84 free, high-quality past-paper-style questions on Forces and momentum for IB Physics HL, each with a full mark scheme. Covers: Impulse and force–time graphs; Connected bodies and contact forces; Friction on an incline; and more. For when you have finished the past papers, or want to practise one topic at a time.A.2 at Standard Level →Read the A.2 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.450 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1A-1A-11Impulse and force–time graphsPaper 1AEasy1 mark
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
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Step 1Impulse equals the area under the force–time graph. The graph is a triangle of base 0.020 s and height 400 N.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Impulse = ½ × 400 × 0.020 = 4.0 N s.
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Step 3Impulse = change in momentum = mv, so v = 4.0/0.25 = 16 m s−1.
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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.
2A-1A-24Connected bodies and contact forcesPaper 1AEasy1 mark
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?
Frictionless horizontal surface
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Step 1Treat the pair as one body: a = 10/5.0 = 2.0 m s−2.
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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.
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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).
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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.
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?
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Step 1The weight of the crate is the gravitational force exerted by the Earth on the crate.
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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.
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Step 3So the partner is the gravitational force exerted by the crate on the Earth.
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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.
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?
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Step 1At rest the elastic force balances the weight: kx = mg = 0.60 × 9.81 = 5.89 N.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Extension x = 0.186 − 0.150 = 0.036 m.
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Step 3k = 5.89/0.036 = 1.6 × 102 N m−1.
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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.
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?
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Step 1The buoyancy force is the loss in reading: Fb = 5.40 − 3.40 = 2.00 N.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Fb = ρVg, so the volume displaced (= volume of the fragment) is V = 2.00/(1000 × 9.81) = 2.04 × 10−4 m3.
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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).
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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.
7A-1A-65Free-body diagrams and translational equilibriumPaper 1AEasy1 mark
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
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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.
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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.
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Step 3T = W/(2 sin θ). As θ becomes small, the tension becomes very large.
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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.
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?
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Step 1Friction provides the centripetal force: F = mv²/r.
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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.
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Step 3New force = F × 4/2 = 2F.
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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.
9IBP2H-004-00-01Vertical circular motionPaper 1AEasy1 mark
During the spin cycle of a washing machine, a wet towel is pressed against the inside of the vertical cylindrical drum. The towel moves in a horizontal circle at constant speed. Which force provides the centripetal force on the towel?
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The drum wall pushes on the towel perpendicular to the wall, and this is horizontal and towards the axis. This normal force is the resultant horizontal force on the towel, so it provides the centripetal force. The weight is vertical, so it cannot provide a horizontal centripetal force. Friction acts along the wall, which is vertical, so it balances the weight. A 'centrifugal force' is not a real force in the ground frame.
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Answer B
Answer: B
Common error: Centripetal force is not an extra force. It is the name for whichever real force, or resultant force, points towards the centre. Find that force on your free-body diagram.
10IBP2H-004-00-08Elastic and inelastic collisionsPaper 1AEasy1 mark
Spring P has spring constant k. Spring Q has spring constant 2k. Each spring is stretched by the same force F, and both obey Hooke's law. What is the ratio \(\frac{\text{elastic energy stored in P}}{\text{elastic energy stored in Q}}\)?
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The extension is x = F/k, so the energy stored is E = ½kx2 = F2/(2k). For the same force, the energy is inversely proportional to k. Therefore EP/EQ = 2k/k = 2. The softer spring stretches twice as far and stores more energy. The distractors come from common mistakes: • ½ assumes the stiffer spring stores more energy, which is only true for equal extensions. • 1 assumes equal forces mean equal energies. • 4 squares the ratio of the spring constants.
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Answer C
Answer: C
Common error: Check whether the force or the extension is fixed. For the same extension, the stiffer spring stores more energy. For the same force, it stores less.
11IBP2H-004-02-01Newton's third lawPaper 1AEasy1 mark
A communications satellite is in a circular orbit around Earth. Which force forms a Newton's third law pair with the gravitational force exerted by Earth on the satellite?
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Newton's third law pairs act on two different bodies, are of the same type and are equal and opposite. Earth pulls the satellite gravitationally, so the satellite pulls Earth gravitationally with an equal and opposite force. The centripetal force is not a separate force: here it is Earth's gravitational pull itself. A 'centrifugal force' does not act in an inertial frame. The Sun's pull involves a different pair of bodies.
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Answer B
Answer: B
Common error: Two forces acting on the same body can never be a third law pair, even when they are equal and opposite.
12IBP2H-004-02-02Connected bodies and contact forcesPaper 1AEasy1 mark
Two ice-dancers stand at rest on frictionless ice and push each other apart. Dancer X has mass 50 kg and dancer Y has mass 75 kg. Which statement is correct while they are pushing?
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By Newton's third law the force X exerts on Y is equal in magnitude and opposite in direction to the force Y exerts on X. Since a = F/m, the lighter dancer X has the larger acceleration (in the ratio 75 : 50 = 3 : 2).
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Answer B
Answer: B
Exam tip: Equal forces do not mean equal effects: apply Newton's second law to each body separately.
13IBP2H-004-01-01Impulse and force–time graphsPaper 1AEasy1 mark
A force sensor in a tennis racket records the force on the ball during one impact. The force increases linearly from zero to 400 N in 2.0 ms, then decreases linearly back to zero in a further 3.0 ms. What is the impulse on the ball?
Force on the ball against time
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Impulse is the area under the force–time graph. The graph is a single triangle with base 5.0 ms and height 400 N, so the area is ½ × 400 × 5.0×10−3 = 1.0 N s. 2.0 N s treats the area as a rectangle. 0.40 N s and 0.60 N s use only the rising or only the falling section.
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Answer C
Answer: C
Exam tip: For a force–time graph made of straight lines, split it into triangles and rectangles. Convert ms to s before multiplying.
14IBP2H-004-01-02Connected bodies and contact forcesPaper 1AEasy1 mark
A stationary object explodes into two fragments. One fragment has mass m and the other has mass 3m. What is \(\frac{\text{kinetic energy of fragment of mass } m}{\text{kinetic energy of fragment of mass } 3m}\)?
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The total momentum is zero, so the fragments have momenta of equal magnitude p. Since Ek = p2/2m, the kinetic energy is inversely proportional to mass for equal p. The lighter fragment therefore has 3 times the kinetic energy. 9 is the ratio of the squared speeds with the masses ignored. 1/3 wrongly assumes energy is shared in proportion to mass. 1 wrongly assumes energy is shared equally.
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Answer C
Answer: C
Exam tip: When the momenta are equal and opposite, use Ek = p²/2m. It gives energy ratios straight away.
15A-1A-03Elastic and inelastic collisionsPaper 1AMedium1 mark
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?
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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.
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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.
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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.)
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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.
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?
Block and pellet before the impact; dashed outline shows the block at its highest point
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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.
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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).
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Step 3mu = (m + M)v gives u = (2.00/0.020) × 1.98 = 198 m s−1 ≈ 2.0 × 102 m s−1.
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Step 4Only about 1 % of the pellet's kinetic energy (3.9 J out of 392 J) remains after the collision.
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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.
19A-1A-38Static and dynamic frictionPaper 1AMedium1 mark
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
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Step 1Maximum static friction = μsFN = 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).
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All 3 steps must be completed — there is no mark for a part-answer.
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 μdFN 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 μsFN (59 N) and also uses it once the crate slides.
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
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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.
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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.
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Step 3Its direction makes tan−1(8.0/12) = 34° with west, towards the south: 34° south of west.
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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.
21IBP2H-004-00-02Elastic and inelastic collisionsPaper 1AMedium1 mark
An engineer models a seat suspension using three identical springs, each of spring constant k. Two of the springs are placed side by side and support a light plate. The third spring connects this plate to the seat, so it is in series with the pair. What is the effective spring constant of the whole arrangement?
Springs of equal spring constant k
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The two springs side by side are in parallel, so they act like one spring of constant 2k. This pair is in series with a spring of constant k. For springs in series, 1/keff = 1/(2k) + 1/k = 3/(2k), so keff = 2k/3. The distractors come from common mistakes: • 3k treats all three springs as parallel. • k/3 treats all three springs as series. • 3k/2 swaps the rules, adding k to k/2.
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Answer B
Answer: B
Exam tip: Ask whether the springs share the load (parallel: constants add) or share the same force (series: reciprocals add). Combine the inner group first.
22IBP2H-004-00-03Newton's second lawPaper 1AMedium1 mark
Gravel falls vertically onto a horizontal conveyor belt at a steady rate of 20 kg s−1. The belt moves at a constant speed of 1.5 m s−1. Friction in the rollers is negligible. What additional horizontal force must the motor exert on the belt to keep its speed constant?
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Each second, 20 kg of gravel gains a horizontal velocity of 1.5 m s−1. The rate of change of horizontal momentum is therefore 20 × 1.5 = 30 kg m s−2. So the belt must exert a 30 N force on the gravel, and the motor must supply this force. The distractors come from common mistakes: • 0 N wrongly assumes that constant belt speed means no force is needed. • 15 N wrongly includes a factor of ½ taken from kinetic energy. • 196 N is the rate of arrival of weight, which acts vertically, not horizontally.
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Answer C
Answer: C
Exam tip: For a steady stream of mass, use F = v(Δm/Δt). Here the velocity is constant but the mass being moved increases.
23IBP2H-004-00-05Impulse and force–time graphsPaper 1AMedium1 mark
A footballer kicks a stationary ball of mass 0.43 kg. A force sensor shows that the force on the ball rises linearly from zero to 1200 N in 3.0 ms. It then falls linearly back to zero over a further 5.0 ms. What is the speed of the ball as it leaves the foot?
Force on the ball against time
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The impulse is the area under the force–time graph. The graph is a triangle with base 8.0 ms and height 1200 N, so the area is ½ × 1200 × 8.0×10−3 = 4.8 N s. This impulse equals the change in momentum, so v = 4.8/0.43 ≈ 11 m s−1. The distractors come from common mistakes: • 22 m s−1 uses a rectangle (1200 × 8.0 ms) instead of a triangle. • 4.2 m s−1 uses only the rising part of the graph. • 7.0 m s−1 uses only the falling part of the graph.
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Answer C
Answer: C
Exam tip: When a force varies, impulse is the whole area under the force–time graph. Do not multiply the peak force by the contact time.
24IBP2H-004-00-07Circular motionPaper 1AMedium1 mark
A test track has a frictionless banked curve of radius r, designed so that cars travelling at speed v need no sideways friction. A new curve is to be built with the same banking angle, for cars travelling at speed 2v. What radius is required?
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With no friction, the vertical component of the normal force balances the weight: N cos θ = mg. The horizontal component provides the centripetal force: N sin θ = mv2/r. Dividing these gives tan θ = v2/(rg). For a fixed angle, r ∝ v2, so doubling the speed requires 4r. The distractors come from common mistakes: • 2r assumes radius is proportional to speed. • √2 r takes a square root instead of squaring. • r/4 inverts the relationship.
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Answer D
Answer: D
Exam tip: The mass cancels in the banked-curve result tan θ = v²/(rg). The design speed therefore does not depend on the vehicle's mass.
25IBP2H-004-02-03Friction on an inclinePaper 1AMedium1 mark
A technician holds a 0.50 kg sensor block stationary against a vertical wall by pressing on it with a horizontal force of 20 N. The coefficient of static friction between the block and the wall is 0.60. What is the magnitude of the frictional force on the block?
Block held at rest against a vertical wall
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The block is in equilibrium, so the upward static friction equals the weight: 0.50 × 9.8 = 4.9 N. The value μsN = 0.60 × 20 = 12 N is only the maximum possible static friction; it is not reached because only 4.9 N is needed.
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Answer A
Answer: A
Common error: F = μsN gives the limiting value of static friction. Unless the body is on the point of slipping, find static friction from equilibrium.
26IBP2H-004-02-04Terminal speed with a speed-dependent resistive forcePaper 1AMedium1 mark
A raindrop falls vertically through still air. The air resistance on it is proportional to the square of its speed, and its terminal speed is vt. Buoyancy is negligible. What is the acceleration of the raindrop when its speed is vt/2?
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At terminal speed kvt2 = mg. At half that speed the drag is k(vt/2)2 = mg/4. The resultant force is mg − mg/4 = 3mg/4, so a = 3g/4. Option A assumes drag is proportional to speed; option C gives the drag-to-weight ratio rather than the acceleration.
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Answer B
Answer: B
Exam tip: Write the drag as a fraction of the weight using the terminal-speed condition. The mass and the constant k then cancel.
27IBP2H-004-02-05Connected bodies and contact forcesPaper 1AMedium1 mark
In a space station, three cargo modules P, Q and R, of masses m, 2m and 3m, float freely in a straight line and in contact. No external forces act on them other than a thruster, which applies a constant force F to P. P pushes Q, which pushes R. What is the ratio \(\frac{\text{force of P on Q}}{\text{force of Q on R}}\)?
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All three modules share the same acceleration, a = F/6m. The force of Q on R is the only horizontal force on R, so it accelerates R alone: 3m × F/6m = F/2. The force of P on Q accelerates Q and R together: 5m × F/6m = 5F/6. The ratio is (5F/6) : (F/2) = 5 : 3, so A is correct. Option B uses the mass ratio of Q to R instead of the masses being accelerated by each contact force. Option C inverts the ratio. Option D wrongly assumes the full force is passed on unchanged.
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Answer A
Answer: A
Exam tip: To find a contact force, isolate everything 'downstream' of the contact. That contact force alone accelerates all of that mass.
28IBP2H-004-02-06Static and dynamic frictionPaper 1AMedium1 mark
A crate rests on the flat bed of a delivery truck. The truck accelerates forwards with acceleration a, and the crate does not slip. The mass of the crate is m and the coefficient of static friction is μs. Which describes the frictional force exerted by the truck bed on the crate?
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Friction is the only horizontal force on the crate, and the crate accelerates forwards with the truck. The friction must therefore act forwards with magnitude ma (Newton's second law). Static friction opposes the relative motion that would happen without it: the crate would slide backwards relative to the bed. It equals μsmg only when the crate is about to slip.
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Answer A
Answer: A
Common error: Friction does not always oppose the motion of the body. It opposes relative sliding between the surfaces, and it can make a body speed up.
29IBP2H-004-01-03Vertical circular motionPaper 1AMedium1 mark
A small ball on a light string moves in a vertical circle. At the top of the circle it has the minimum speed needed to complete the circle. Which statement about the ball at the top is correct?
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At the minimum speed the string just becomes slack, so the tension is zero. The weight alone provides the centripetal force: mg = mv2/r. The acceleration is therefore g, directed downwards towards the centre. Centripetal force is not an extra force, and the resultant force cannot be zero for circular motion.
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Answer B
Answer: B
Common error: Do not draw 'centripetal force' as a separate force on a free-body diagram. It is the resultant of the real forces, such as weight and tension.
30IBP2H-004-01-04Vertical circular motionPaper 1AMedium1 mark
A roller-coaster car passes through the lowest point of a vertical loop of radius 8.0 m. A passenger feels a normal force from the seat equal to three times her weight. What is the speed of the car at this point?
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At the bottom, N − mg = mv2/r. With N = 3mg, this gives 2mg = mv2/r, so v = √(2gr) = √(2 × 9.8 × 8.0) = 12.5 m s−1. 15.3 m s−1 sets N = mv2/r and ignores the weight. 17.7 m s−1 adds the weight instead of subtracting it. 8.9 m s−1 uses v = √(gr).
✓ 1
Answer B
Answer: B
Exam tip: Always write the resultant force towards the centre first: (force towards centre) − (force away from centre) = mv²/r.
31IBP2H-004-01-06Impulse and force–time graphsPaper 1AMedium1 mark
Two identical balls hit a rigid wall at right angles with the same speed. Ball X sticks to the wall. Ball Y rebounds with the same speed. The contact time of Y with the wall is twice that of X. What is \(\frac{\text{average force on Y}}{\text{average force on X}}\)?
Show mark scheme
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Mark
Notes
The change in momentum for X is mv. For Y it is 2mv, because the direction reverses. The average force is Δp/Δt, so FY/FX = (2mv/2t)/(mv/t) = 1. 0.5 ignores the reversal of direction. 2 ignores the longer contact time. 4 doubles both effects in the wrong sense.
✓ 1
Answer B
Answer: B
Common error: A rebound reverses the velocity, so the change in momentum is m(v + u), not m(v − u).
32IBP2H-004-01-08Elastic and inelastic collisionsPaper 1AMedium1 mark
Trolley A moves along a frictionless horizontal track and collides with stationary trolley B. They stick together. Which statements are correct? I. The total momentum of the two trolleys is conserved. II. The total kinetic energy of the two trolleys is conserved. III. The impulse on A has the same magnitude as the impulse on B.
Show mark scheme
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Notes
No external horizontal force acts, so momentum is conserved (I). By Newton's third law, the forces on A and B are equal and opposite and act for the same time, so the impulses have equal magnitude (III). When the trolleys stick together the collision is perfectly inelastic, so kinetic energy is not conserved (II is false).
✓ 1
Answer B
Answer: B
Common error: Momentum is conserved in every collision of an isolated system. Kinetic energy is conserved only in elastic collisions.
33A-1A-67Elastic and inelastic collisionsPaper 1AHard1 mark
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?
Frictionless air track
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.
34IBP2H-004-00-04Vertical circular motionPaper 1AHard1 mark
A small ball of mass m is attached to a light string and moves in a vertical circle. At the top of the circle, it has the minimum speed needed to keep the string just taut. Air resistance is negligible. What is the tension in the string when the ball is at the bottom of the circle?
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Notes
At the top, the tension is zero, so mg = mvt2/r. This gives vt2 = gr. Between the top and the bottom, the ball falls 2r. Energy conservation gives vb2 = gr + 4gr = 5gr. At the bottom, T − mg = m(5gr)/r, so T = 6mg. The distractors come from common mistakes: • 5mg forgets that the tension must also balance the weight. • 2mg assumes the speed is the same at the bottom as at the top. • mg assumes the tension only supports the weight.
✓ 1
Answer D
Answer: D
Common error: In a vertical circle, the speed is not constant. Use energy conservation to link the speeds at different points before applying F = mv²/r.
35IBP2H-004-00-06Newton's second lawPaper 1AHard1 mark
A rocket of total mass 3.0×104 kg stands vertically on its launch pad. At ignition, it ejects exhaust gas at 150 kg s−1, with a speed of 3.0×103 m s−1 relative to the rocket. What is the initial upward acceleration of the rocket?
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Notes
The thrust is (rate of mass ejection) × (exhaust speed) = 150 × 3.0×103 = 4.5×105 N. The weight is 3.0×104 × 9.8 = 2.94×105 N. The resultant force is 1.56×105 N upwards, so a = 1.56×105/3.0×104 = 5.2 m s−2. The distractors come from common mistakes: • 15 m s−2 ignores the weight. • 25 m s−2 adds the weight instead of subtracting it. • 'Zero' uses ½ × (mass rate) × (exhaust speed) for the thrust, which wrongly brings in a kinetic-energy factor.
✓ 1
Answer A
Answer: A
Common error: Thrust is not the resultant force. Draw the free-body diagram with thrust up and weight down, then apply F = ma to the resultant.
36IBP2H-004-02-07Viscous drag, buoyancy and terminal speedPaper 1AHard1 mark
Two solid spheres made of the same material, of radii r and 2r, fall through the same viscous oil. Both reach terminal speed, and the drag obeys Stokes' law. What is the ratio \(\frac{\text{terminal speed of larger sphere}}{\text{terminal speed of smaller sphere}}\)?
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Notes
At terminal speed the weight minus the buoyancy, (4/3)πr3(ρs − ρf)g, equals the drag 6πηrv. Therefore v ∝ r3/r = r2. Doubling the radius multiplies the terminal speed by 4. Option C forgets that the drag also grows with r; option D assumes all objects fall alike, as they would in a vacuum.
✓ 1
Answer B
Answer: B
Exam tip: For scaling questions, write each force in terms of r and cancel before substituting any numbers.
37IBP2H-004-02-08Connected bodies and contact forcesPaper 1AHard1 mark
A 3.0 kg block on a horizontal bench is connected by a light, inextensible cord over a frictionless pulley to a hanging 1.0 kg mass. The coefficient of kinetic friction between the block and the bench is 0.20. The system is released from rest and moves. Take g = 10 m s−2. What is the tension in the cord?
Light inextensible cord over a frictionless pulley
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Notes
The friction on the block is 0.20 × 3.0 × 10 = 6.0 N. For the whole system, a = (10 − 6.0)/4.0 = 1.0 m s−2. For the hanging mass, mg − T = ma, so T = 1.0 × (10 − 1.0) = 9.0 N. Option C assumes the system is in equilibrium; option A equates the tension with the friction; option D is the resultant force on the whole system.
✓ 1
Answer B
Answer: B
Common error: In an accelerating system the tension is not equal to the hanging weight. It must be less than the weight, otherwise the hanging mass could not accelerate downwards.
38IBP2H-004-01-05Elastic and inelastic collisionsPaper 1AHard1 mark
In a reactor moderator, a neutron of mass m makes a head-on elastic collision with a stationary carbon nucleus of mass 12m. What fraction of the neutron's initial kinetic energy is transferred to the carbon nucleus?
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Notes
Momentum conservation gives u = vn + 12vC. In an elastic collision the relative speed of approach equals the relative speed of separation, so u = vC − vn. Solving these gives vC = 2u/13 and vn = −11u/13. The neutron keeps (11/13)2 = 121/169 of its kinetic energy. The carbon nucleus therefore gains 1 − 121/169 = 48/169. As a check, ½(12m)(2u/13)2 ÷ ½mu2 = 48/169. Why the other options are wrong: 2/13 and 11/13 are speed ratios, not energy fractions. 121/169 is the fraction of kinetic energy the neutron keeps, not the fraction it transfers.
✓ 1
Answer C
Answer: C
Exam tip: For one-dimensional elastic collisions, pair momentum conservation with 'speed of approach = speed of separation'. This avoids solving quadratic energy equations.
39IBP2H-004-01-07Vertical circular motionPaper 1AHard1 mark
A mass m on a light string moves in a vertical circle. Air resistance is negligible and the string stays taut throughout. What is the tension at the lowest point minus the tension at the highest point?
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Notes
At the bottom: Tb − mg = mvb2/r. At the top: Tt + mg = mvt2/r. Energy conservation over a height drop of 2r gives vb2 = vt2 + 4gr. Subtracting the two force equations gives Tb − Tt = 2mg + 4mg = 6mg, whatever the speed. 2mg ignores the energy change. 4mg ignores the weight terms. The speed cancels, so the last option is wrong.
✓ 1
Answer C
Answer: C
Exam tip: For vertical circles, combine Newton's second law at each point with energy conservation between the points.
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)
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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)
41A-1B-13Static and dynamic frictionPaper 1BEasy7 marks
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 FN 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.
FN / 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 μs and the coefficient of dynamic friction μd.
(2)
(ii)
Another student calculates μs 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
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Mark
Notes
Part (a)(i)
μs = gradient of the peak-force line, e.g. (5.54 − 1.10)/(12.0 − 2.0) = 0.44
✓ 1
Accept 0.42–0.46.
μd = 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 = μsmg cos θ, so tan θ = μs
✓ 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 μd < μs.
Answers: (a)(i) μs = 0.44; μd = 0.30 · (b)(i) 24° (the remaining parts are explanations — see the table above)
42A-1B-14Momentum and variable massPaper 1BEasy7 marks
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)
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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)
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
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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)
44A-1B-04Circular motion and orbitsPaper 1BMedium7 marks
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)
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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)
45A-1B-08Impulse from a force–time graphPaper 1BMedium6 marks
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)
Fav = 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)
46A-1B-11Viscous drag, buoyancy and terminal speedPaper 1BMedium7 marks
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 Fb = ρfVg and viscous drag Fd = 6πηrv gives v = 2r²(ρs − ρf)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.
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(ρs − ρf)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, (ρs − ρf) is replaced by ρs: 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)
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)
48A-1B-06Terminal speed with a speed-dependent resistive forcePaper 1BHard7 marks
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)
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)
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)
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)
52IBP2H-004-02-09Connected bodies and contact forcesPaper 2Easy5 marks
A sprinter of mass 68 kg starts a race from rest in starting blocks. For 0.32 s the blocks exert an average horizontal force of 820 N on the sprinter. Ignore all other horizontal forces.
(a)
State the Newton's third law partner of the force exerted by the blocks on the sprinter.
(1)
(b)
Calculate the average horizontal acceleration of the sprinter.
(2)
(c)
Calculate the horizontal speed of the sprinter on leaving the blocks.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The horizontal force exerted by the sprinter on the blocks. It is equal in magnitude (820 N) and opposite in direction, acting backwards
✓ 1
Part (b)
a = F/m = 820/68
✓ 1
a = 12 m s−2
✓ 1
Part (c)
Impulse = FΔt = 820 × 0.32 = 262 N s, or v = at using the acceleration from the previous part
✓ 1
v = 262/68 = 3.9 m s−1; accept 3.8 m s−1 from a = 12 m s−2
✓ 1
Allow error carried forward from the previous part.
—
Exam tip: Impulse = FΔt = Δp gives the speed directly, without first finding the acceleration, and it avoids carrying a rounded value forward.
53IBP2H-004-01-09Impulse and force–time graphsPaper 2Easy6 marks
A footballer kicks a stationary ball of mass 0.45 kg. A force plate inside a test boot records the force on the ball. Assume the force varies linearly between readings.
t / ms: 0, 2.0, 4.0, 6.0, 8.0, 10.0 F / N: 0, 300, 600, 600, 300, 0
The ball leaves the boot at t = 10.0 ms.
Force on the ball against time (straight lines between readings)
(a)
State what the area under a force–time graph represents.
(1)
(b)
Determine the impulse on the ball.
(2)
(c)
Calculate the average force on the ball during the kick.
(1)
(d)
Calculate the speed of the ball as it leaves the boot.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Impulse / change in momentum
✓ 1
Part (b)
Sums the trapezium areas, with time in seconds: 0.30 + 0.90 + 1.20 + 0.90 + 0.30 N s
✓ 1
Impulse = 3.6 N s
✓ 1
Part (c)
F = 3.6 / 0.010 = 360 N (allow follow-through from the impulse found earlier)
✓ 1
Part (d)
v = impulse / m = 3.6 / 0.45
✓ 1
v = 8.0 m s−1 (allow follow-through from the impulse found earlier)
✓ 1
Exam tip: Convert milliseconds to seconds before multiplying. An answer a thousand times too big is the most common slip with force–time data.
54A-2-09Elastic and inelastic collisionsPaper 2Medium14 marks
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 mA couples to a stationary wagon of mass mB, the fraction of the kinetic energy transferred to other forms is mB/(mA + mB). 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)
vB = 7.2 × 103/8.0 × 103 = 0.90 m s−1
✓ 1
Allow ECF from (a)(i).
A receives an equal and opposite impulse: vA = (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.
Common velocity mAu/(mA + mB); final Ek = mA²u²/2(mA + mB), so final/initial = mA/(mA + mB) and the fraction lost is mB/(mA + mB)
✓ 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)
55A-2-21Vertical circular motionPaper 2Medium14 marks
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.
Rope swing (not to scale)
(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)
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.
Frictionless air track; spring held compressed by a thread
(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 EkA/EkB = mB/mA.
(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)
Ep = 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 Ek = p²/2m
✓ 1
Symbolic working required.
EkA/EkB = (p²/2mA)/(p²/2mB) = mB/mA
✓ 1
Allow ECF from (b)(i) (equal and opposite momenta).
Part (b)(iii)
EkA = 0.25 × 0.60/1.00 = 0.15 J and EkB = 0.10 J
✓ 1
Allow ECF from (a)(i) and (b)(ii).
vA = √(2 × 0.15/0.40) = 0.87 m s−1; vB = √(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): EkNp/Ekα = mα/mNp ≈ 4/237
✓ 1
Allow ECF from (b)(ii).
EkNp = 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) vA = 0.87 m s−1, vB = 0.58 m s−1 · (c)(i) 93 keV (the remaining parts are explanations — see the table above)
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)
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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)
58IBP2H-004-00-10Circular motionPaper 2Medium6 marks
A medical laboratory centrifuge separates blood samples. It spins at 3000 revolutions per minute. The sample tubes are held so that each sample moves in a horizontal circle of radius 0.12 m about the axis.
(a)
Show that the angular speed of the rotor is about 310 rad s−1.
(2)
(b)
Calculate the centripetal acceleration of a sample, and express it as a multiple of g.
(2)
(c)
A tube and its contents have a total mass of 15 g. Calculate the magnitude of the resultant force on the tube, and state its direction.
(2)
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Part (a)
Frequency f = 3000/60 = 50 Hz .
✓ 1
ω = 2πf = 2π × 50 = 314 rad s−1, which is about 310 rad s−1 .
✓ 1
Part (b)
a = ω2r = 3142 × 0.12 = 1.18×104 m s−2 . Using 310 rad s−1 gives 1.15×104 m s−2, which is also accepted.
✓ 1
Divide by g = 9.81 m s−2: a ≈ 1.2×103g . Using g = 9.8 or 10 m s−2 is also accepted.
✓ 1
Part (c)
F = ma = 0.015 × 1.18×104 = 178 N . Accept 170–180 N. Allow error carried forward from the previous part.
✓ 1
The resultant force is directed horizontally towards the axis of rotation .
✓ 1
Exam tip: Convert revolutions per minute to rad s⁻¹ by dividing by 60 and then multiplying by 2π. In a 'show that' question, show both steps and give the answer to more significant figures than the value quoted in the question.
59IBP2H-004-00-11Momentum and variable massPaper 2Medium7 marks
A firefighter holds a hose nozzle horizontally. Water of density 1000 kg m−3 leaves the nozzle at a speed of 25 m s−1, with a mass flow rate of 12 kg s−1. Assume the water speed inside the hose, just before the nozzle, is negligible.
(a)
Calculate the horizontal force the nozzle must exert on the water.
(2)
(b)
Explain, with reference to Newton's laws, why the firefighter must push forwards on the nozzle to hold it still.
(3)
(c)
Determine the diameter of the nozzle opening.
(2)
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Part (a)
F = (Δm/Δt) × Δv = 12 × 25 .
✓ 1
F = 300 N .
✓ 1
Part (b)
The nozzle exerts a forward force on the water to increase its momentum (Newton's second law) .
✓ 1
By Newton's third law, the water exerts an equal backward force of 300 N on the nozzle .
✓ 1
For the nozzle to stay in equilibrium (zero resultant force), the firefighter must supply a forward force of 300 N .
✓ 1
Part (c)
A = (Δm/Δt)/(ρv) = 12/(1000 × 25) = 4.8×10−4 m2 .
✓ 1
d = √(4A/π) = 0.025 m (2.5 cm) .
✓ 1
ECF for a mass flow rate used incorrectly.
—
Exam tip: For a fluid jet, the mass flow rate is ρAv. Combining this with F = v(Δm/Δt) gives the useful result F = ρAv².
60IBP2H-004-00-12Newton's second lawPaper 2Medium8 marks
A delivery drone of mass 2.4 kg carries a package of mass 0.80 kg. The package hangs below the drone on a light vertical cable. The rotors provide a constant upward thrust of 40 N. Air resistance is negligible. Take g = 9.8 m s−2.
(a)
Describe the free-body diagram for the package while the drone accelerates upwards, stating the relative sizes of the forces.
(2)
(b)
Calculate the acceleration of the drone and package.
(2)
(c)
Calculate the tension in the cable.
(2)
(d)
The cable suddenly snaps while the thrust stays at 40 N. Calculate the acceleration of the drone immediately afterwards.
(2)
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Part (a)
There are two forces: the weight acting downwards and the cable tension acting upwards .
✓ 1
The tension is greater than the weight because the package accelerates upwards .
✓ 1
Part (b)
Resultant force = 40 − (3.2 × 9.8) = 8.64 N .
✓ 1
a = 8.64/3.2 = 2.7 m s−2 upwards .
✓ 1
Part (c)
T − 0.80 × 9.8 = 0.80 × 2.7 .
✓ 1
T = 10 N .
✓ 1
Allow an error carried forward from (b).
—
Part (d)
Resultant force on the drone = 40 − 2.4 × 9.8 = 16.5 N .
✓ 1
a = 16.5/2.4 = 6.9 m s−2 upwards .
✓ 1
Exam tip: For connected bodies, find the acceleration using the whole system. Then isolate one body to find the internal force, such as the tension.
61IBP2H-004-00-16Static and dynamic frictionPaper 2Medium8 marks
A student places a coin on a horizontal turntable at a distance r from the axis. The student slowly increases the angular speed until the coin begins to slide, and records the angular speed ω at that moment. This is repeated for several values of r.
r / m: 0.05, 0.10, 0.15, 0.20 ω at slipping / rad s−1: 11.1, 7.8, 6.4, 5.5
(a)
Explain why the coin starts to slide when ω exceeds a certain value.
(2)
(b)
Show that the data are consistent with ω2r being constant.
(2)
(c)
Determine the coefficient of static friction between the coin and the turntable.
(2)
(d)
Outline how a straight-line graph could be used to find the coefficient of friction from these data.
(2)
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Part (a)
Static friction provides the centripetal force mω2r, and the force required increases as ω increases .
✓ 1
Friction has a maximum value μmg, because the normal reaction equals mg. When mω2r exceeds this maximum, friction cannot supply the centripetal force and the coin slides .
✓ 1
Part (b)
Calculates ω2r for the data sets: 6.16, 6.08, 6.14 and 6.05 m s−2 .
✓ 1
States that these values agree to within about 2%, so ω2r is constant within experimental uncertainty .
✓ 1
Part (c)
At slipping, μmg = mω2r, so μg = ω2r. Uses the mean value of ω2r, 6.11 m s−2 .
✓ 1
μ = 6.11/9.8 = 0.62 .
✓ 1
Allow error carried forward from the values found in the previous part.
—
Part (d)
Plot ω2 against 1/r, which should give a straight line through the origin. Alternatively, plot r against 1/ω2 .
✓ 1
The gradient equals μg, so μ = gradient/g .
✓ 1
Common error: The condition μmg = mω²r applies only at the instant the coin starts to slip. Below that speed, the friction force is smaller than μmg.
62IBP2H-004-02-10Static and dynamic frictionPaper 2Medium6 marks
A hospital bed of total mass 120 kg has its wheels locked, so it can only slide on the floor. The coefficient of static friction between the rubber wheels and the floor is 0.70, and the coefficient of kinetic friction is 0.55. A nurse pushes horizontally on the bed.
(a)
Calculate the minimum horizontal force needed to start the bed sliding.
(2)
(b)
The nurse keeps pushing with the force found in (a) after the bed starts to slide. Determine the acceleration of the bed.
(2)
(c)
Explain why the bed is harder to start moving if the nurse pushes along a line directed partly downwards rather than horizontally.
(2)
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Part (a)
N = mg = 120 × 9.8 = 1176 N and F = μsN
✓ 1
F = 0.70 × 1176 = 823 N (accept 820 N)
✓ 1
Part (b)
Kinetic friction = 0.55 × 1176 = 647 N, so resultant force = 823 − 647 = 176 N
✓ 1
a = 176/120 = 1.47 m s−2 (accept answers from 1.4 to 1.5 m s−2; allow ECF from (a))
✓ 1
Part (c)
The downward component of the push adds to the weight, so the normal reaction increases and the maximum static friction μsN increases
✓ 1
Only the horizontal component of the push acts against friction, and it is smaller than the full push, so a larger total push is needed to start the bed sliding
✓ 1
Common error: Do not assume N = mg whenever an applied force has a vertical component. Resolve the forces vertically first.
63IBP2H-004-02-11Newton's second lawPaper 2Medium9 marks
A delivery drone releases a parcel attached to a small parachute. Wind-tunnel tests give the drag force F on the parcel and parachute at different air speeds v:
v / m s−1: 2.0, 4.0, 6.0, 8.0 F / N: 0.50, 2.0, 4.5, 8.0
The parcel and parachute have a total mass of 2.4 kg. Buoyancy is negligible. Take g = 9.81 m s−2.
(a)
Show that the data are consistent with the relationship F = kv2.
(2)
(b)
State the value of k, including its unit.
(1)
(c)
Calculate the terminal speed of the parcel.
(2)
(d)
The parcel is released from rest. Describe how its speed and acceleration change with time until it reaches terminal speed.
(2)
(e)
Calculate the acceleration of the parcel when its speed is 7.0 m s−1.
(2)
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Part (a)
Calculates F/v2 for at least three data pairs, e.g. 0.50/4.0, 4.5/36 and 8.0/64
✓ 1
All give 0.125, so F/v2 is constant. Alternatively, shows that doubling v (2.0→4.0 or 4.0→8.0) quadruples F
✓ 1
Part (b)
k = 0.125 N s2 m−2 (or kg m−1)
✓ 1
Part (c)
At terminal speed mg = kvt2, so vt = √(2.4 × 9.81/0.125)
✓ 1
vt = 14 m s−1 (accept 13.7). Allow error carried forward from (b)
✓ 1
Part (d)
Speed increases from zero with initial acceleration g (9.81 m s−2), and the acceleration decreases continuously as drag increases
✓ 1
Speed approaches the terminal speed from (c) gradually (asymptotically) without exceeding it, and the acceleration tends to zero
✓ 1
Part (e)
Drag = 0.125 × 7.02 = 6.1 N (6.125 N) and weight = 2.4 × 9.81 = 23.5 N
✓ 1
a = (23.5 − 6.1)/2.4 = 7.3 m s−2 (accept 7.2–7.3). Allow error carried forward from (b)
✓ 1
Exam tip: For a 'show that' with tabulated data, test the proposed relationship on several rows, not just one, and state the conclusion clearly.
64IBP2H-004-02-12Elastic and inelastic collisionsPaper 2Medium7 marks
An electric locomotive of mass 60 × 103 kg pulls two wagons along a straight, level track. Wagon A (mass 25 × 103 kg) is coupled directly to the locomotive, and wagon B (mass 15 × 103 kg) is coupled behind A. The track exerts a forward driving force of 48 kN on the locomotive. Each vehicle experiences a resistive force equal to 0.010 times its own weight.
Driving force acts on the locomotive
(a)
Calculate the acceleration of the train.
(2)
(b)
Determine the tension in the coupling between wagon A and wagon B.
(2)
(c)
Determine the tension in the coupling between the locomotive and wagon A.
(2)
(d)
State the magnitude and direction of the force exerted by wagon A on the locomotive.
Treats A and B as one body of mass 40 × 103 kg: T′ = 40 × 103 × (0.382 + 0.098)
✓ 1
T′ = 1.9 × 104 N
✓ 1
ECF from (a)
—
Part (d)
1.9 × 104 N, directed backwards (opposite to the motion), by Newton's third law
✓ 1
ECF from (c)
—
Exam tip: When resistance is proportional to mass, each coupling carries the fraction (mass behind it)/(total mass) of the driving force. Use this as a quick check: 0.40 × 48 kN = 19 kN.
65IBP2H-004-02-14Static and dynamic frictionPaper 2Medium8 marks
In a tug-of-war, team A (total mass 420 kg) pulls against team B (total mass 380 kg) using a light, horizontal rope. The coefficients of friction between team B's shoes and the ground are μs = 0.80 and μk = 0.60. Take g = 9.8 m s−2.
(a)
Explain why the rope exerts forces of equal magnitude on the two teams, even when the rope is accelerating.
(2)
(b)
Calculate the maximum static frictional force the ground can exert on team B.
(1)
(c)
Team B starts to slide. Both teams then move in the direction from team B towards team A, with an acceleration of 0.25 m s−2. Determine the tension in the rope.
(2)
(d)
Determine the horizontal frictional force exerted by the ground on team A.
(2)
(e)
State the Newton's third law partner of the frictional force on team A that you determined in the previous part.
(1)
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Part (a)
The rope is light, so its mass is negligible
✓ 1
By F = ma the resultant force on the rope is then (approximately) zero, so the tension, and hence the pull on each team, is the same at both ends
✓ 1
Part (b)
F = 0.80 × 380 × 9.8 = 3.0 × 103 N
✓ 1
Part (c)
For team B: T − μkmBg = mBa
✓ 1
T = 380 × (0.60 × 9.8 + 0.25) = 2.3 × 103 N
✓ 1
Part (d)
For team A: FA − T = mAa
✓ 1
FA = 2330 + 420 × 0.25 = 2.4 × 103 N
✓ 1
Allow error carried forward from the tension calculated in the previous part.
—
Part (e)
The frictional force exerted by team A's feet on the ground, of magnitude 2.4 × 103 N, acting towards team B
✓ 1
Exam tip: A tug-of-war is won by the team that receives the larger horizontal force from the ground, not by the team pulling harder on the rope. The rope tension is the same for both teams.
A satellite of mass 150 kg is released from an upper rocket stage of mass 450 kg (not including the satellite) in deep space. Before release, the two are at rest relative to an inertial observer moving with them. Compressed springs push the satellite away. After release, the satellite moves at 0.60 m s−1 relative to this observer. Assume all the elastic potential energy stored in the springs is transferred to the kinetic energy of the satellite and the stage.
(a)
State why the total momentum of the satellite and stage is conserved.
(1)
(b)
Calculate the velocity of the rocket stage after release.
(2)
(c)
Determine the elastic potential energy that was stored in the springs.
(2)
(d)
The springs act for 0.15 s. Calculate the average force exerted by the springs on the satellite.
(2)
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Part (a)
No external force acts on the system; the spring forces are internal
67IBP2H-004-01-11Vertical circular motionPaper 2Medium7 marks
A jet aircraft flies a vertical loop of radius 600 m. The pilot has a mass of 75 kg. At the lowest point of the loop the aircraft's speed is 140 m s−1.
(a)
Outline why the pilot feels heavier at the lowest point of the loop than in level flight.
(2)
(b)
Calculate the force exerted by the seat on the pilot at the lowest point.
(2)
(c)
Determine the ratio of the seat force to the pilot's weight at the lowest point.
(1)
(d)
The aircraft later passes through the top of the same loop. Calculate the speed at the top of the loop at which the seat force on the pilot becomes zero.
(2)
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Notes
Part (a)
The resultant force on the pilot must act upwards, towards the centre of the loop, to provide the centripetal acceleration
✓ 1
So the normal force from the seat must be greater than the weight (in level flight it equals the weight), and the sensation of weight comes from this normal force
✓ 1
Part (b)
N − mg = mv2/r, so N = 75 × 9.8 + 75 × 1402/600
✓ 1
N = 735 + 2450 ≈ 3.2×103 N
✓ 1
Part (c)
3185/735 ≈ 4.3 (accept 4.3–4.4; allow error carried forward from the seat force calculated in the previous part)
✓ 1
Part (d)
At the top, the weight alone provides the centripetal force: mg = mv2/r
✓ 1
v = √(9.8 × 600) ≈ 77 m s−1
✓ 1
Exam tip: At the top of a loop the weight acts towards the centre. At the bottom it acts away from the centre. Check the signs at each point before writing Newton's second law.
68IBP2H-004-01-13Elastic and inelastic collisionsPaper 2Medium8 marks
In a crash test, a car of mass 1200 kg moving at 15 m s−1 hits a stationary van of mass 1800 kg. The vehicles lock together. Friction with the road is negligible during the collision.
Before the collision (not to scale)
(a)
Calculate the velocity of the vehicles immediately after the collision.
(2)
(b)
Determine the fraction of the initial kinetic energy that is transferred to other forms.
(3)
(c)
Outline what happens to the kinetic energy that is lost.
(1)
(d)
A crash-test dummy of mass 70 kg, held by a seat belt, slows from 15 m s−1 to the common velocity in 0.12 s. Calculate the average force on the dummy.
(2)
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Part (a)
Conservation of momentum: 1200 × 15 = (1200 + 1800) v
✓ 1
v = 6.0 m s−1
✓ 1
Part (b)
Initial Ek = ½ × 1200 × 152 = 1.35×105 J
✓ 1
Final Ek = ½ × 3000 × 6.02 = 5.4×104 J (allow ECF from the velocity calculated in the first part)
✓ 1
Fraction = (1.35×105 − 5.4×104)/1.35×105 = 0.60
✓ 1
Part (c)
Any one of: transferred to internal (thermal) energy of the vehicles; used to do work permanently deforming the vehicles; transferred as sound
✓ 1
Part (d)
F = mΔv/Δt = 70 × (15 − 6.0)/0.12 (allow ECF from the velocity calculated in the first part)
✓ 1
F = 5.25×103 N ≈ 5.3×103 N (accept either)
✓ 1
Common error: The dummy does not stop. It slows only to the common velocity, so Δv = 9.0 m s⁻¹, not 15 m s⁻¹.
69IBP2H-004-01-16Collisions in two dimensionsPaper 2Medium8 marks
A 2.0 kg firework shell is moving horizontally at 30 m s−1 at the top of its flight when it explodes into two fragments. Immediately after the explosion, a 0.50 kg fragment moves vertically upwards at 60 m s−1.
Top of the flight of the shell, before and just after the explosion (not to scale)
(a)
Explain why momentum can be treated as conserved during the explosion, even though gravity acts on the shell.
(2)
(b)
Determine the speed and direction of the 1.5 kg fragment immediately after the explosion.
(4)
(c)
Calculate the energy released by the explosion as kinetic energy.
(2)
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Part (a)
The explosive forces are internal and much larger than the weight
✓ 1
The explosion lasts a very short time, so the impulse of the weight is negligible
✓ 1
Part (b)
Initial momentum = 2.0 × 30 = 60 kg m s−1, horizontal
✓ 1
The 1.5 kg fragment has horizontal momentum 60 kg m s−1 and vertical momentum 30 kg m s−1 downwards
✓ 1
p = √(602 + 302) = 67 kg m s−1, so v = 67/1.5 ≈ 45 m s−1
✓ 1
Direction: tan−1(30/60) ≈ 27° below the horizontal, in the original direction of motion
✓ 1
Part (c)
Ek after = ½ × 0.50 × 602 + ½ × 1.5 × 44.72 = 900 + 1500 = 2400 J. Using 45 m s−1 is accepted, and an error carried forward from the previous part is allowed
✓ 1
Energy released = 2400 − ½ × 2.0 × 302 = 1.5×103 J
✓ 1
Exam tip: Momentum is a vector. Conserve the horizontal and vertical components separately, then combine them with Pythagoras.
70A-2-01Collisions in two dimensionsPaper 2Hard14 marks
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.
Plan view of the air table; before and after the collision
(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)
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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
vB = √(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 = vA + vB as vectors, so the three velocities form a closed triangle
✓ 1
No numbers are required.
Kinetic energy: u² = vA² + vB², which is Pythagoras' theorem for that triangle, so the angle between vA and vB is 90°
✓ 1
Award the second mark only with a reason linking Pythagoras (or a zero cross-term 2vA·vB = 0) to the right angle.
Part (c)(i)
F = Δp/Δt = 0.69/(5.0 × 10−3)
✓ 1
Allow ECF from (a)(i): Δp = 0.30vB.
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 = vB²/(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)
71A-2-07Momentum and variable massPaper 2Hard12 marks
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)
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)
Ek = 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(ωi + ωf)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)
73A-2-20Static and dynamic frictionPaper 2Hard14 marks
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: FN = mg cos 12° − T sin 20°
✓ 1
The hidden step: the rope lifts the skier and reduces FN.
Parallel: T cos 20° = mg sin 12° + 0.080FN
✓ 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)
FN = 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 FN = 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)
Model sketch
Free-body diagram of the skier (arrow lengths approximate; slope at 12°)
74A-2-22Momentum and variable massPaper 2Hard13 marks
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 Ek = 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].
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)
75A-2-37Terminal speed with a speed-dependent resistive forcePaper 2Hard14 marks
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/vt²).
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
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 ρ4 km/ρground = (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)
76A-2-42Forces, work and power in an aircraft launchPaper 2Hard20 marks
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)
77IBP2H-004-00-13Friction on an inclinePaper 2Hard9 marks
A velodrome track is banked at 40° to the horizontal at a bend of radius 25 m. A cyclist and bicycle have a combined mass of 80 kg. Treat the cyclist and bicycle as a particle moving in a horizontal circle.
Cross-section of the banked bend (not to scale)
(a)
Show that the speed at which no sideways friction is needed is about 14 m s−1.
(3)
(b)
The cyclist rides the bend at 18 m s−1. Explain why the friction force on the tyres acts down the slope.
(1)
(c)
Determine the friction force at 18 m s−1.
(3)
(d)
Deduce the minimum coefficient of static friction needed between the tyres and the track.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Vertical: N cos 40° = mg .
✓ 1
Horizontal: N sin 40° = mv2/r .
✓ 1
v = √(rg tan 40°) = √(25 × 9.8 × 0.839) = 14.3 m s−1 .
✓ 1
Part (b)
Above the no-friction speed, the horizontal component of the normal force alone cannot provide the larger centripetal force needed. Friction therefore acts down the slope so that its horizontal component points towards the centre .
✓ 1
Part (c)
A correct equation, either along the slope, f + mg sin θ = (mv2/r) cos θ, or the horizontal and vertical pair N sin θ + f cos θ = mv2/r and N cos θ − f sin θ = mg .
✓ 1
v2/r = 182/25 = 12.96 m s−2 .
✓ 1
f = 80(12.96 × 0.766 − 9.8 × 0.643) ≈ 290 N .
✓ 1
Part (d)
Perpendicular to the slope: N = m(g cos θ + (v2/r) sin θ) = 80(7.51 + 8.33) ≈ 1.27×103 N .
✓ 1
μ = f/N = 290/1270 ≈ 0.23 .
✓ 1
Allow error carried forward from the friction force found in the previous part.
—
Common error: The centripetal acceleration v²/r is horizontal, not along the slope. If you resolve along and perpendicular to the slope, include its components: (v²/r) cos θ along the slope and (v²/r) sin θ perpendicular to it.
78IBP2S-003-06-16Momentum and variable massPaper 2Hard8 marks
A spacecraft in deep space, far from any significant gravitational field, has an initial total mass of 2400 kg. Its engine ejects propellant at a constant rate of 12 kg s−1 with a speed of 2.8 km s−1 relative to the spacecraft. The engine fires for 60 s.
(a)
Show that the thrust on the spacecraft is about 3.4 × 104 N.
(2)
(b)
Calculate the acceleration of the spacecraft at the start and at the end of the 60 s burn.
(2)
(c)
By taking the acceleration during the burn to be the mean of the start and end values you calculated, estimate the increase in speed of the spacecraft during the burn.
(2)
(d)
Explain whether your estimate of the increase in speed is an overestimate or an underestimate.
(2)
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Part (a)
Thrust = (Δm/Δt) × vrel
✓ 1
= 12 × 2800 = 3.36 × 104 N ≈ 3.4 × 104 N
✓ 1
Part (b)
Start: a = 3.36 × 104 / 2400 = 14 m s−2
✓ 1
End: mass = 2400 − 12 × 60 = 1680 kg, so a = 3.36 × 104 / 1680 = 20 m s−2
✓ 1
Allow error carried forward from the thrust value.
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Part (c)
Mean acceleration ≈ (14 + 20) / 2 = 17 m s−2
✓ 1
Δv ≈ 17 × 60 ≈ 1.0 × 103 m s−1
✓ 1
Allow error carried forward from the accelerations.
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Part (d)
a = F/m with m decreasing linearly, so the acceleration increases at an increasing rate. The acceleration–time curve bends upwards and lies below the straight line joining its end points.
✓ 1
The true mean acceleration is therefore less than the mean of the end values, so the estimate is an overestimate.
✓ 1
Exam tip: When you estimate using a changing quantity, state the approximation you are making, such as the mean of the start and end values. Then judge the direction of the error from the shape of the variation.
79IBP2H-004-02-13Viscous drag, buoyancy and terminal speedPaper 2Hard9 marks
In a diagnostic test, polystyrene microspheres of radius 5.0 × 10−6 m and density 1050 kg m−3 settle through blood plasma. The plasma has density 1025 kg m−3 and viscosity 1.2 × 10−3 Pa s. The drag on a microsphere obeys Stokes' law.
(a)
Outline the forces acting on a microsphere when it falls at terminal speed.
(2)
(b)
Show that the terminal speed is about 1.1 × 10−6 m s−1.
(3)
(c)
Calculate the time taken for a microsphere to settle through 1.0 cm at terminal speed.
(1)
(d)
To speed up the test, the sample is spun in a centrifuge at a radius of 0.080 m. Assume the analysis above still applies with g replaced by ω2r. Determine the angular speed needed for the microspheres to settle through 1.0 cm in 30 s.
(3)
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Part (a)
Weight acts downwards; buoyancy (upthrust) and viscous drag both act upwards
✓ 1
At terminal speed the resultant force is zero: weight = buoyancy + drag
v = 1.28 × 10−13/(6π × 1.2 × 10−3 × 5.0 × 10−6) = 1.13 × 10−6 m s−1
✓ 1
Part (c)
t = 0.010/1.13 × 10−6 = 8.8 × 103 s (about 2.5 h)
✓ 1
Part (d)
Terminal speed ∝ g, so the effective g must increase by a factor of 8.8 × 103/30 ≈ 290 (or: required terminal speed = 0.010/30 = 3.3 × 10−4 m s−1)
✓ 1
ω2r = 290 × 9.8 ≈ 2.9 × 103 m s−2
✓ 1
ω = √(2.9 × 103/0.080) ≈ 190 rad s−1
✓ 1
Allow error carried forward from the time calculated in the previous part.
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Common error: Forgetting buoyancy here gives an answer about 40 times too large, because the densities of the sphere and the fluid are very close. Always use the density difference.
80IBP2H-004-02-15Circular motionPaper 2Hard8 marks
An autonomous warehouse robot carries a box on its flat, horizontal top tray. The coefficient of static friction between the box and the tray is 0.35. The robot moves on a horizontal floor.
(a)
The robot follows a circular arc at constant speed. Explain why the frictional force on the box acts towards the centre of the arc.
(2)
(b)
Calculate the minimum radius of arc the robot can follow at 1.6 m s−1 without the box slipping.
(2)
(c)
The robot enters an arc of radius 0.90 m at 1.6 m s−1 and brakes while turning. Determine the maximum rate at which it can slow down at that instant without the box slipping.
(3)
(d)
Outline the direction of the frictional force on the box during the braking described in the previous part.
(1)
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Part (a)
The box moves in a circle, so it needs a resultant (centripetal) force directed towards the centre
✓ 1
Static friction from the tray is the only horizontal force on the box, so it must point towards the centre
✓ 1
Part (b)
μsmg = mv2/r, so the mass cancels
✓ 1
r = 1.62/(0.35 × 9.8) = 0.75 m
✓ 1
Part (c)
Centripetal acceleration = 1.62/0.90 = 2.84 m s−2
✓ 1
Maximum resultant acceleration = μsg = 3.43 m s−2
✓ 1
at = √(3.432 − 2.842) = 1.9 m s−2
✓ 1
Part (d)
It has a component towards the centre and a component opposite to the velocity. So it points inwards and backwards, at about 34° to the radius, tilted towards the rear of the robot
✓ 1
Common error: In non-uniform circular motion, the limiting friction must supply the vector sum of the centripetal and tangential accelerations. Do not compare each component separately with μsg.
81IBP2H-004-02-16Static and dynamic frictionPaper 2Hard8 marks
A film crew uses a camera dolly of mass 18 kg that runs on a smooth, level track (friction between the dolly and the track is negligible). A 6.0 kg camera rests unclamped on the flat top of the dolly. Between the camera and the dolly, μs = 0.30 and μk = 0.25. A horizontal pulling force is applied to the dolly only, starting from rest. Take g = 9.8 m s−2.
Smooth, level track; the camera is not clamped
(a)
Determine the maximum pulling force for which the camera does not slip on the dolly.
(3)
(b)
The pulling force is 100 N. Determine the accelerations of the camera and of the dolly.
(3)
(c)
Determine the time taken for the camera to slide 0.40 m backwards relative to the dolly.
(2)
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Part (a)
Only friction accelerates the camera, so its maximum acceleration is μsg = 0.30 × 9.8 = 2.94 m s−2
✓ 1
Without slipping, camera and dolly share this acceleration and can be treated as one 24 kg body
✓ 1
F = 24 × 2.94 = 71 N
✓ 1
Part (b)
100 N exceeds the maximum found in the previous part, so the camera slips and kinetic friction acts: camera a = μkg = 2.45 m s−2
✓ 1
By Newton's third law, the camera exerts a backward friction force of 0.25 × 6.0 × 9.8 = 14.7 N on the dolly
✓ 1
Dolly: a = (100 − 14.7)/18 = 4.7 m s−2 (4.74 m s−2)
✓ 1
Part (c)
Relative acceleration = 4.74 − 2.45 = 2.29 m s−2 (accept 2.3 m s−2, or a value consistent with the previous part)
✓ 1
0.40 = ½ × 2.29 × t2, so t = √(2 × 0.40/2.29) = 0.59 s (accept 0.59–0.60 s)
✓ 1
Exam tip: Check first whether the bodies move together: find the largest acceleration friction can give the top body. Treat the bodies as one system only if the actual acceleration is below this value.
82IBP2H-004-01-12Collisions in two dimensionsPaper 2Hard7 marks
In a golf-swing test rig, a club head of mass 0.20 kg moving at 45 m s−1 strikes a stationary golf ball of mass 0.046 kg head-on. Model the collision as elastic.
(a)
State what is meant by an elastic collision.
(1)
(b)
Determine the speed of the ball and of the club head immediately after the collision.
(4)
(c)
The contact time is 0.50 ms. Calculate the average force on the ball.
(2)
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Part (a)
Total kinetic energy is the same before and after the collision
✓ 1
Part (b)
Momentum: 0.20 × 45 = 0.20 v1 + 0.046 v2
✓ 1
Elastic: relative speed of approach = relative speed of separation, so 45 = v2 − v1 (OR uses kinetic energy conservation)
✓ 1
Solves: 9.0 = 0.20(v2 − 45) + 0.046 v2, so v2 = 73 m s−1
✓ 1
v1 = 28 m s−1, in the original direction
✓ 1
Part (c)
F = (0.046 × 73.2)/(0.50×10−3)
✓ 1
F ≈ 6.7×103 N (ECF from (b))
✓ 1
Exam tip: 'Speed of approach = speed of separation' is valid only for elastic collisions. It turns a quadratic energy equation into a linear one.
83IBP2H-004-01-14Vertical circular motionPaper 2Hard8 marks
A stunt motorcyclist rides around the inside of a vertical circular loop of radius 4.0 m. The combined mass of rider and motorcycle is 250 kg. Treat them as a point mass and ignore friction and air resistance unless told otherwise. The engine is switched off inside the loop.
(a)
Explain why there is a minimum speed at the top of the loop for the motorcycle to stay in contact with the track.
(2)
(b)
Show that the minimum speed at the bottom of the loop is about 14 m s−1.
(3)
(c)
Calculate the normal force on the motorcycle at the bottom of the loop when it moves at this minimum speed.
(2)
(d)
In practice there are resistive forces. Outline the effect this has on the speed needed at the bottom of the loop.
(1)
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Part (a)
At the top, both the weight and the normal force act downwards, towards the centre, and together provide the centripetal force mv2/r
✓ 1
The normal force cannot be less than zero (it is zero at the minimum speed), so mv2/r ≥ mg, giving v ≥ √(gr)
✓ 1
Part (b)
vtop2 = gr = 9.8 × 4.0 = 39.2 m2 s−2
✓ 1
Energy conservation: ½mvb2 = ½mvt2 + mg(2r)
✓ 1
vb2 = 39.2 + 4 × 9.8 × 4.0 = 196 m2 s−2, so vb = 14.0 m s−1
✓ 1
Part (c)
N − mg = mv2/r, so N = 250 × 9.8 + 250 × 196/4.0
✓ 1
N = 1.47×104 N ≈ 1.5×104 N (= 6mg)
✓ 1
Part (d)
Energy is dissipated on the way up, so the speed at the bottom must be greater than 14 m s−1 for the motorcycle to reach the top with at least √(gr)
✓ 1
Exam tip: For a 'show that' answer, give a value to one more significant figure than the one quoted, and show every step.
A sports scientist studies a vertical jump. An athlete stands still on a force plate, then pushes off. Assume the force varies linearly between readings. The athlete leaves the plate at t = 0.40 s.
t / s: 0.00, 0.10, 0.20, 0.30, 0.40 F / N: 588, 1000, 1400, 1000, 0
Take g = 9.8 m s−2.
Force from the plate on the athlete against time (straight lines between readings)
(a)
Deduce the mass of the athlete.
(1)
(b)
Determine the impulse exerted by the force plate on the athlete between t = 0 and t = 0.40 s.
(2)
(c)
Determine the speed of the athlete at take-off.
(3)
(d)
Calculate the height that the athlete's centre of mass rises after take-off.
(2)
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Part (a)
At t = 0 the athlete is at rest, so F = weight: m = 588/9.8 = 60 kg
✓ 1
Part (b)
Sum of trapezium areas: 79.4 + 120 + 120 + 50 (N s)
✓ 1
Impulse ≈ 369 N s (≈ 3.7×102 N s)
✓ 1
Part (c)
Impulse of weight = 588 × 0.40 = 235 N s
✓ 1
Net impulse = 369 − 235 = 134 N s (allow error carried forward from the impulse found earlier)
✓ 1
v = 134/60 ≈ 2.24 m s−1 (accept 2.2 m s−1)
✓ 1
Part (d)
Uses h = v2/2g (or energy conservation ½mv2 = mgh)
✓ 1
h = 2.242/(2 × 9.8) ≈ 0.26 m (accept 0.25–0.26 m; allow error carried forward from the take-off speed)
✓ 1
Common error: The plate's impulse is not the change in momentum. You must subtract the impulse of the weight to get the net impulse.
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