Forces and momentum

Topic A: Space, Time and Motion

Lesson A.2 Forces and Momentum | IB Physics HL

Kinematics describes motion; dynamics explains how interactions change it. This lesson connects free-body diagrams and Newton’s laws to momentum, impulse, collisions, explosions and circular motion. Always identify the object or system before deciding which forces and momenta to include.

Learning objectives

Describe forces as interactions and distinguish contact forces from field forces.
Draw free-body diagrams and resolve forces in one and two dimensions.
Explain and apply Newton’s three laws, including the identification of third-law pairs.
Use weight, normal force, static and dynamic friction, tension, Hooke’s law, buoyancy and viscous drag.
Apply momentum and impulse, including the area under a force–time graph.
Analyse collisions and explosions, considering both momentum and kinetic energy.
Conserve momentum independently in perpendicular directions in HL problems.
Apply circular-motion relations in horizontal and vertical planes.
Evaluate an investigation of resultant force and acceleration.

IB scope: A.2 has shared SL and HL content. HL quantitative collision and explosion questions may involve two dimensions. Circular motion is part of A.2, including non-uniform motion in vertical circles at the top and bottom. Quantitative banked-surface problems and simultaneous solution of momentum and energy equations for collisions are not required.

Notation and assumptions

Use $F_N$ for normal force, $F_f$ for friction, $\mu_s$ for static friction coefficient and $\mu_d$ for dynamic or sliding friction coefficient. Use $F_H$ for elastic restoring force, $F_d$ for viscous drag and $F_b$ for buoyancy. In this lesson, tension is written $F_T$ so that T can represent the period of circular motion.

$g=9.81\,\mathrm{m\,s^{-2}}$

Choose positive directions, retain extra digits during calculations and include directions for vector answers. Strings are light and inextensible and pulleys are frictionless with negligible inertia only when these assumptions are stated.

A.2.1 — Forces, resultant force and translational equilibrium

A force is an interaction that can change an object’s velocity or deform it. Force is a vector. The resultant force is the vector sum of the forces acting on the selected object or system.

$\mathbf{F}_{\mathrm{net}}=\sum\mathbf{F},\quad 1\,\mathrm{N}=1\,\mathrm{kg\,m\,s^{-2}}$

Translational equilibrium means zero resultant force and zero acceleration. An object may be stationary or move with constant velocity. Constant speed alone does not establish equilibrium: a body moving around a circle changes velocity direction.

$\sum F_x=0,\quad\sum F_y=0$

For a force of magnitude F at angle θ above a horizontal axis:

$F_x=F\cos\theta,\quad F_y=F\sin\theta$

Add components using their signs. For perpendicular resultant components:

$F_{\mathrm{net}}=\sqrt{F_x^2+F_y^2},\quad\tan\phi=\frac{F_y}{F_x}$

Use both component signs to select the correct direction and quadrant.

A.2.2 — Free-body diagrams

A free-body diagram shows all external forces on one selected body or system. Represent the body by a point or simple box. Draw labelled arrows for forces; acceleration and velocity may be shown separately, but they are not extra forces.

Choose the body or system boundary.
Identify the physical interaction responsible for each external force.
Draw the force arrows and label them.
Choose convenient axes and resolve forces.
Apply Newton’s second law along each axis.

Equal vertical arrows represent zero vertical acceleration. The longer pull arrow represents a resultant horizontal force to the right.

Avoid double counting: Resolve a force into components for calculation, but do not count the original force and its components as separate interactions. Do not add a separate “centripetal force” arrow alongside the actual forces supplying the inward resultant.

A.2.3 — Newton’s three laws

First law: inertia

In an inertial reference frame, an object remains at rest or moves with constant velocity unless a resultant external force acts. Mass measures inertia: a larger mass requires a larger resultant force for the same acceleration.

Second law: force changes momentum

The average resultant external force equals the change in momentum per unit time. For constant mass this reduces to the familiar force–acceleration relation.

$\mathbf{F}_{\mathrm{net,avg}}=\frac{\Delta\mathbf{p}}{\Delta t},\quad\mathbf{F}_{\mathrm{net}}=m\mathbf{a}\quad\text{(constant mass)}$

If a body exchanges mass, such as a rocket ejecting exhaust, include the momentum carried by the exchanged material when defining the system. Simply inserting a changing mass into $F=ma$ is not a complete analysis.

Third law: interaction pairs

If body A exerts a force on body B, body B simultaneously exerts an equal-magnitude force in the opposite direction on body A. The pair involves the same interaction and acts on different bodies.

$\mathbf{F}_{A\to B}=-\mathbf{F}_{B\to A}$

For a book resting on a table, Earth’s gravitational force on the book is paired with the book’s gravitational force on Earth. The table’s normal force on the book is paired with the book’s contact force on the table. Weight and normal force both act on the book and are not a third-law pair.

System boundary matters: Third-law forces do not cancel on either individual body. When both interacting bodies are included in one system, their interaction forces are internal and cancel in the system’s total-force sum.

A.2.4 — Contact and field forces

Forces and their models
Force	Direction and meaning	Model or condition
Weight	Gravitational force toward Earth locally	$F_g=mg$
Normal force	Perpendicular contact force from a surface	Find using the perpendicular force equation; not always mg.
Static friction	Opposes relative slipping that would otherwise occur at contact	$0\leq F_f\leq\mu_sF_N$
Dynamic friction	Opposes relative sliding at contact	$F_f=\mu_dF_N$
Tension	Pulls along a taut string, away from the body	Equal along an ideal light string over an ideal pulley.
Elastic restoring force	Opposes extension or compression from natural length	$F_H=-kx$ within the linear regime.
Buoyancy	Upward force from the pressure difference in a fluid	$F_b=\rho Vg$ ; V is displaced fluid volume.
Viscous drag on a small sphere	Opposes motion relative to the fluid	$F_d=6\pi\eta rv$ in the low-Reynolds-number regime.

The friction coefficients are dimensionless. Static friction adjusts to the required value until its limiting value is reached; it is not always equal to $\mu_sF_N$ . Friction opposes relative slipping at the contact, so it can act in the direction of a body’s motion, as when the ground pushes a walking person forward.

In Hooke’s law, k is the spring constant in $\mathrm{N\,m^{-1}}$ and x is the signed extension or compression. In Stokes’ drag law, η is dynamic viscosity in $\mathrm{Pa\,s}$ , r is sphere radius and v is speed relative to the fluid. This linear drag law is not a general model for every object moving through air or water.

Field-force connections

Gravitational, electric and magnetic forces can act without direct material contact. The following expressions connect this lesson to Topic D; detailed field calculations and direction rules are developed there.

$F_g=\frac{Gm_1m_2}{r^2},\quad F_e=k_e\frac{|q_1q_2|}{r^2}$

$\mathbf{F}_m=q(\mathbf{v}\times\mathbf{B}),\quad F_m=|q|vB\sin\theta$

$F_m=BIL\sin\theta$

The gravitational formula applies to point masses or suitable spherical bodies; r is centre-to-centre separation. Coulomb’s law describes point charges; like charges repel and unlike charges attract. Magnetic force on a moving charge is perpendicular to both its velocity and the field, with direction reversed for negative charge. For a straight current-carrying segment in a uniform field, L is its length and θ is the angle between conventional current and the field. The cross product records the magnetic-force direction. Here $k_e$ denotes the Coulomb constant, distinct from spring constant k.

Worked example 1 — Resultant force and acceleration

A 2.5 kg trolley is pulled horizontally with a force of 18 N. Friction opposing the motion is 6.0 N. Determine resultant force and acceleration.

Show worked answer

Given: Mass 2.5 kg, horizontal pull 18 N and opposing friction 6.0 N. Choose the pull direction as positive.

Step 1 — Find the resultant horizontal force. Subtract the opposing friction from the pull.

$F_{\mathrm{net}}=18-6.0=12\,\mathrm{N}$

Step 2 — Apply Newton’s second law.

$a=\frac{F_{\mathrm{net}}}{m}=\frac{12}{2.5}=4.8\,\mathrm{m\,s^{-2}}$

Final answers: Resultant force 12 N; acceleration 4.8 m/s². Both are in the direction of the pull.

Check: With zero vertical acceleration, normal force balances weight. The horizontal force used in Newton’s law is the resultant 12 N.

Worked example 2 — Sliding friction

A 12 kg crate slides across a horizontal floor under a horizontal pull. The coefficient of dynamic friction is 0.30. Calculate the normal force and friction.

Show worked answer

Given: Mass 12 kg and dynamic friction coefficient 0.30. The pull is horizontal and vertical acceleration is zero.

Step 1 — Use vertical force balance.

$F_N-mg=0$

$F_N=(12)(9.81)=117.72\,\mathrm{N}$

Step 2 — Calculate dynamic friction.

$F_f=\mu_dF_N=(0.30)(117.72)=35.3\,\mathrm{N}$

Final answers: Normal force approximately 118 N upward; friction approximately 35.3 N opposite the sliding direction.

Check: The normal force equals weight here because the pull has no vertical component. Values are shown to three significant figures for comparison; use the precision requested in an exam.

Worked example 3 — Hooke’s law

A spring has constant $k=250\,\mathrm{N\,m^{-1}}$ and is stretched by 0.080 m within its linear regime. Calculate the restoring-force magnitude.

Show worked answer

Given: Spring constant 250 N/m and extension 0.080 m. The spring remains within its linear regime. Choose positive x along the extension.

Step 1 — Apply Hooke’s law.

$F_H=-kx$

Step 2 — Substitute the extension.

$F_H=-(250)(0.080)=-20.0\,\mathrm{N}$

Final answers: Restoring-force magnitude 20.0 N, directed opposite the extension.

Check: The negative sign describes the restoring direction; the force magnitude is positive.

Worked example 4 — Buoyancy

A fully submerged object displaces $2.5\times10^{-3}\,\mathrm{m^3}$ of water of density $1000\,\mathrm{kg\,m^{-3}}$ . Calculate the buoyant force.

Show worked answer

Given: Displaced water volume 0.0025 m³ and water density 1000 kg/m³. The object is fully submerged.

Step 1 — Apply Archimedes’ principle. Buoyancy equals the weight of displaced fluid.

$F_b=\rho Vg$

Step 2 — Substitute the data.

$F_b=(1000)(2.5\times10^{-3})(9.81)=24.5\,\mathrm{N}$

Final answers: Buoyant force approximately 24.5 N upward.

Check: To decide whether the object rises or falls, compare buoyancy with weight and any other forces.

A.2.5 — Applying Newton’s laws

Inclined surfaces

For a slope at angle θ to the horizontal, choose axes parallel and perpendicular to the slope. Weight has components $mg\sin\theta$ down the slope and $mg\cos\theta$ into the slope. If there are no other perpendicular forces and no perpendicular acceleration:

$F_N=mg\cos\theta$

For a body sliding down the slope with dynamic friction and no other parallel forces:

$ma=mg\sin\theta-\mu_dmg\cos\theta$

Lifts and apparent weight

A scale measures its normal contact force. With upward positive:

$F_N-mg=ma_y$

Upward acceleration gives a larger scale reading; downward acceleration gives a smaller one. The direction of velocity does not decide the reading. An upward-moving lift can have downward acceleration while slowing.

Connected bodies

Draw a diagram for each body. Use a common acceleration magnitude only when the string constraint permits it. Alternatively, treat the connected bodies as one system to eliminate internal tensions, then use an individual-body equation to find the tension.

Worked example 5 — A crate sliding down a slope

A 5.00 kg crate slides down a 30.0° incline with $\mu_d=0.200$ . No other forces act. Find normal force, friction and acceleration down the slope.

Show worked answer

Given: Mass 5.00 kg, slope angle 30.0° and dynamic friction coefficient 0.200. Choose down the slope as positive.

Step 1 — Resolve weight perpendicular to the slope. There is no perpendicular acceleration.

$F_N=mg\cos\theta=(5.00)(9.81)\cos30.0^\circ=42.5\,\mathrm{N}$

Step 2 — Find friction. The crate slides downward, so friction acts up the slope.

$F_f=\mu_dF_N=0.200F_N=8.50\,\mathrm{N}$

Step 3 — Apply Newton’s law along the slope.

$ma=mg\sin\theta-F_f$

$a=g(\sin30.0^\circ-0.200\cos30.0^\circ)=3.21\,\mathrm{m\,s^{-2}}$

Final answers: Normal force 42.5 N; friction 8.50 N up the slope; acceleration 3.21 m/s² down the slope.

Check: Normal force balances only the perpendicular component of weight. Keep unrounded values when calculating acceleration.

Worked example 6 — A scale in an accelerating lift

A 60.0 kg person stands on a scale in a lift accelerating upward at $1.50\,\mathrm{m\,s^{-2}}$ . Calculate the scale’s force reading.

Show worked answer

Given: Person’s mass 60.0 kg and upward acceleration 1.50 m/s². Choose upward as positive.

Step 1 — Identify the forces on the person. The scale pushes upward with normal force; weight acts downward.

$F_N-mg=ma_y$

Step 2 — Rearrange and substitute.

$F_N=m(g+a_y)$

$F_N=60.0(9.81+1.50)=679\,\mathrm{N}$

Final answers: Scale force reading 679 N.

Check: The person’s weight is still approximately 589 N. The larger contact force produces upward acceleration.

Worked example 7 — Two connected hanging masses

Masses of 3.00 kg and 2.00 kg hang on opposite sides of a light, inextensible string over a frictionless pulley of negligible inertia. Find acceleration and tension.

Show worked answer

Given: Hanging masses 3.00 kg and 2.00 kg. The string and pulley are ideal, so tension is equal on both sides and acceleration magnitudes are equal.

Step 1 — Choose a positive direction for each mass. Take downward as positive for the 3.00 kg mass and upward as positive for the 2.00 kg mass.

$3.00g-F_T=3.00a$

$F_T-2.00g=2.00a$

Step 2 — Add the equations to eliminate tension.

$(3.00-2.00)g=(3.00+2.00)a$

$a=\frac{(3.00-2.00)(9.81)}{5.00}=1.96\,\mathrm{m\,s^{-2}}$

Step 3 — Find tension from the lighter mass. Use unrounded acceleration.

$F_T=2.00(g+a)=23.5\,\mathrm{N}$

Final answers: Acceleration magnitude 1.96 m/s²: the 3.00 kg mass moves downward and the 2.00 kg mass upward. Tension 23.5 N.

Check: Tension lies between the two weights. Equal tension follows from the ideal string and pulley assumptions.

Terminal speed with buoyancy and viscous drag

For a small dense sphere falling slowly through a stationary liquid in the Stokes regime, terminal speed occurs when upward drag and buoyancy balance weight.

$mg=\rho Vg+6\pi\eta rv_t$

$v_t=\frac{mg-\rho Vg}{6\pi\eta r}$

This downward-terminal-speed expression requires $mg>\rho Vg$ . The sphere continues moving, although acceleration and resultant force are zero. If buoyancy is negligible, the balance simplifies to drag equals weight.

A.2.6 — Momentum and impulse

Linear momentum is mass multiplied by velocity. It is a vector, so its sign or direction matters.

$\mathbf{p}=m\mathbf{v},\quad[p]=\mathrm{kg\,m\,s^{-1}}$

Impulse is the change in momentum caused by a resultant force. For a constant force, or the average resultant force over the interval:

$\mathbf{J}=\Delta\mathbf{p}=\mathbf{F}_{\mathrm{net,avg}}\Delta t$

$1\,\mathrm{N\,s}=1\,\mathrm{kg\,m\,s^{-1}}$

The signed area under a resultant force–time graph gives the corresponding momentum change. Find rectangular, triangular or trapezium areas as appropriate. If the force changes, do not automatically multiply its maximum value by the duration.

For the same momentum change, a longer stopping time gives a smaller average force. This helps explain airbags, crash barriers and bending the knees on landing.

Worked example 8 — A ball rebounding from a wall

A 0.15 kg ball strikes a wall horizontally at $12\,\mathrm{m\,s^{-1}}$ and rebounds at $8.0\,\mathrm{m\,s^{-1}}$ in the opposite direction. Contact lasts 0.020 s. Find horizontal impulse and average horizontal force on the ball.

Show worked answer

Given: Mass 0.15 kg, initial speed 12 m/s, rebound speed 8.0 m/s and contact time 0.020 s. Choose the initial direction as positive.

Step 1 — Assign signed velocities.

$u=+12\,\mathrm{m\,s^{-1}},\quad v=-8.0\,\mathrm{m\,s^{-1}}$

Step 2 — Find the momentum change.

$J_x=m(v-u)=0.15(-8.0-12)=-3.0\,\mathrm{N\,s}$

Step 3 — Divide impulse by contact time.

$F_{\mathrm{net,avg},x}=\frac{J_x}{\Delta t}=\frac{-3.0}{0.020}=-150\,\mathrm{N}$

Final answers: Horizontal impulse 3.0 N s and average horizontal force 150 N, both opposite the initial motion.

Check: Reversal requires signed velocities. Neglecting other horizontal forces, the wall produces this average force and experiences an equal and opposite force from the ball.

A.2.7 — Conservation of momentum and collision energy

The total momentum of a system is conserved if the resultant external impulse during the event is zero or negligible. Internal forces can change individual momenta while leaving their vector sum unchanged.

$\sum\mathbf{p}_{\mathrm{before}}=\sum\mathbf{p}_{\mathrm{after}}$

For two bodies in one dimension:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

Short collision durations can make external impulse negligible even when some external forces are present. Check the system and the interval rather than assuming momentum is always conserved for every object.

Collisions in a system with negligible external impulse
Type	Momentum	Kinetic energy
Elastic	Conserved	Total kinetic energy conserved
Ordinary inelastic collision	Conserved	Some kinetic energy becomes internal energy, deformation or sound
Perfectly inelastic	Conserved; bodies stick together	Greatest kinetic-energy loss for fixed initial conditions in a two-body collision without energy release

$E_k=\frac12mv^2$

Total energy is conserved even when kinetic energy is not. In explosions, stored internal energy can become kinetic energy. Momentum conservation does not imply kinetic-energy conservation.

Worked example 9 — Carts sticking together

A 0.40 kg cart moving at $2.5\,\mathrm{m\,s^{-1}}$ strikes a stationary 0.60 kg cart and sticks to it. External horizontal impulse is negligible. Find their common velocity and the kinetic-energy decrease.

Show worked answer

Given: First cart: 0.40 kg at 2.5 m/s. Second cart: 0.60 kg at rest. The carts stick together and external horizontal impulse is negligible.

Step 1 — Conserve system momentum. Choose the first cart’s initial direction as positive.

$m_1u_1+m_2u_2=(m_1+m_2)v$

$(0.40)(2.5)+(0.60)(0)=(1.00)v$

$v=1.0\,\mathrm{m\,s^{-1}}$

Step 2 — Calculate kinetic energy before and after.

$E_{k,i}=\frac12(0.40)(2.5)^2=1.25\,\mathrm{J}$

$E_{k,f}=\frac12(1.00)(1.0)^2=0.50\,\mathrm{J}$

Step 3 — Find the decrease.

$E_{k,i}-E_{k,f}=1.25-0.50=0.75\,\mathrm{J}$

Final answers: Common velocity 1.0 m/s in the first cart’s initial direction; kinetic-energy decrease 0.75 J.

Check: This is a perfectly inelastic collision. The decrease becomes other forms of energy; total energy is conserved.

Worked example 10 — An explosion from rest

A 3.0 kg object initially at rest separates into 1.0 kg and 2.0 kg pieces. The 1.0 kg piece moves right at $6.0\,\mathrm{m\,s^{-1}}$ . External impulse is negligible. Find the other piece’s velocity and the total final kinetic energy.

Show worked answer

Given: A stationary 3.0 kg object splits into 1.0 kg and 2.0 kg pieces. The lighter piece moves right at 6.0 m/s; external impulse is negligible. Choose right as positive.

Step 1 — Conserve total momentum. Initial momentum is zero.

$0=(1.0)(6.0)+(2.0)v_2$

$v_2=-3.0\,\mathrm{m\,s^{-1}}$

Step 2 — Add the kinetic energies of both pieces.

$E_{k,f}=\frac12(1.0)(6.0)^2+\frac12(2.0)(3.0)^2$

$E_{k,f}=18+9.0=27\,\mathrm{J}$

Final answers: The 2.0 kg piece moves left at 3.0 m/s. Total final kinetic energy is 27 J.

Check: Equal and opposite momenta have different speeds when masses differ. Stored internal energy supplies the kinetic energy.

A.2.8 — Two-dimensional momentum: HL application

Momentum is conserved separately along perpendicular axes when the corresponding external impulses are negligible. Resolve each velocity before adding momenta.

$\sum m u_x=\sum m v_x,\quad\sum m u_y=\sum m v_y$

For a body moving at angle θ to the positive x-axis, use $v_x=v\cos\theta$ and $v_y=v\sin\theta$ . Components can be negative. Do not conserve scalar speeds or add momentum magnitudes as if all vectors point in the same direction.

Worked example 11 — An oblique collision

A 0.200 kg particle moves east at $4.00\,\mathrm{m\,s^{-1}}$ and collides with a stationary 0.500 kg particle. Afterwards, the first particle has velocity components $v_{1x}=1.50\,\mathrm{m\,s^{-1}}$ east and $v_{1y}=2.00\,\mathrm{m\,s^{-1}}$ north. Find the second particle’s velocity and determine whether the collision is elastic. External impulse is negligible.

Show worked answer

Given: First particle: 0.200 kg initially at 4.00 m/s east. Second particle: 0.500 kg initially at rest. After collision, the first has components 1.50 m/s east and 2.00 m/s north. External impulse is negligible.

Step 1 — Conserve east–west momentum. Choose east as positive x and north as positive y.

$(0.200)(4.00)=(0.200)(1.50)+(0.500)v_{2x}$

$v_{2x}=1.00\,\mathrm{m\,s^{-1}}$

Step 2 — Conserve north–south momentum.

$0=(0.200)(2.00)+(0.500)v_{2y}$

$v_{2y}=-0.800\,\mathrm{m\,s^{-1}}$

Step 3 — Find speed and direction.

$v_2=\sqrt{1.00^2+0.800^2}=1.28\,\mathrm{m\,s^{-1}}$

$\phi=\tan^{-1}(0.800/1.00)=38.7^\circ$

Step 4 — Compare total kinetic energies. Use the squared component sum for each particle.

$E_{k,i}=\frac12(0.200)(4.00)^2=1.60\,\mathrm{J}$

$E_{k,f}=\frac12(0.200)(1.50^2+2.00^2)+\frac12(0.500)(1.00^2+0.800^2)$

$E_{k,f}=1.035\,\mathrm{J}$

Final answers: Second particle: 1.28 m/s at 38.7° south of east. The collision is inelastic; kinetic energy decreases by 0.565 J.

Check: Both momentum components are conserved, while kinetic energy decreases.

A.2.9 — Circular motion

In uniform circular motion, speed remains constant but velocity changes direction. The acceleration and resultant force point toward the centre.

$v=\frac{2\pi r}{T}=\omega r,\quad\omega=\frac{2\pi}{T}$

$a_c=\frac{v^2}{r}=\omega^2r=\frac{4\pi^2r}{T^2}$

$F_{\mathrm{net,radial}}=ma_c=\frac{mv^2}{r}$

T is period in seconds and ω is angular speed in $\mathrm{rad\,s^{-1}}$ . “Centripetal” describes the inward resultant, which may be supplied by tension, friction, gravity, a normal force or a combination. It is not an additional physical interaction.

Horizontal circles

For a car on a level road, static friction can supply the horizontal inward force while the normal force balances weight. At the limiting speed before slipping:

$\frac{mv^2}{r}=\mu_smg,\quad v_{\max}=\sqrt{\mu_sgr}$

This expression assumes a level surface, no additional vertical forces and the simple static-friction model.

Non-uniform and vertical circular motion

If speed changes, there is also a tangential acceleration. The inward radial component remains $v^2/r$ . In a vertical circle, gravitational potential energy and speed generally change, so do not use one constant speed for the entire motion.

For a mass on a taut string in a vertical circle, choose inward as positive at each point. At the top, both tension and weight point inward; at the bottom, tension points inward and weight points outward.

$\text{top:}\quad F_{T,\mathrm{top}}+mg=\frac{mv_{\mathrm{top}}^2}{r}$

$\text{bottom:}\quad F_{T,\mathrm{bottom}}-mg=\frac{mv_{\mathrm{bottom}}^2}{r}$

A string can pull but cannot push. The limiting speed for remaining taut at the top follows from zero limiting tension:

$v_{\mathrm{top,min}}=\sqrt{gr}$

This is the minimum speed at the top, not the minimum speed at the bottom. Energy relationships from A.3 can connect speeds at different heights when the relevant energy-loss assumptions hold.

Worked example 12 — Required centripetal force

A 0.50 kg mass moves in a horizontal circle of radius 0.80 m at constant speed $4.0\,\mathrm{m\,s^{-1}}$ . Calculate centripetal acceleration and the required inward resultant force.

Show worked answer

Given: Mass 0.50 kg, circle radius 0.80 m and constant speed 4.0 m/s.

Step 1 — Calculate centripetal acceleration.

$a_c=\frac{v^2}{r}=\frac{(4.0)^2}{0.80}=20\,\mathrm{m\,s^{-2}}$

Step 2 — Find the required radial resultant force.

$F_{\mathrm{net,radial}}=ma_c=(0.50)(20)=10\,\mathrm{N}$

Final answers: Acceleration 20 m/s² and radial resultant force 10 N, both toward the centre.

Check: Constant speed does not mean constant velocity. The physical setup determines which interactions provide the inward resultant.

Worked example 13 — Tension at the top and bottom

A 0.200 kg mass moves on a taut 0.800 m string in a vertical circle. Its speed is $3.00\,\mathrm{m\,s^{-1}}$ at the top and $6.36\,\mathrm{m\,s^{-1}}$ at the bottom. Find tension at each point.

Show worked answer

Given: Mass 0.200 kg, string length and radius 0.800 m, top speed 3.00 m/s and bottom speed 6.36 m/s. Choose inward as positive at each point.

Step 1 — Write the radial equation at the top. Weight and tension both point inward.

$F_{T,\mathrm{top}}+mg=\frac{mv_{\mathrm{top}}^2}{r}$

$F_{T,\mathrm{top}}=\frac{(0.200)(3.00)^2}{0.800}-(0.200)(9.81)=0.288\,\mathrm{N}$

Step 2 — Write the radial equation at the bottom. Tension points inward; weight points outward.

$F_{T,\mathrm{bottom}}-mg=\frac{mv_{\mathrm{bottom}}^2}{r}$

$F_{T,\mathrm{bottom}}=\frac{(0.200)(6.36)^2}{0.800}+(0.200)(9.81)=12.1\,\mathrm{N}$

Final answers: Tension at the top 0.288 N; tension at the bottom 12.1 N.

Check: Both tensions are positive. Weight helps supply the inward force at the top; at the bottom, tension must overcome weight and provide the radial resultant.

A.2.10 — Common misconceptions and exam technique

“Motion requires a forward resultant force.” Constant velocity requires zero resultant force.
“Weight and normal force are always equal.” Apply the relevant force equation; lifts, slopes and angled pulls often change the normal force.
“Static friction always equals its limiting value.” It adjusts up to that limit.
“Third-law forces cancel on one body.” They act on different bodies.
“Momentum conservation means kinetic energy is conserved.” These are different conditions.
“Rebound impulse uses the difference of speed magnitudes.” Use signed velocities; reversal can produce a large momentum change.
“Centripetal force is an extra force.” It is the inward resultant of actual interactions.

Problem-solving checklist

Choose the body or system and state the interval.
Draw a labelled free-body diagram or momentum diagram.
Set positive directions and resolve vectors.
Decide whether to use Newton’s law, impulse or momentum conservation.
State assumptions: friction model, external impulse, string/pulley idealization or circular constraint.
Check units, directions and whether the result is physically possible.

A.2.11 — IB-style practice

These are original practice questions. Expand each answer after attempting the question.

Multiple choice

Multiple choice 1 [1 mark]

A box remains at rest while pushed horizontally with 4.0 N. Its maximum static friction is 10 N. What is its actual friction force?

0 N.
4.0 N opposite the push.
10 N opposite the push.
14 N opposite the push.

Show answer and explanation

Answer: B. Static friction adjusts to balance the 4.0 N push; it has not reached its 10 N limit.

Multiple choice 2 [1 mark]

Which force is the third-law partner of Earth’s gravitational force on a falling ball?

Air resistance on the ball.
The ball’s gravitational force on Earth.
The resultant force on the ball.
The ball’s force on the surrounding air.

Show answer and explanation

Answer: B. A third-law pair involves the same interaction on different bodies: Earth attracts the ball and the ball attracts Earth.

Multiple choice 3 [1 mark]

A resultant force–time pulse is triangular, with peak 10 N and duration 0.040 s. What impulse does it produce?

0.100 N s.
0.400 N s.
0.200 N s.
250 N s.

Show answer and explanation

Answer: C. The impulse is the triangular area: one-half × 10 × 0.040 = 0.200 N s.

Multiple choice 4 [1 mark]

Two carts collide and stick together with negligible external impulse. Which statement is correct?

Momentum and total kinetic energy are both conserved.
Momentum is conserved, but total kinetic energy decreases unless there was no relative motion.
Kinetic energy is conserved but momentum is not.
Both momentum and total energy disappear.

Show answer and explanation

Answer: B. The collision is perfectly inelastic. Momentum is conserved for the system; kinetic energy transfers to other forms.

Multiple choice 5 [1 mark]

A body moves at constant speed around a circle. What is the direction of its acceleration?

Along its velocity.
Away from the centre.
It has no acceleration.
Toward the centre.

Show answer and explanation

Answer: D. Velocity changes direction continuously, giving inward centripetal acceleration even though speed is constant.

Multiple choice 6 [1 mark]

A dense sphere falls through a liquid at terminal speed. Which relation applies?

Weight equals drag plus buoyancy.
Weight is zero.
Drag and buoyancy are both zero.
The resultant downward force equals weight.

Show answer and explanation

Answer: A. At terminal speed, upward drag and buoyancy balance downward weight; resultant force and acceleration are zero.

Extended-response practice

Show working, units and directions. Mark allocations are suggested guides.

Practice 1 — Resolving a pulling force [5 marks]

A 10.0 kg crate slides right on a horizontal floor. A 40.0 N pull acts 30.0° above horizontal, and $\mu_d=0.200$ . Find normal force, friction and horizontal acceleration. Vertical acceleration is zero.

Show worked answer

Vertical balance gives $F_N+40.0\sin30.0^\circ-mg=0$ . [1]

$F_N=98.1-20.0=78.1\,\mathrm{N}$

Correct normal force. [1]

$F_f=0.200(78.1)=15.6\,\mathrm{N}$

Correct friction opposite sliding. [1]

Use $ma_x=40.0\cos30.0^\circ-F_f$ . [1]

$a_x=1.90\,\mathrm{m\,s^{-2}}\text{ right}$

Correct acceleration. [1]

Practice 2 — Third-law pairs [3 marks]

A book rests on a table. Identify the third-law partner of its weight and the third-law partner of the table’s normal force on it. Explain why weight and normal force are not a third-law pair.

Show worked answer

The book’s gravitational force on Earth is paired with Earth’s gravitational force on the book. [1] The book’s contact force on the table is paired with the table’s normal force on the book. [1] Weight and normal force act on the same body and arise from different interactions. [1]

Practice 3 — Rebound impulse [4 marks]

A 0.200 kg ball moves right at $5.00\,\mathrm{m\,s^{-1}}$ , then rebounds left at $3.00\,\mathrm{m\,s^{-1}}$ . The collision lasts 0.0400 s. Find horizontal impulse and average resultant horizontal force.

Show worked answer

Use right-positive signs, $u=+5.00$ and $v=-3.00$ . [1]

$J_x=(0.200)(-3.00-5.00)=-1.60\,\mathrm{N\,s}$

Correct impulse. [1] Use $F_{\mathrm{avg},x}=J_x/\Delta t$ . [1]

$F_{\mathrm{avg},x}=-40.0\,\mathrm{N}$

Force is leftward. [1]

Practice 4 — A collision with opposite initial velocities [5 marks]

A 0.300 kg cart moves right at $4.00\,\mathrm{m\,s^{-1}}$ and a 0.200 kg cart moves left at $1.00\,\mathrm{m\,s^{-1}}$ . They stick together. External horizontal impulse is negligible. Find common velocity and kinetic-energy decrease.

Show worked answer

Take right positive and conserve signed momentum. [1]

$(0.300)(4.00)+(0.200)(-1.00)=(0.500)v$

$v=2.00\,\mathrm{m\,s^{-1}}\text{ right}$

Correct velocity. [1]

$E_{k,i}=\frac12(0.300)(4.00)^2+\frac12(0.200)(1.00)^2=2.50\,\mathrm{J}$

Correct initial kinetic energy. [1]

$E_{k,f}=\frac12(0.500)(2.00)^2=1.00\,\mathrm{J}$

Correct final kinetic energy. [1] Decrease is 1.50 J. [1]

Practice 5 — Two-dimensional explosion: HL [5 marks]

A stationary object separates into three pieces with negligible external impulse. A 1.00 kg piece moves east at $3.00\,\mathrm{m\,s^{-1}}$ , and a 2.00 kg piece moves north at $2.00\,\mathrm{m\,s^{-1}}$ . Find the velocity of the third piece, whose mass is 2.00 kg.

Show worked answer

Total initial momentum is zero. Conserve each component. [1]

$0=3.00+2.00v_{3x},\quad v_{3x}=-1.50\,\mathrm{m\,s^{-1}}$

Correct x-component. [1]

$0=4.00+2.00v_{3y},\quad v_{3y}=-2.00\,\mathrm{m\,s^{-1}}$

Correct y-component. [1]

$v_3=\sqrt{1.50^2+2.00^2}=2.50\,\mathrm{m\,s^{-1}}$

Correct speed. [1] Direction is $\tan^{-1}(2.00/1.50)=53.1^\circ$ south of west. [1]

Practice 6 — Circular motion from the period [4 marks]

A 0.250 kg mass moves uniformly in a circle of radius 0.600 m, completing one revolution in 1.20 s. Find its speed and inward resultant force.

Show worked answer

Use $v=2\pi r/T$ . [1]

$v=\frac{2\pi(0.600)}{1.20}=3.14\,\mathrm{m\,s^{-1}}$

Correct speed. [1] Use $F=mv^2/r$ . [1]

$F=4.11\,\mathrm{N}\text{ inward}$

Correct inward force, using unrounded speed. [1]

Practice 7 — Viscous drag and terminal speed [4 marks]

A small sphere falls in the Stokes regime through a liquid at terminal speed. Its weight is $3.00\times10^{-4}\,\mathrm{N}$ , buoyant force is $1.00\times10^{-4}\,\mathrm{N}$ and radius is $1.00\times10^{-3}\,\mathrm{m}$ . The liquid viscosity is $1.00\,\mathrm{Pa\,s}$ . Find drag and terminal speed.

Show worked answer

Use terminal force balance $F_d+F_b=mg$ . [1]

$F_d=(3.00-1.00)\times10^{-4}=2.00\times10^{-4}\,\mathrm{N}$

Drag acts upward. [1] Use $v_t=F_d/(6\pi\eta r)$ . [1]

$v_t=\frac{2.00\times10^{-4}}{6\pi(1.00)(1.00\times10^{-3})}=1.06\times10^{-2}\,\mathrm{m\,s^{-1}}$

Speed is 0.0106 m/s, with downward velocity. [1]

Practice 8 — A vertical-circle limiting condition [4 marks]

A mass on a light string moves in a vertical circle of radius 0.500 m. Determine the minimum speed at the top for the string to remain taut. Explain why a negative calculated tension is not physically possible for a string.

Show worked answer

At the top, $F_T+mg=mv^2/r$ . [1] The limiting condition is $F_T=0$ . [1]

$v_{\mathrm{top,min}}=\sqrt{gr}=\sqrt{(9.81)(0.500)}=2.21\,\mathrm{m\,s^{-1}}$

Correct limiting speed. [1] A string cannot push: a negative tension would mean the assumed taut circular motion is impossible and the string becomes slack. [1]

A.2.12 — Lab investigation: force and acceleration

Research question

How does the acceleration of a trolley depend on the applied horizontal force when its total moving mass is held constant?

Use a level track, a motion sensor and a horizontal force sensor or calibrated pulling system. Measure the force on the trolley while it moves. Keep trolley mass and surface conditions fixed. Repeat trials, check sensor zero and avoid runs in which the trolley hits the end of the track.

Illustrative mean measurements
Applied horizontal force / N	Acceleration / m s⁻²
0.600	0.800
1.000	1.600
1.400	2.400
1.800	3.200
2.200	4.000

Model the opposing friction force as approximately constant over these trials.

Derive a linear relationship for applied force plotted vertically against acceleration horizontally.
Use the first and last points to estimate trolley mass and opposing friction.
Explain why the intercept is non-zero even if Newton’s second law is correct.
Describe how to estimate uncertainty in the mass from a graph.
Identify a systematic error and a realistic improvement.

Show investigation answers

(a) Along the direction of motion:

$F_{\mathrm{applied}}-F_f=ma,\quad F_{\mathrm{applied}}=ma+F_f$

Gradient is mass in kg; vertical intercept is opposing friction in N.

(b)

$m=\frac{2.200-0.600}{4.000-0.800}=0.500\,\mathrm{kg}$

$F_f=0.600-(0.500)(0.800)=0.200\,\mathrm{N}$

The other points agree with this idealized model. In a real experiment, use a best-fit line through all data with uncertainty bars.

(c) Applied force is not the resultant force. Some of it balances friction. The law relates acceleration to the resultant $F_{\mathrm{applied}}-F_f$ , so a positive intercept is consistent with Newton’s second law.

(d) Include uncertainty bars on force and acceleration. Draw the steepest and shallowest acceptable straight lines consistent with them. An approximate gradient uncertainty is:

$\Delta m\approx\frac{m_{\max}-m_{\min}}{2}$

No numerical uncertainty can be calculated from this table alone because measurement uncertainties have not been supplied.

(e) A tilted track adds a component of weight along the track; a force-sensor offset shifts readings. Level the track and zero/calibrate the sensors. Repeats assess random scatter. If using a hanging mass to pull the trolley, do not equate its weight with string tension while it accelerates; use the full connected-system model or measure tension directly.

A.2.13 — Strength, care and responsibility

Scripture — Psalm 18:2, KJV

“The LORD is my rock, and my fortress, and my deliverer.”

Creation: The study of forces reveals an ordered world in which interactions can be investigated carefully. Christians receive the ability to reason, experiment and understand as gifts from God, to be used with gratitude and responsibility.

The Fall: Human strength and knowledge can be used selfishly or to harm others. Scripture reminds us that sin affects our choices and relationships. Honest data, fair treatment of classmates and care for those who need help are practical ways to resist that selfishness.

Redemption: Jesus Christ gave His life for our sins and rose again. His saving love invites us to trust Him as our Saviour and Lord. We do not earn God’s forgiveness through strength or academic achievement; we receive His grace and learn to serve others.

Restoration: God’s promise of renewal gives hope and purpose to our work. Use your learning to protect life, solve problems and build up your community. Read the Bible, pray and connect with other believers as you grow in faith.

Reflection: Where can you use your abilities to help someone rather than simply advance yourself? How can you practise integrity when a result differs from your expectation?

Prayer: Lord, thank You for being our strength and deliverer. Help me study carefully, act honestly and serve others with the abilities You have given me. Lead me to trust Jesus and grow in His love. Amen.

Lesson summary and readiness checklist

Draw forces on the selected body and add them as vectors.
Use $\mathbf{F}_{\mathrm{net}}=m\mathbf{a}$ for constant mass.
Identify third-law pairs by naming both bodies and the interaction.
Distinguish adjustable static friction from dynamic friction.
Impulse is momentum change and the signed force–time area.
Conserve system momentum only when external impulse is negligible.
Compare kinetic energies separately to classify collisions.
For HL collisions and explosions, conserve x- and y-momentum independently.
In circular motion, the actual forces must supply the inward resultant.
At the top and bottom of a vertical circle, use the correct inward direction and local speed.

Next lesson: A.3 — Work, Energy and Power develops the energy approach to motion and interactions.

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Forces and momentum

Topic A: Space, Time and Motion

Learning objectives

Notation and assumptions

A.2.1 — Forces, resultant force and translational equilibrium

A.2.2 — Free-body diagrams

A.2.3 — Newton’s three laws

First law: inertia

Second law: force changes momentum

Third law: interaction pairs

A.2.4 — Contact and field forces

Field-force connections

Worked example 1 — Resultant force and acceleration

Worked example 2 — Sliding friction

Worked example 3 — Hooke’s law

Worked example 4 — Buoyancy

A.2.5 — Applying Newton’s laws

Inclined surfaces

Lifts and apparent weight

Connected bodies

Worked example 5 — A crate sliding down a slope

Worked example 6 — A scale in an accelerating lift

Worked example 7 — Two connected hanging masses

Terminal speed with buoyancy and viscous drag

A.2.6 — Momentum and impulse

Worked example 8 — A ball rebounding from a wall

A.2.7 — Conservation of momentum and collision energy

Worked example 9 — Carts sticking together

Worked example 10 — An explosion from rest

A.2.8 — Two-dimensional momentum: HL application

Worked example 11 — An oblique collision

A.2.9 — Circular motion

Horizontal circles

Non-uniform and vertical circular motion

Worked example 12 — Required centripetal force

Worked example 13 — Tension at the top and bottom

A.2.10 — Common misconceptions and exam technique

Problem-solving checklist

A.2.11 — IB-style practice

Multiple choice

Multiple choice 1 [1 mark]

Multiple choice 2 [1 mark]

Multiple choice 3 [1 mark]

Multiple choice 4 [1 mark]

Multiple choice 5 [1 mark]

Multiple choice 6 [1 mark]

Extended-response practice

Practice 1 — Resolving a pulling force [5 marks]

Practice 2 — Third-law pairs [3 marks]

Practice 3 — Rebound impulse [4 marks]

Practice 4 — A collision with opposite initial velocities [5 marks]

Practice 5 — Two-dimensional explosion: HL [5 marks]

Practice 6 — Circular motion from the period [4 marks]

Practice 7 — Viscous drag and terminal speed [4 marks]

Practice 8 — A vertical-circle limiting condition [4 marks]

A.2.12 — Lab investigation: force and acceleration

Research question

A.2.13 — Strength, care and responsibility

Scripture — Psalm 18:2, KJV

Lesson summary and readiness checklist

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