law) and leaves through the hole	✓ 1	The idea of no centripetal force at the hole and tangential motion is needed.
Part (c)(i)
α = 2π × (20 − 8)/20 = 3.77 rad s⁻²	✓ 1	Accept 3.8 rad s⁻². 0.60 rad s⁻² (revolutions per second squared) scores 0.
Revolutions = area under the graph = ½ × 10 × 8 + 20 × 8 + ½ × (8 + 20) × 20 = 480	✓ 1	Accept 470–490.
Part (c)(ii)
The speed of the sock is increasing, so it has a tangential acceleration as well as the centripetal acceleration; the resultant force therefore has a component along the direction of motion	✓ 1	Award only for a link between the increasing speed and a tangential component of force or acceleration.

Answers: (a)(i) 126 rad s⁻¹ · (a)(ii) 402g · (a)(iii) 474 N · (a)(iv) 2.35 N · (c)(i) 3.77 rad s⁻²; 480 revolutions (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — that the motion along a circular trajectory can be described in terms of the angular velocity ω which is related to the linear speed v by the equation as given by v = 2πr/T = ωr; that bodies moving along a circular trajectory at a constant speed experience an acceleration that is directed radially towards the centre of the circle — known as a centripetal acceleration — as given by a = v²/r = ω²r = 4π²r/T²; that circular motion is caused by a centripetal force acting perpendicular to the velocity; that a centripetal force causes the body to change direction even if its magnitude of velocity may remain constant (guidance: non-uniform circular motion; forces in a vertical circle at the top and the bottom); Newton’s first law; A.4 — angular velocity and angular acceleration; the area under a frequency–time graph as the number of revolutions Command term: Determine

57A-2-67

Forces, work and power in an aircraft launch·A.2 Forces and momentum

Paper 2Hard20 marks

Short answer & extended response13 steps to full marksDetermine

An aircraft of mass 2.5 × 10⁴ kg is launched from the deck of a ship by an electromagnetic catapult. The aircraft starts from rest and leaves the catapult after 90 m. Throughout the launch its engines provide a constant forward thrust of 1.2 × 10⁵ N. The graph shows how the force exerted by the catapult on the aircraft varies with the displacement of the aircraft. Ignore air resistance and friction.

Force F exerted by the catapult on the aircraft against displacement s along the deck (drawn to scale).

(a)

(i)

Determine the work done by the catapult on the aircraft.

(2)

(ii)

Show that the speed of the aircraft as it leaves the catapult is about 73 m s⁻¹.

(2)

(iii)

Estimate the time taken for the launch.

(1)

(iv)

State the assumption made in your estimate and suggest whether the actual time is longer or shorter.

(1)

(b)

(i)

An unpowered drone of mass one fifth of the aircraft's mass is launched by the same catapult with the same force–displacement graph. Assuming uniform acceleration, deduce the ratios of the drone's launch speed and launch time to those of an unpowered aircraft of the original mass.

(2)

(c)

(i)

The catapult force is produced by 160 straight conductors, each of length 0.60 m, that carry equal currents perpendicular to a magnetic field of flux density 1.2 T. Calculate the current in each conductor while the force is 7.0 × 10⁵ N.

(2)

(ii)

Use F = BIL to express the tesla in fundamental SI units.

(1)

(iii)

Determine the maximum power delivered by the catapult to the aircraft.

(3)

(d)

In a second launch the ship steams at a constant velocity of 14 m s⁻¹, in the direction of the launch, through still air. The catapult and the engines act as before.

(i)

Explain why the aircraft still leaves the catapult at the speed found in (a)(ii) relative to the deck.

(1)

(ii)

The aircraft needs a speed of at least 82 m s⁻¹ relative to the air to fly. Deduce whether it can fly as it leaves the deck.

(2)

(iii)

Determine the minimum speed of the ship for which the aircraft could fly as it leaves the deck.

(1)

(e)

(i)

On landing, an aircraft of mass 2.0 × 10⁴ kg touches down at 65 m s⁻¹ relative to the deck and is stopped by an arresting cable in 2.2 s. Assuming a uniform deceleration, determine the average force exerted by the cable on the aircraft and the distance the aircraft travels along the deck while stopping.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Work = area under the graph = 7.0 × 10⁵ × 70 + ½ × 7.0 × 10⁵ × 20	✓ 1
= 5.6 × 10⁷ J	✓ 1	Accept 5.5–5.7 × 10⁷ J.
Part (a)(ii)
Total work = 5.6 × 10⁷ + 1.2 × 10⁵ × 90 = 6.68 × 10⁷ J	✓ 1	Allow ECF from (a)(i). The work done by the engines must be included: 66.9 m s⁻¹ (engines ignored) scores [1 max].
v = √(2 × 6.68 × 10⁷/2.5 × 10⁴) = 73.1 m s⁻¹	✓ 1	Must see 73.1 or full substitution.
Part (a)(iii)
t ≈ 2s/v = 2 × 90/73 = 2.5 s	✓ 1	Allow ECF from (a)(ii). Accept 2.4–2.6 s.
Part (a)(iv)
Assumption: the acceleration is uniform (average speed = half the final speed); the force is in fact smaller over the last 20 m, so the aircraft spends more of the stroke at high speed and the actual time is slightly shorter	✓ 1	Allow ECF from (a)(iii). Assumption required; accept the direction with a valid reason.
Part (b)(i)
The same work W is done: ½mv² = W so v ∝ 1/√m: speed ratio √5 (≈ 2.2)	✓ 1	Number-free reasoning required.
t = 2s/v with the same s, so t ∝ √m: time ratio 1/√5 (≈ 0.45)	✓ 1
Part (c)(i)
F = 160BIL, so I = 7.0 × 10⁵/(160 × 1.2 × 0.60)	✓ 1
I = 6.1 × 10³ A	✓ 1	Accept 6.1 × 10³ A.
Part (c)(ii)
T = N A⁻¹ m⁻¹ = kg m s⁻² A⁻¹ m⁻¹ = kg s⁻² A⁻¹	✓ 1	Answer in fundamental units required.
Part (c)(iii)
Maximum Fv occurs at s = 70 m, where the force is still 7.0 × 10⁵ N and the speed is greatest	✓ 1	The hidden step: beyond 70 m F falls faster than v rises.
½mv² = (7.0 × 10⁵ + 1.2 × 10⁵) × 70, v = 67.8 m s⁻¹	✓ 1	Allow ECF from (a)(ii) method.
P = 7.0 × 10⁵ × 67.8 = 4.7 × 10⁷ W	✓ 1	Accept 4.6–4.8 × 10⁷ W. 5.1 × 10⁷ W (using 73 m s⁻¹) scores [1 max].
Part (d)(i)
The ship moves at constant velocity, so it is an inertial frame; Newton’s laws are the same in all inertial frames (Galilean relativity), so the work–energy calculation of (a) applies unchanged relative to the deck	✓ 1	Reference to constant velocity / inertial frame is needed.
Part (d)(ii)
Speed relative to the air = speed relative to the sea = 73.1 + 14 = 87 m s⁻¹	✓ 1	Galilean velocity addition. Allow ECF from (a)(ii); 73 m s⁻¹ (ship speed ignored) scores 0.
87 m s⁻¹ > 82 m s⁻¹, so it can fly	✓ 1	The conclusion must follow from the candidate’s value.
Part (d)(iii)
Minimum ship speed = 82 − 73.1 = 8.9 m s⁻¹	✓ 1	Allow ECF from (a)(ii). Accept 9 m s⁻¹.
Part (e)(i)
Average force = Δp/Δt = 2.0 × 10⁴ × 65/2.2 = 5.9 × 10⁵ N	✓ 1
Distance = average speed × time = ½ × 65 × 2.2 = 72 m	✓ 1	Accept 71–72 m. Equivalent: s = v²/2a with a = 29.5 m s⁻².

Answers: (a)(i) 5.6 × 10⁷ J · (a)(iii) 2.5 s · (c)(i) 6.1 × 10³ A · (c)(ii) kg s⁻² A⁻¹ · (c)(iii) 4.7 × 10⁷ W · (d)(ii) 87 m s⁻¹; it can fly · (d)(iii) 8.9 m s⁻¹ · (e)(i) 5.9 × 10⁵ N; 72 m (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — Newton's second law; that the applied external impulse equals the change in momentum of the system; A.3 — work done by a variable force as the area under a force–displacement graph; kinetic energy; power P = Fv; A.1 — uniformly accelerated motion; A.5 — that Newton’s laws of motion are the same in all inertial reference frames (Galilean relativity); the velocity addition equation u′ = u − v; D.3 — the magnetic force on a current-carrying conductor F = BIL sin θ; Tools — fundamental SI units Command term: Determine

A.2 Forces and momentum: IB Physics HL exam-style questions

30 practice questions on A.2 Forces and momentum

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