Mechanics — IB Physics SL Study Guide

For: IB Physics SL candidates sitting IB Physics SL — Topic 2 (Mechanics).

Covers: 1D kinematics, projectile motion, Newton’s three laws, static/kinetic friction and tension, linear momentum conservation, impulse, work-energy theorem, and power/efficiency for the 2024 IB DP Physics SL syllabus.

You should already know: Basic algebra, vectors as arrows, free-body diagrams.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Physics SL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Mechanics?

Mechanics is the foundational branch of physics that describes the motion of objects, the forces acting on them, and the relationship between motion and force. It is split into two core subfields: kinematics (the description of motion without reference to forces) and dynamics (the analysis of how forces cause or change motion). This topic forms the base of all IB Physics Theme A (Space, time and motion) content and is applied across later units including electromagnetism, thermal physics, and energy production.

2. Kinematics in 1D — displacement, velocity, acceleration

Kinematics in 1 dimension describes motion along a straight line, using three core vector quantities:

• Displacement ($\Delta s$): The vector change in position of an object, calculated as $\Delta s=s_f-s_i$ where $s_f$ is final position and $s_i$ is initial position. Units are meters (m). It is distinct from distance, a scalar that measures total path length travelled.
• Velocity ($v$): The rate of change of displacement. Average velocity is $v_{avg}=\frac{\Delta s}{\Delta t}$, while instantaneous velocity is the limit of average velocity as $\Delta t$ approaches 0. Units are meters per second (m/s). It is distinct from speed, a scalar measure of total distance travelled over time.
• Acceleration ($a$): The rate of change of velocity, calculated as $a=\frac{\Delta v}{\Delta t}$. Units are meters per second squared (m/s²). Acceleration can be zero even if speed is constant (e.g. an object moving at constant speed changing direction, though this does not occur in 1D motion).

For motion with constant acceleration, you can use the four SUVAT equations: $$v=u+at$$ $$s=ut+\frac{1}{2}a{t}^2$$ $$v^2=u^2+2as$$ $$s=\frac{(u+v)}{2}t$$ Where $u$ = initial velocity, $t$ = time elapsed.

Worked Example

A cyclist accelerates from rest at $1.5{\text{ m/s}}^2$ for 8 seconds. Calculate their final velocity and total displacement.

1. Identify known values: $u=0$, $a=1.5{\text{ m/s}}^2$, $t=8\text{ s}$
2. Calculate final velocity: $v=0+(1.5)(8)=12\text{ m/s}$
3. Calculate displacement: $s=0+\frac{1}{2}(1.5)(8{)}^2=48\text{ m}$

Exam tip: Always explicitly define a positive direction at the start of any kinematics problem (e.g. upwards = positive for vertical motion) so you assign the correct sign to acceleration due to gravity ($g=-9.81{\text{ m/s}}^2$ if upwards is positive).

3. Projectile motion — independence of horizontal and vertical components

The core principle of projectile motion is that horizontal and vertical motion are completely independent of one another, with no cross-effect between the two axes (when air resistance is ignored, the standard assumption for IB SL exams):

• Horizontal axis: No acceleration acts, so horizontal velocity remains constant for the full flight of the projectile.
• Vertical axis: Constant acceleration due to gravity ($g=9.81{\text{ m/s}}^2$ downwards) acts, so you use standard SUVAT equations for vertical motion.

To solve projectile problems, split the initial launch velocity into components using the launch angle $\theta$ above the horizontal: $$v_{x,initial}=u\cos \theta$$ $$v_{y,initial}=u\sin \theta$$

Worked Example

A ball is launched at $16\text{ m/s}$ at an angle of $45^{\circ }$ above the horizontal, from ground level. Calculate its maximum height and total range.

1. Split initial velocity: $v_x=16\cos 45^{\circ }=11.31\text{ m/s}$, $v_{y,initial}=16\sin 45^{\circ }=11.31\text{ m/s}$
2. Maximum height: At the peak of flight, vertical velocity = 0. Use $v^2=u^2+2as$ for vertical motion: $0=(11.31{)}^2+2(-9.81)s_y$ → $s_y=6.52\text{ m}$
3. Total flight time: Time to reach peak = $\frac{11.31}{9.81}=1.15\text{ s}$, total flight time = $2\times 1.15=2.30\text{ s}$
4. Range: Horizontal displacement = $v_x\times t=11.31\times 2.30=26.0\text{ m}$

4. Newton's three laws — force, mass, action-reaction

First, define core terms: A force is a vector push or pull, measured in Newtons ($1\text{ N}=1{\text{ kg m/s}}^2$). Mass is a scalar measure of an object’s inertia (resistance to change in motion), measured in kilograms.

1. Newton’s First Law: An object remains at rest or moving at constant velocity unless acted on by a net external force. If net force = 0, acceleration = 0.
2. Newton’s Second Law: The net force acting on an object equals its mass multiplied by its acceleration, for constant mass: $F_{net}=ma$. The vector direction of acceleration matches the direction of net force.
3. Newton’s Third Law: If object A exerts a force on object B, object B exerts a force of equal magnitude and opposite direction on object A. These forces act on different objects, and are always the same type (e.g. both gravitational, both contact forces).

Worked Example

A 3kg block is pulled to the right with a force of 12N, with a 3N frictional force acting left. Calculate its acceleration.

1. Calculate net force: $F_{net}=12-3=9\text{ N}$ right
2. Apply second law: $a=\frac{F_{net}}{m}=\frac{9}{3}=3{\text{ m/s}}^2$ right

Exam tip: Never include action-reaction pairs on the same free-body diagram, as they act on different objects and do not cancel each other out for a single system.

5. Friction (static and kinetic) and tension

Friction

Friction is a contact force that opposes relative motion between two surfaces, caused by intermolecular interactions between the surfaces:

• Static friction ($f_s$): Acts when surfaces are not sliding relative to each other. It increases to match the applied force up to a maximum value: $f_{s,max}={\mu }_{s}N$, where ${\mu }_s$ is the dimensionless coefficient of static friction, and $N$ is the normal contact force between the surfaces.
• Kinetic friction ($f_k$): Acts when surfaces are sliding past each other. It has a constant value: $f_k={\mu }_{k}N$, where ${\mu }_k$ is the coefficient of kinetic friction. ${\mu }_k<{\mu }_s$ for all surface pairs.

Tension

Tension is a pulling force transmitted along a massless string, rope or cable. For a massless string (the standard IB SL assumption), tension is identical throughout the string, and always acts along the string, pulling on the objects attached to both ends.

Worked Example

A 4kg box rests on a horizontal table, with ${\mu }_s=0.5$ and ${\mu }_k=0.3$. What minimum force is required to start moving the box, and what is its acceleration if you apply this force after it starts moving?

1. Normal force $N=mg=4\times 9.81=39.24\text{ N}$
2. Minimum force to move: $f_{s,max}=0.5\times 39.24=19.62\text{ N}$
3. Kinetic friction when moving: $f_k=0.3\times 39.24=11.77\text{ N}$
4. Net force: $19.62-11.77=7.85\text{ N}$
5. Acceleration: $a=\frac{7.85}{4}=1.96{\text{ m/s}}^2$

6. Linear momentum — $p=mv$ and conservation

Linear momentum ($p$) is a vector quantity equal to the product of an object’s mass and velocity: $$p=mv$$ Units are kilogram-meters per second ($\text{kg m/s}$), and its direction matches the direction of velocity. Newton’s original second law is written in terms of momentum: $F_{net}=\frac{\Delta p}{\Delta t}$.

The conservation of linear momentum states that for a closed, isolated system (no external forces acting on the system), total linear momentum is constant: $$\sum p_{initial}=\sum p_{final}$$ This applies to all collisions and explosions. Collisions are classified as elastic (kinetic energy is conserved) or inelastic (kinetic energy is lost; maximum loss occurs when objects stick together, called perfectly inelastic collision).

Worked Example

A 1.5kg cart moving right at 4 m/s collides with a 2kg cart moving left at 1.5 m/s. The carts stick together after collision. Find their final velocity.

1. Define right as positive. Total initial momentum: $(1.5\times 4)+(2\times -1.5)=6-3=3\text{ kg m/s}$
2. Total mass after collision: $1.5+2=3.5\text{ kg}$
3. Final velocity: $3=3.5v$ → $v=0.86\text{ m/s}$ right

7. Impulse and force-time graphs

Impulse ($J$) is the change in momentum of an object, equal to the product of the average force acting on the object and the time interval the force acts for: $$J=F_{avg}\Delta t=\Delta p=m(v_f-v_i)$$ Units are Newton-seconds (Ns), equivalent to $\text{kg m/s}$.

For a force-time ($F-t$) graph, impulse is equal to the total area under the graph, regardless of the shape of the force curve. For a triangular $F-t$ graph, impulse is $\frac{1}{2}\times \text{base}\times \text{height}$.

Worked Example

A 0.1kg tennis ball hits a wall moving left at 30 m/s, and bounces off moving right at 25 m/s. The contact time with the wall is 0.005s. Calculate the average force exerted by the wall on the ball.

1. Define right as positive. Initial velocity $v_i=-30\text{ m/s}$, final velocity $v_f=25\text{ m/s}$
2. Change in momentum: $\Delta p=0.1(25-(-30))=0.1\times 55=5.5\text{ kg m/s}$
3. Average force: $F_{avg}=\frac{\Delta p}{\Delta t}=\frac{5.5}{0.005}=1100\text{ N}$ right

8. Work, kinetic energy, and the work-energy theorem

Work ($W$) is a scalar measure of energy transferred by a force, calculated as: $$W=Fs\cos \theta$$ Where $F$ is the magnitude of the force, $s$ is the magnitude of displacement, and $\theta$ is the angle between the force and displacement vectors. Units are Joules ($1\text{ J}=1\text{ N m}$). Work is positive if energy is added to the object ($\theta <90^{\circ }$), negative if energy is removed ($\theta >90^{\circ }$), and zero if the force is perpendicular to displacement ($\theta =90^{\circ }$, e.g. normal force on a horizontally moving object does no work).

Kinetic energy ($E_k$) is the energy of motion of an object: $$E_k=\frac{1}{2}m{v}^2$$ It is a scalar, measured in Joules.

The work-energy theorem states that the net work done on an object equals its change in kinetic energy: $$W_{net}=\Delta E_k=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$$

Worked Example

A 5kg block is pulled 4m along a horizontal surface by a 15N force at 30° above the horizontal, with a 4N frictional force acting. Calculate the net work done on the block, and its final speed if it starts from rest.

1. Work done by applied force: $15\times 4\times \cos 30^{\circ }=52.0\text{ J}$
2. Work done by friction: $4\times 4\times \cos 180^{\circ }=-16\text{ J}$
3. Work done by normal force and gravity: 0, as they are perpendicular to displacement
4. Net work: $52.0-16=36.0\text{ J}$
5. Final speed: $36.0=0.5\times 5\times v^2$ → $v^2=14.4$ → $v=3.79\text{ m/s}$

9. Power, efficiency, and energy transfer

Power ($P$) is the rate of energy transfer, or the rate of doing work. It is a scalar quantity measured in Watts ($1\text{ W}=1\text{ J/s}$). Two common formulas for power: $$P=\frac{W}{\Delta t}$$ $$P=Fv\cos \theta$$ Where $v$ is instantaneous velocity, and $\theta$ is the angle between force and velocity.

Efficiency ($\eta$) is the ratio of useful output energy (or power) to total input energy (or power). It is always less than 1 (or 100%) due to energy losses from friction, heat and sound: $$\eta =\frac{\text{Useful energy output}}{\text{Total energy input}}=\frac{\text{Useful power output}}{\text{Total power input}}$$

Worked Example

A 600W electric motor lifts an 8kg mass 3m vertically upwards in 2 seconds. Calculate the efficiency of the motor.

1. Total input energy: $600\times 2=1200\text{ J}$
2. Useful output energy (gravitational potential energy gained): $mgh=8\times 9.81\times 3=235.4\text{ J}$
3. Efficiency: $\frac{235.4}{1200}=0.196=19.6\%$

Exam tip: Examiners almost always ask for efficiency as a percentage, so multiply your decimal value by 100 and include the % sign to avoid losing marks.

10. Common Pitfalls (and how to avoid them)

• Sign errors for gravitational acceleration: Students often use $+9.81{\text{ m/s}}^2$ for acceleration when they have defined upwards as positive, leading to incorrect suvat results. Fix: Always explicitly state your positive axis direction at the start of any kinematics problem, and assign the correct sign to $g$.
• Including action-reaction pairs on free-body diagrams: Students incorrectly assume third-law forces cancel out, but they act on different objects. Fix: Only include forces acting on the object you are analyzing on its FBD, not forces the object exerts on other systems.
• Mixing up static and kinetic friction: Students use ${\mu }_k$ to calculate the force needed to start moving an object, leading to an under-estimate. Fix: Use ${\mu }_s$ for stationary objects, and only use ${\mu }_k$ once the object is sliding.
• Treating momentum as a scalar in collision problems: Students add magnitudes of momentum instead of using signed velocity values, leading to incorrect final velocity calculations. Fix: Assign a positive direction, and give velocities negative signs if they point opposite to this direction before summing initial momenta.
• Using the wrong angle for work calculations: Students use the angle of the force to the horizontal instead of the angle between the force and displacement vectors. Fix: Always measure the angle directly between the force arrow and displacement arrow when calculating $W=Fs\cos \theta$.

11. Practice Questions (IB Physics SL Style)

Question 1

A student throws a ball vertically upwards at 14 m/s from a height of 1.2m above the ground. Air resistance is negligible. (a) Calculate the maximum height of the ball above the ground. (b) Calculate the total time taken for the ball to hit the ground.

Solution

(a) Define upwards as positive. At maximum height, $v=0$. Use $v^2=u^2+2as$: $0=14^2+2(-9.81)s$ → $s=\frac{196}{19.62}=9.99\text{ m}$. Add initial height: total height = $9.99+1.2=11.2\text{ m}$. (b) Displacement when hitting the ground: $s=-1.2\text{ m}$. Use $s=ut+\frac{1}{2}a{t}^2$: $-1.2=14t-4.905t^2$ → $4.905t^2-14t-1.2=0$. Quadratic solution: $t=\frac{14+\sqrt{196+23.54}}{9.81}=2.94\text{ s}$. (Discard negative time solution.)

Question 2

A 1100kg car moving east at 16 m/s collides with a 900kg car moving west at 12 m/s. The collision is perfectly inelastic. (a) State what is meant by a perfectly inelastic collision. (b) Calculate the velocity of the combined cars immediately after the collision.

Solution

(a) A perfectly inelastic collision is one where the colliding objects stick together after impact, and kinetic energy is not conserved (maximum possible kinetic energy is lost). (b) Define east as positive. Total initial momentum = $(1100\times 16)+(900\times -12)=17600-10800=6800\text{ kg m/s}$. Total mass after collision = $2000\text{ kg}$. Final velocity = $\frac{6800}{2000}=3.4\text{ m/s}$ east.

Question 3

A crane lifts a 250kg load vertically upwards at a constant speed of 0.7 m/s. (a) Calculate the useful power output of the crane. (b) If the crane’s motor has a total power input of 2200W, calculate the efficiency of the crane as a percentage.

Solution

(a) Constant speed → upward force equals weight: $F=mg=250\times 9.81=2452.5\text{ N}$. Power $P=Fv=2452.5\times 0.7=1716.75\text{ W}\approx 1720\text{ W}$. (b) Efficiency = $\frac{1716.75}{2200}\times 100=78.0\%$.

12. Quick Reference Cheatsheet

13. What's Next

Mastery of IB Physics SL Topic 2 Mechanics is essential for all subsequent units in the syllabus. It is the direct foundation for Theme B (Mechanics and materials), which covers circular motion, gravitation, and solid deformation, as well as Theme D (Energy production), where work, power, and efficiency are core to calculating energy yields from renewable and non-renewable sources. You will also apply Newton’s laws and kinematics to electromagnetism topics, such as calculating the force on a charged particle in an electric or magnetic field, and to practical experimental work across all units.

If you have gaps in your understanding, or want to practice more exam-style questions tailored to your weak spots, you can ask Ollie, our AI tutor, for personalized support at any time. You can also find more study guides and practice materials for other IB Physics SL topics on the homepage to prepare fully for your exams.