IB Physics HL A.2 Forces and momentum Questions

SLSL HLHL

Practise IB Physics A.2 by resolving forces in diagrams, applying force models, tracing momentum changes in collisions and explaining the forces needed for circular motion.

387 questions in this syllabus scope, with question parts counted together.

Exam points & syllabus

Exam points

Select one body before drawing forces: show real interactions, identify the force source and distinguish a third-law pair from balanced forces.
Resolve forces along useful axes and use the resultant; equilibrium means a zero vector sum, rather than absence of forces.
Use F = ma with the mass of the selected body or full connected system, including resistance when finding a pulling force.
For a mass suspended in an accelerating vehicle, combine vertical balance and horizontal acceleration to obtain a = g tanθ.
Distinguish adjusting static friction from its limit μsN; equate it to μsN only at impending slip.
For sliding or no-slip acceleration, select the correct body to find friction; a greater push need not increase Coulomb kinetic friction.
On an incline, use the correct normal component and friction direction before applying force balance or stopping-energy calculations.
At a static threshold, compare the required force with available friction; for rolling hoops use the actual demand, not μsN at every angle.
In a Rotor, the wall normal supplies radial acceleration while vertical static friction supports weight; convert rpm before testing no slip.
Use tension components for the actual string angle; shallow symmetric supporting strings can each carry tension greater than the weight.
For a spring, evaluate restoring force at the stated displacement; distinguish extra extension from equilibrium from the full spring extension.
Explain buoyancy from the fluid-pressure difference and use displaced volume, rather than automatically using the object’s total volume.
Use buoyant-force balance to analyse apparent weight, internal filling fraction, spring support and changes caused by floating ice melting.
For falling or rising motion in a fluid, show how increasing drag reduces the resultant until all forces balance at terminal speed.
Verify terminal speed from a constant position–time gradient; keep the symbolic Stokes sphere model separate from unverified numerical data.
State Newton’s second law as force equal to momentum-change rate; read this force from the gradient of a momentum–time graph.
Use momentum flux for ejected or captured material, keeping mass per second, velocity change and retained load separate.
For a vehicle losing a package or changing propellant flow, reconsider its supported weight, remaining mass and momentum flux.
Explain post-ejection ion spreading without confusing internal exhaust interactions with a change in the stated spacecraft thrust.
Define momentum as a vector and conserve the total only when resultant external force is zero; specify the system and direction signs.
For recoil, use equal and opposite momenta; convert inverse-mass kinetic-energy ratios into a total-energy fraction only when asked.
In collision calculations, retain every final moving mass and distinguish bodies that stick from bodies that leave at different speeds.
For rocket recoil, use the rocket plus exhaust system; estimate speed with the stated fuel allocation and the declared mass approximation.
Find impulse from force–time area and divide by the actual contact interval for mean force; a triangular pulse has peak twice its mean.
For reversals, add the opposing speed magnitudes in |Δp|; the change in momentum is different from the change in its magnitude.
Separate net force from contact force: add weight for vertical ground reaction, but do not add it to an already stated resultant.
Explain protective nets or packaging with the same momentum change, a longer stopping time and a smaller average force.
Convert impact times and mass units carefully, and distinguish impulse units from force units and order-of-magnitude answers.
Draw momentum through a collision with gradient matching the force, maintaining the correct sign change and monotonic direction.
Check elasticity by comparing total kinetic energies; explain inelastic losses through deformation or internal-energy transfer.
Use energy before and after a sticking collision separately; conserving mechanical energy through impact can give a wrong rebound height.
For a pivoted collision in HL, include the attached particle in the final moment of inertia before calculating rotational energy loss.
Distinguish tangent velocity from inward acceleration; constant speed can coexist with a changing velocity and a nonzero radial resultant.
Identify the real inward interaction in circular motion, and explain zero orbital work or apparent weightlessness without setting gravity to zero.
At the top or bottom of a vertical circle, combine weight with tension or normal force using the correct inward sign.
Use zero tension or zero normal force only at a contact threshold, then combine the threshold with energy where necessary.
For uniform vertical rotation driven by a rod, distinguish the varying rod force from the constant-magnitude inward resultant.
Use force ratios and limiting friction with the stated mass, speed, radius and angular-speed constraints.
Relate period and angular speed to linear speed using v = 2πr/T = ωr; points on one rigid platform share ω rather than v.
For circular-force calculations, identify the physical interaction and inertial mass rather than substituting a central body’s mass into mv²/r.

Covered syllabus points

A.2.1—Newton’s three laws of motion A.2.2—Forces as interactions between bodies A.2.3—Free-body diagrams A.2.4—Resultant force from diagrams A.2.5—Contact forces A.2.6—Normal force A.2.7—Frictional force A.2.8—Tension A.2.9—Hooke’s law restoring force A.2.10—Viscous drag A.2.11—Buoyancy A.2.12—Field forces A.2.13—Weight A.2.14—Electric force Fe A.2.15—Magnetic force Fm A.2.16—Linear momentum conservation A.2.17—Impulse A.2.18—Impulse-momentum change A.2.19—Newton’s second law forms A.2.20—Elastic and inelastic collisions A.2.21—Explosions A.2.22—Collision energy A.2.23—Centripetal acceleration A.2.24—Centripetal force A.2.25—Direction change in circular motion A.2.26—Angular and linear speed

Exam points

Link constant velocity or hovering to zero resultant force; this does not mean that no individual forces act.
Identify the force component that produces acceleration or deceleration; a law name or bare F = ma does not replace the physical explanation.
Name both bodies in a third-law pair and state equal magnitudes and opposite directions; “the same force” or “reaction” alone is insufficient.
Pair Earth’s force on a book with the book’s force on Earth; weight and normal reaction on the same book are not a third-law pair.
Third-law force magnitudes remain equal for unequal charges or a body held fixed; equality does not imply equal accelerations.
Distinguish an interaction force from a body’s resultant force: rope tension can balance resistance only when the forward acceleration is zero.
Explain fan or rocket thrust through backward momentum transfer to air or gas and the opposite force on the vehicle.
Explain a powered boat’s limiting speed by the balance of increasing resistance and forward thrust, giving zero resultant force and acceleration.
A decelerating passenger needs a rearward restraint force; connect the greater seat-belt tension to the change in the passenger’s motion.
Use inertia to explain the outward sensation in a turning car: unrestrained motion would be tangential, while the restraint supplies an inward force.
Explain orbital weightlessness by the astronaut and spacecraft sharing free fall; gravity can remain nonzero.
For straight constant-speed travel up an incline, a freely hanging string stays vertically downward because its forces balance.

[Maximum number: 2]

This question is about the forces on a skier.

A skier is pulled up a hill by a rope at a steady velocity. The hill makes an angle of $12^{\circ}$ with the horizontal. The mass of the skier and skis is 73 kg . The diagram below shows three of the forces acting on the skier.

Explain, using Newton's first law of motion, why the resultant force on the skier must be zero.

Exam points

Name the sources of the girl’s vertical forces: Earth pulls downward and the sledge supports upward; no vertical acceleration requires balance.
Isolate block M to find the contact force Ma; its equal and opposite force on m differs from the external force accelerating both blocks.

[Maximum number: 3]

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

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

Exam points

Draw forces acting on the chosen body and label them as forces; use weight, W or mg rather than the bare word “gravity”.
Draw weight vertically downward and the normal force perpendicular to the supporting surface, including on an incline.
Use the resultant and the stated motion to judge relative arrow lengths; constant velocity requires balanced components.
With an upward angled pull, include its vertical component: the normal force need not equal the weight.
For a released sliding stone, draw friction at the contact interface opposite motion and omit a continuing driving force.
Choose friction direction from the actual contact behaviour: a block pulled uphill has downhill friction, while the rolling hoop’s friction is uphill.
When application points are assessed, put weight at the centre of mass and contact forces at the relevant support or surface.
For a ball moving horizontally in a bowl, draw weight equal in length to the normal force’s vertical component.
At a curved hilltop, weight exceeds the upward normal force; at the bottom of a pendulum swing, upward tension exceeds weight.
Represent centripetal acceleration through the resultant of real forces; do not add a separate “centripetal force” arrow.
In the Rotor, draw weight downward, friction upward and the wall’s normal force inward towards the rotation axis.
For a tray accelerating with a waiter, include forward contact friction as well as its weight and normal force.
A hanging ball has weight and string tension; balance vertical components only when the stated motion has no vertical acceleration.
Immediately after release from rest in a fluid, omit drag and any hand force; compare buoyancy and weight using the stated displacement from equilibrium.
For the equatorial asteroid probe, use inward gravity minus outward normal force as mω²R and derive the normal-force expression from the diagram.
At the highest swing position, draw tension along the string and weight vertically down; a momentary stop does not establish equilibrium.
When an inclined lift arrow is supplied, complete the stated diagram with weight down and drag opposite motion; omit a released launch cable.

[Maximum number: 2]

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

Draw the free-body diagram for the sledge at the position shown on the snow slope.

arrow vertically downwards labelled weight/gravitational force $F_{\mathrm g}$ ;
arrow perpendicular to the snow slope labelled reaction/normal contact force $F_{\mathrm N}$ ;
friction force acting up the slope

Exam points

Use zero vector resultant as the condition for translational equilibrium; individual forces can be nonzero and the body need not be at rest.
Check that force components cancel independently along both axes; three coplanar forces in equilibrium are not all mutually perpendicular.
Bound the possible resultant by vector geometry: it cannot exceed the sum of the force magnitudes, and a closed triangle can give zero.
For a frictionless incline, the normal cancels the perpendicular weight component and the resultant is W sinθ down the slope.
In the bowl, balance N sinθ with mg when θ is measured from the horizontal; the remaining horizontal resultant is mg/tanθ.
Test whether an assumed circular path can maintain vertical balance: a horizontal normal at the bowl rim cannot support the ball’s weight.
For a thread at angle θ to the vertical, balance components to find the horizontal force mg tanθ; at 45°, it equals the weight.
Distinguish an individual horizontal force from the horizontal resultant; opposing forces can be nonzero while their sum is zero.
Use horizontal equilibrium to equate a ladder’s wall normal force with ground friction; these are balancing forces, not a third-law pair.
A pendulum’s resultant is nonzero at its turning points and at the bottom; zero tangential force at the bottom does not remove radial acceleration.
Combine perpendicular force magnitudes by Pythagoras: given F and 2F at right angles, the resultant is √5F.

[Maximum number: 3]

A vertical wall carries a uniform positive charge on its surface. This produces a uniform horizontal electric field perpendicular to the wall. A small, positively-charged ball is suspended in equilibrium from the vertical wall by a thread of negligible mass.

The thread makes an angle of $30^{\circ}$ with the vertical wall. The ball has a mass of 0.025 kg .

Determine the horizontal force that acts on the ball.

Horizontal force on ball: $F = T\sin 30$ .
$T = \frac{mg}{\cos 30}$ .
$F = mg\tan 30 = 0.025 \times 9.8 \times \tan 30 = 0.14\ \text{N}$ .

Allow $g=10\ \text{N kg}^{-1}$ .
Award [3] marks for a bald correct answer.
Award [1 max] for an answer of zero if the horizontal force is interpreted as the horizontal component of the net force.

Exam points

Isolate block M to find the contact force Ma; its opposite force on m differs from the external force accelerating both blocks.

[Maximum number: 1]

The diagram shows two blocks of mass m and M in contact on a frictionless surface. A force F is applied on the block of mass m and causes the blocks to move with an acceleration a.

What is the force that the block of mass M is exerting on the block of mass m ?

m a

(M+m) a

$\frac{m}{M+m} a$

Exam points

With a rope parallel to the slope and no perpendicular acceleration, use N = mg cosθ rather than setting N = mg.

[Maximum number: 2]

Calculate the magnitude of the normal reaction acting on the skier.

$\quad(N=) m g \cos \theta /$ correct substitution;
$\left(=73 \times 9.81 \times \cos 12^{\circ}=\right) 700 \mathrm{~N}$ ;

Exam points

Distinguish actual static friction from its limit: on a stationary slope Ff = mg sinθ, while Ff ≤ μs mg cosθ.
At impending sliding, equate the required friction to μsN; use μs = tanθ only for the simple unforced incline.
Explain acceleration after a fixed push exceeds limiting static friction: kinetic friction is smaller, leaving a forward resultant.
For a horizontal sliding block, subtract ma from the pull before using Ff = μdmg to find the coefficient.
Apply F = ma to the trolley alone to find friction; divide by the sliding block’s normal force to obtain μd.
Test whether a book slips on an accelerating table using ma ≤ μsmg, giving amax = μsg.
For uphill sliding, include both downslope friction and mg sinθ; give the acceleration direction as well as its magnitude.
For constant-speed motion up a ramp, use T = mg sinθ + Ff to calculate friction down the slope.
Use stopping distance to link frictional work to kinetic-energy loss, or find deceleration first and then μd.
For sliding bounces with equal contact time and normal force, greater friction lowers horizontal rebound speed and increases its angle.
For a rolling hoop, compare required friction mg sinθ/2 with μsmg cosθ; slipping begins at tanθ = 2μs.
In a Rotor, use N = mω²R and μsN ≥ mg; convert rpm to rad s⁻¹ before comparing with the no-slip threshold.
When checking tipping versus sliding, compare the tipping force with maximum static friction, not with a presumed actual μsN.
At constant speed down a rough incline, use μdN = mg sinθ and select the inverse-sine expression for θ.
Under the Coulomb model, doubling a horizontal pull changes neither μd nor kinetic friction when N stays unchanged.

[Maximum number: 5]

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

(a)

The sledge, without the girl on it, now travels up a snow slope that makes an angle of $6.5^{\circ}$ to the horizontal. At the start of the slope, the speed of the sledge is $4.2 \mathrm{~m} \mathrm{~s}^{-1}$ . The coefficient of dynamic friction of the sledge on the snow is 0.11 .

[ 3 ]

(i)

Show that the acceleration of the sledge is about $-2 \mathrm{~m} \mathrm{~s}^{-2}$ .

[ 3 ]

friction force down slope $=\mu mg\cos(6.5^\circ)=5.9\ \mathrm N$

component of weight down slope $=mg\sin(6.5^\circ)=6.1\ \mathrm N$

acceleration $=\frac{12}{5.5}=2.2\ \mathrm{m\,s^{-2}}$

(b)

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

[ 2 ]

maximum frictional force $=\mu R=7.5\ \mathrm N$ ;
the sledge will not move because the maximum static friction force is greater than the component of weight down the slope

Exam points

For two symmetric strings at a shallow angle θ to the horizontal, use 2T sinθ = W; each tension can exceed the weight.
For a rope parallel to a slope pulling at constant speed, calculate T = Ff + mg sinθ, including both downslope forces.

[Maximum number: 2]

The total frictional force acting is 65 N . Determine the tension in the rope.

tension = frictional force + component of weight parallel to slope /
tension $=65+m g \sin \theta$ ;
$214 / 210 \mathrm{~N}$ ;

Exam points

Find the maximum spring force at amplitude: use Fmax = kA, with k = mω² if needed, and convert centimetres to metres.

[Maximum number: 2]

An object attached to a spring oscillates with simple harmonic motion in a horizontal plane. The graph shows how the kinetic energy of the object varies with the displacement x.

The mass of the object is 0.15 kg .

Calculate the maximum force exerted on the object by the spring.

ALTERNATIVE 1
$\begin{aligned} & a_{\max }=8.6^{2} \times 0.060 \text { OR }\left(\frac{2 \pi}{0.73}\right)^{2} \times 0.060<=4.44 \mathrm{~m} \mathrm{~s}^{-2}> & F_{\max }=0.15 \times 4.44=0.67<\mathrm{N} \gg \end{aligned}$

ALTERNATIVE 2
$\begin{aligned} & k=8.6^{2} \times 0.15 \text { OR }\left(\frac{2 \pi}{0.73}\right)^{2} \times 0.15<=11.1 \mathrm{~N} \mathrm{~m}^{-1}> & F_{\max }=11.1 \times 0.060=0.67<\mathrm{N}> & \end{aligned}$

MP1 find a or k MP2 F=m a / F=k x

Award [2] for BCA

Exam points

For a falling sphere at terminal speed, balance weight with both viscous drag and upthrust; repeating “terminal speed” earns no extra mark.
Explain a rising droplet’s decreasing acceleration: increasing speed increases downward drag and reduces the upward resultant.
For a rising droplet at constant velocity, set buoyancy equal to weight plus drag, rather than equating drag with weight.
Verify terminal speed in a sphere-viscosity experiment by identifying a straight section of its position–time graph with constant gradient.
In the Stokes bubble model with negligible weight, equate 6πηRv to 4πR³ρg/3 and rearrange to v = 2R²ρg/(9η).

[Maximum number: 4]

A solid sphere is released from rest below the surface of a fluid and begins to fall.

(a)

Explain why the sphere will reach a terminal speed.

[ 2 ]

(b)

The weight of the sphere is 6.16 mN and the radius is $5.00 \times 10^{-3} \mathrm{~m}$ . For a fluid of density $8.50 \times 10^{2} \mathrm{~kg} \mathrm{~m}^{-3}$ , the terminal speed is found to be $0.280 \mathrm{~ms}^{-1}$ . Calculate the viscosity of the fluid.

[ 2 ]

any evidence (numerical or algebraic) of a realisation that
$\begin{aligned}6\pi\eta rv+\rho gV&=W \\ \eta&=\frac{6.16\times10^{-3}-4.366\times10^{-3}}{6\pi\times0.005\times0.280} \\ &=0.0680\ \mathrm{Pa\,s}\end{aligned}$

Exam points

Explain buoyancy using greater fluid pressure at the lower surface; link the pressure difference and surface area to an upward resultant.
To neglect a bubble’s weight, calculate W/Fb = ρgas/ρliquid and state why the small ratio justifies the approximation.
For a floating cube, use submerged fraction = ρobject/ρfluid, then subtract from one if the fraction above the surface is requested.
Explain unchanged water level when floating ice melts: the meltwater volume equals the volume of water originally displaced.
Read fluid density at the stated temperature and use Fb = ρfluidg(4πr³/3) for a fully immersed thermometer sphere.
For a neutrally buoyant hollow sphere with negligible glass mass, equate displaced-fluid weight to the weight of its liquid filling.
For a floating sun bed, choose upthrust equal to the bed’s weight plus the downward contact force from the person.
For a stationary submerged sphere supported by a spring, use Fspring = mg − ρfluidVg; there is no viscous drag at rest.
For a floating cylinder, equate ρobjectAHg and ρfluidADg to derive D/H = ρobject/ρfluid.
Calculate buoyancy from the change in balance reading: Fb = (mair − mimmersed)g, converting grams to kilograms.

[Maximum number: 1]

A group of students is trying to determine the density and the viscosity of a liquid.

To determine the density, they use a balance to read the mass m of a sphere in air and immersed in the liquid.

They use a sphere of volume $V=1.827 \times 10^{-7} \mathrm{~m}^{3}$ .
The readings are $m_{\text {air }}=1.427 \mathrm{~g}$ in air and $m_{\text {immersed }}=1.208 \mathrm{~g}$ in the liquid.
The readings are different due to buoyancy. The buoyancy force $F_{\mathrm{b}}$ is given by

F_{\mathrm{b}}=\rho V g

where V is the volume of the sphere and $\rho$ is the density of the liquid.

Show that $F_{\mathrm{b}}$ is about 2 mN .

$(0.001427-0.001208) \cdot 9.8$ OR 2.1 mN

Allow the calculation to give a result
in mN or N . Look for full substitution
or answer with at least two SF.

Exam points

Identify fields as the concept used to describe action at a distance, rather than wave motion, atomic theory or tunnelling.

[Maximum number: 1]

Which new concept was required to understand action at a distance?

Wave motion

Atomic theory

Tunnelling

Fields

Exam points

Calculate weight using W = mg and the local value of g, finding mass from density and cube dimensions.
Resolve the component of weight down a slope as mg sin θ, using the angle to the horizontal.

[Maximum number: 1]

A girl rides a bicycle that is powered by an electric motor. A battery transfers energy to the electric motor. The emf of the battery is 16 V and it can deliver a charge of 43 kC when discharging completely from a full charge.

The bicycle and the girl have a total mass of 66 kg . The girl rides up a slope that is at an angle of $3.0^{\circ}$ to the horizontal.

Calculate the component of weight for the bicycle and girl acting down the slope.

$66 \mathrm{~g} \sin \left(3^{\circ}\right)=34$ « N »

Exam points

In E = FE/q, define FE as the force on a small positive test charge; “electric force” alone does not earn the mark.

[Maximum number: 1]

This question is in two parts

The magnitude of gravitational field strength g is defined from the equation shown below.

g=\frac{F_{\mathrm{g}}}{m}

The magnitude of electric field strength E is defined from the equation shown below.

E=\frac{F_{\mathrm{E}}}{q}

For each of these defining equations, state the meaning of the symbols

$F_{\mathrm{E}}$ .

Exam points

Define linear momentum as mass × velocity; if writing p = mv, identify the symbols and retain the velocity direction.
State that total momentum remains constant when the resultant external force is zero; include both the conserved total and the condition.
Explain why a ball’s momentum changes in a floor collision using external force, impulse or transfer to the floor, rather than energy loss.
For connected bodies with equal speeds, compare momentum directions and masses: equal speed does not imply equal momentum.
For a stationary parent or spring-separated pair, use equal and opposite momenta to obtain recoil speeds inversely proportional to mass.
Use equal momentum magnitudes and KE = p²/(2m) to compare recoil energies; the lighter product carries more kinetic energy.
Distinguish a recoil-energy ratio from a fraction of total released energy; in alpha decay use Eα/Etotal = Mdaughter/(Mdaughter + Mα).
For a rocket, choose rocket plus exhaust as the isolated system and describe their equal and opposite momentum changes.
For a ball thrown on frictionless ice, link equal and opposite forces on ball and person to opposite recoils and zero total momentum change.
Estimate ion-spacecraft speed using only the fuel allocated to acceleration; state the constant-mass approximation when using initial acceleration.
When a person leaves a moving sledge with zero final horizontal velocity, include both initial masses but only the sledge’s final momentum.
For objects that stick together, conserve signed total momentum using the sum of final masses; solve for the unknown initial or common speed.
When a load falls onto a trolley, conserve horizontal momentum: mu = (m + M)v, so the added mass is M = m(u − v)/v.
For a pellet embedding in a ball, include the pellet in the final moving mass; omitting it can lose a mark even if the speeds are close.
Use opposite signs for products travelling in opposite directions; add their momenta to recover the parent momentum.
In the stated equal-and-opposite annihilation setup, explain why a single photon cannot conserve the initial zero total momentum.
For a photoelectron, subtract the work function from photon energy, then use p = √(2mKE) and report momentum in kg m s⁻¹.
When two colliding bodies separate, retain both final momenta: subtract the ball’s gained momentum to find the club head’s final speed.

[Maximum number: 2]

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

When the sledge is moving on the horizontal region of the snow, the girl jumps off the sledge. The girl has no horizontal velocity after the jump. The velocity of the sledge immediately after the girl jumps off is $4.2 \mathrm{~m} \mathrm{~s}^{-1}$ . The mass of the girl is 55 kg and the mass of the sledge is 5.5 kg . Calculate the speed of the sledge immediately before the girl jumps from it.

mention of conservation of momentum

OR

$5.5\times4.2=(55+5.5)v$ $v=0.38\ \mathrm{m\,s^{-1}}$

Exam points

Identify impulse as the change in momentum, J = Δp; use average resultant force in J = FavgΔt for a varying force.
Find impulse from every section of a force–time graph, then use v = Δp/m for an object initially at rest.
For a triangular force pulse, equate ½FmaxΔt to the momentum change; the peak force is twice the mean force.
Read the collision duration as end time minus start time on the graph; divide impulse by this interval to find mean force.
For a rebound, reverse the velocity sign: |Δp| = m(u + v), even when the supplied graph plots only positive speeds.
Distinguish ground contact force from resultant force: during vertical stopping or rebound, N = Δp/Δt + mg.
When the stated force is already the resultant, use Favg = Δp/Δt directly without adding weight again.
Explain protective surfaces with the same momentum change, longer stopping time and smaller mean force; state the full causal chain.
Convert collision times from ms or μs to seconds before calculating mean force; use the mass of the body whose momentum changes.
For an accelerating car with no reversal, use the difference of velocities and convert cm s⁻¹ and mN s consistently.
State stopping impulse in N s or kg m s⁻¹; force in newtons is a different quantity.
Draw momentum through impact with a gradient matching the force: reverse its sign at mid-impact and keep the gradient sign consistent.
For a car slowed by a collision, use that car’s mass and speed change; preserve units and avoid early rounding of the initial speed.
Use Newton’s third law to transfer the impact-force or impulse magnitude from ball to wall or paddle, reversing its direction.
For a descending probe braked by a rocket, subtract weight from thrust before using Δt = m|Δv|/Fnet.
Estimate an impact-force order of magnitude with mΔv/Δt after converting grams and milliseconds to SI units.

[Maximum number: 3]

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

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

same change in momentum/impulse;
the time taken to stop would be greater with the snow

$F=\frac{\Delta p}{\Delta t}$ , therefore F is smaller with the snow

OR

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

Exam points

Explain why a flexible net reduces injury: the same momentum change takes longer, so the average stopping force is smaller.
Read velocities on both sides of a bounce, use N − mg = m(vafter − vbefore)/Δt, then round the ground force to the stated precision.

[Maximum number: 3]

A toy rocket is made from a plastic bottle that contains some water.

Air is pumped into the vertical bottle until the pressure inside forces water and air out of the bottle. The bottle then travels vertically upwards.

The air-water mixture is called the propellant.
The variation with time of the vertical velocity of the bottle is shown.

The bottle reaches its highest point at time $T_{1}$ on the graph and returns to the ground at time $T_{2}$ . The bottle then bounces. The motion of the bottle after the bounce is shown as a dashed line.

The bottle bounces when it returns to the ground.

The mass of the bottle is 27 g and it is in contact with the ground for 85 ms .

Determine the average force exerted by the ground on the bottle. Give your answer to an appropriate number of significant figures.

Attempt to use force = momentum change ÷ time
$\frac{(4.5+10) \times 0.027}{85 \times 10^{-3}}=4.6$

Force $=4.6+0.3=4.9\ \mathrm{N}$

Any answer to 2 sf

Marking guidance:

Accept ± 0.5 velocity values from graph

Exam points

State Newton’s second law as resultant force equal to the rate of momentum change; define all symbols if giving only an equation.
Define one newton as the resultant force that accelerates 1 kg at 1 m s⁻² in the force direction.
For constant acceleration, require both constant resultant force and constant mass; constant force alone is insufficient for changing mass.
For a hanging mass driving a frictionless horizontal mass, divide the hanging weight by the combined accelerating mass.
Resolve a tilted suspended mass into T cosθ = mg and T sinθ = ma to obtain a = g tanθ for angle θ to the vertical.
To find a pulling tension, add resistance to ma; use the acceleration from the actual motion rather than substituting g.
For downward acceleration, use T = m(g − a), selecting the mass of the suspended body or full lift as appropriate.
Calculate thrust as mass-flow rate × velocity change, then divide the resultant force by the instantaneous vehicle mass.
Distinguish ions spreading after ejection from a change at the nozzle: mutual forces between departed ions do not change the stated thrust.
Use the initial tangent of a speed–time graph with the initial thrust to infer mass; an interval-average acceleration can lose a mark.
For gravel landing without rebound, find vertical force from mass-flow rate × impact speed, then add the weight retained on the belt.
Equal and opposite forces can give zero centre-of-mass acceleration while still producing rotation; separate resultant force from torque.
Use F = qE followed by a = F/m for the electron’s initial acceleration; the field strength itself is not a force.
Read maximum acceleration from the graph before applying Fmax = mamax, rather than using the period or displacement directly.
For extra displacement from a vertical spring’s equilibrium, use the extra restoring force kx as the resultant; weight is already balanced.
Add both restoring forces in a two-spring system and give the acceleration direction towards equilibrium.
Read force as the gradient of a momentum–time graph; a horizontal line at nonzero momentum still means zero resultant force.
Subtract opposing forces before finding acceleration or momentum rate: falling weight minus drag, or cyclist drive minus air resistance.
Use zero horizontal resultant to justify constant horizontal momentum in flight, even though gravity changes vertical momentum.
For a charged ball released from rest in a uniform horizontal field, combine electric force and weight to predict constant acceleration.
Compare propellants at the stated equal exhaust speed using mass expelled per second: a larger flow gives a larger thrust in this model.
For hovering, equate downward air momentum per second to the total supported weight, including the package.
After a package is released at unchanged lift, subtract the aircraft’s remaining weight and divide by its remaining mass.
For a spacecraft collecting stationary dust, use Δm = ρSvΔt and Δp = vΔm to derive opposing drag F = ρSv².
For a fixed combined mass stopping over distance d, use a = −u²/(2d) and Fnet = ma to obtain its rate of momentum change.
For normally reflected photons, use momentum change 2h/λ per photon; divide momentum rate by area to derive P = 2I/c.
Convert radiation pressure to force with F = PA, then use a = F/m for the acceleration due to solar radiation alone.

[Maximum number: 2]

One end of a light spring is attached to a rigid horizontal support.

An object W of mass 0.15 kg is suspended from the other end of the spring. The extension x of the spring is proportional to the force F causing the extension. The force per unit extension of the spring k is $18 \mathrm{Nm}^{-1}$ .

A student pulls W down such that the extension of the spring increases by 0.040 m . The student releases W and as a result W performs simple harmonic motion (SHM).

Determine the maximum acceleration of W .

$F=(18 \times 0.04=) 0.72 \mathrm{~N}$ ;
acceleration $=\frac{0.72}{0.15}=4.8 \mathrm{~m} \mathrm{~s}^{-2}$ ;

Exam points

For an isolated sticking collision, conserve total momentum while allowing total kinetic energy to decrease.
Test a proposed collision outcome using signed momentum first; an inelastic outcome can involve two bodies separating at different speeds.
In a fall–collision–rise problem, use energy on either side of impact and momentum during sticking; conserving energy through impact can fail.
Use the combined mass after catching or coupling; choose the speed immediately before impact rather than an earlier speed.
To verify elasticity, find the unknown recoil speed from momentum and compare the sum of both bodies’ kinetic energies before and after.
After a sticking collision, calculate KE using the common speed and combined mass; state the energy decrease rather than only the final speed.
Explain embedding losses using deformation and transfer of kinetic energy to internal or elastic energy; the label “inelastic” alone is insufficient.

[Maximum number: 2]

This question is in two parts. Part 1 is about momentum and energy. Part 2 is about the de Broglie hypothesis and radioactive decay.

Jane and Joe are two ice skaters initially at rest on a horizontal skating rink. They are facing each other and Jane is holding a ball. Jane throws the ball to Joe who catches it. The speed at which the ball leaves Jane, measured relative to the ground, is $8.0 \mathrm{~ms}^{-1}$ . The following data are available.

\begin{aligned} & \text { Mass of Jane }=52 \mathrm{~kg} \\ & \text { Mass of Joe }=74 \mathrm{~kg} \\ & \text { Mass of ball }=1.3 \mathrm{~kg} \end{aligned}

Use the data to calculate the

speed V of Joe relative to the ground immediately after he catches the ball.

Exam points

For a particle sticking to a pivoted rod, subtract ½Icombinedω² from its initial ½mv²; include the attached particle’s moment of inertia.
For trucks that stick together, calculate the final kinetic energy with their combined mass and subtract it from the initial total.
Explain that the moving truck’s kinetic energy both gives the other truck motion and increases internal energy in the couplings.

[Maximum number: 3]

A horizontal rigid bar of length 2 R is pivoted at its centre. The bar is free to rotate in a horizontal plane about a vertical axis through the pivot. A point particle of mass M is attached to one end of the bar and a container is attached to the other end of the bar.

A point particle of mass $\frac{M}{3}$ moving with speed v at right angles to the rod collides with the container and gets stuck in the container. The system then starts to rotate about the vertical axis.

The mass of the rod and the container can be neglected.

Determine in terms of M and v the energy lost during the collision.

$\text { initial } K E=\frac{M v^{2}}{6}$

final $K E=\frac{M v^{2}}{24}$

energy loss $=\frac{M v^{2}}{8}$

Exam points

Distinguish tangential velocity from acceleration towards the centre when choosing or drawing vectors for uniform circular motion.
Explain acceleration at constant speed by a change in velocity direction; constant speed does not mean constant velocity.
For riders sharing an angular speed, find the speed at the rider’s own radius before using a = v²/r.

[Maximum number: 7]

Aibhe and Euan are sitting on opposite sides of a merry-go-round, which is rotating at constant speed around a fixed centre. The diagram below shows the view from above.

Aibhe is moving at speed $1.0 \mathrm{~m} \mathrm{~s}^{-1}$ relative to the ground.

(a)

Determine the magnitude of the velocity of Aibhe relative to

[ 2 ]

(i)

Euan.

[ 1 ]

2.0 or $0\left(\mathrm{~ms}^{-1}\right)$ ;

(ii)

the centre of the merry-go-round.

[ 1 ]

1.0 or $0\left(\mathrm{~ms}^{-1}\right)$ ;

(b)

Outline why Aibhe is accelerating even though she is moving at constant speed.

[ 2 ]

(c)

Draw an arrow on the diagram on page 29 to show the direction in which Aibhe is accelerating.

[ 1 ]

(d)

Aibhe moves so that she is sitting at a distance of 0.75 m from the centre of the merry-go-round, as shown below.

Euan pushes the merry-go-round so that he is again moving at $1.0 \mathrm{~ms}^{-1}$ relative to the ground.

[ 2 ]

(i)

Calculate the magnitude of Aibhe's acceleration.
Part 2 Charge-coupled devices (CCDs)

[ 2 ]

$\quad a\left(=\frac{v^{2}}{r}\right)=\frac{0.5^{2}}{0.75}$ ;
$a=0.33\left(\mathrm{~ms}^{-2}\right)$ ;

Marking guidance:

Allow ECF from (d)(i).
Award [2] for a bald correct answer.

Part 2 Charge-coupled devices (CCDs)

Exam points

Identify the real interaction supplying centripetal force: gravitational attraction in an orbit, tyre friction or the platform’s contact force.
Explain that an inward force changes velocity direction; without it an orbiting body would continue along a straight tangent.
For horizontal motion inside a bowl, direct the resultant towards the centre of the horizontal circle, not the centre of the bowl.
In a circular orbit the radial force is perpendicular to velocity, so it does no work despite remaining nonzero.
Explain orbital weightlessness using equal inward acceleration of astronaut and spacecraft, giving zero contact reaction despite gravity.
At the top of a vertical string circle, use T + mg = mv²/R; hence KE = R(T + mg)/2.
At the bottom of a vertical circle, add weight to mv²/R to obtain the string tension.
For the minimum bottom energy completing a vertical string circle, combine zero top tension with the 2R height gain.
At a convex crest, use mg − N = mv²/r; contact is lost at N = 0, giving vmax = √(gr).
At the top of a loop, use N + mg = mv²/r and justify continued contact with N > 0 or v > √(gr).
For constant-speed vertical motion driven by a rod, its force must balance weight as well as provide the inward resultant.
A swinging pendulum is not in equilibrium at its centre: its nonzero speed requires a radial resultant even when tangential acceleration is zero.
Use F = 2KE/r for force ratios when kinetic energy and circle radius both change.
At fixed tension and mass, use F = mω²r: doubling angular speed requires a quarter of the original radius.
For the same breaking tension and rope radius, keep mω² fixed; four times the mass requires half the angular speed.
At the same speed, scale circular force as mass divided by radius: 1.5 times the mass and twice the radius gives 0.75 times the force.
In the stated hanging-mass model, set radial tension Mg equal to mv²/r, or replace v using the rotation period.
On a level bend, set maximum static friction μmg equal to mv²/R to find vmax = √(μgR).
For an orbital-force calculation, use the masses of both attracting bodies and their centre-to-centre separation in GMm/r².
For an electron orbiting an alpha particle, equate 2ke²/r² to mev²/r, using electron mass and the alpha particle’s +2e charge.

[Maximum number: 1]

Aibhe and Euan are sitting on opposite sides of a merry-go-round, which is rotating at constant speed around a fixed centre. The diagram below shows the view from above.

Aibhe is moving at speed $1.0 \mathrm{~m} \mathrm{~s}^{-1}$ relative to the ground.

Identify the force that is causing Aibhe to move in a circle.

Exam points

Explain apparent outward motion using tangential inertia: without the inward force needed to turn, the upper body initially continues straight.

[Maximum number: 2]

Aibhe and Euan are sitting on opposite sides of a merry-go-round, which is rotating at constant speed around a fixed centre. The diagram below shows the view from above.

Aibhe is moving at speed $1.0 \mathrm{~m} \mathrm{~s}^{-1}$ relative to the ground.

The diagram below shows a side view of Aibhe and Euan on the merry-go-round.

Explain why Aibhe feels as if her upper body is being "thrown outwards", away from the centre of the merry-go-round.

Exam points

Use one circumference per revolution to write v = 2πr/T and rearrange for the period, converting the radius to metres.
Riders on a rigid rotating platform share angular speed and period, so halving the radius halves linear speed.
Calculate angular velocity from ω = 2π/T; convert a period in hours to seconds before reporting rad s⁻¹.

[Maximum number: 3]

Aibhe and Euan are sitting on opposite sides of a merry-go-round, which is rotating at constant speed around a fixed centre. The diagram below shows the view from above.

Aibhe is moving at speed $1.0 \mathrm{~m} \mathrm{~s}^{-1}$ relative to the ground.

Aibhe moves so that she is sitting at a distance of 0.75 m from the centre of the merry-go-round, as shown below.

Euan pushes the merry-go-round so that he is again moving at $1.0 \mathrm{~ms}^{-1}$ relative to the ground.

Determine Aibhe's speed relative to the ground.

Aibhe's period of revolution is the same as before;
from $v=\frac{2 \pi r}{T}$ , since r is halved, v is halved;
$v=0.5\left(\mathrm{~ms}^{-1}\right) ;$

Marking guidance:

Award [3] for a bald correct answer.

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A.2.15—Magnetic force FmNo questions yet

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