Mechanics
Kinematics, dynamics, work and energy, systems of particles, collisions, moments of inertia and gravitation. Every topic with step-by-step theory, interactive diagrams and solved exercises.
Complete Theory
7Physical quantities fall into two categories: fundamental, defined by convention (mass kg, length m, time s, temperature K, current A, amount of substance mol, luminous intensity cd), and derived, obtained by combining fundamental ones (velocity m/s, force N = kg·m/s², energy J = N·m, power W = J/s).
Dimensional analysis is a powerful tool for checking the formal correctness of a physical formula. The principle is simple: every term in an equation must have the same physical dimensions. You cannot add metres to seconds.
We write dimensions of fundamental quantities as for length, for time, for mass. Then:
Dimensional analysis cannot determine dimensionless constants (like ), but it is the first check to perform on any formula: if the dimensions do not match, the formula is certainly wrong.
Dimensional analysis is a powerful tool for checking the formal correctness of a physical formula. The principle is simple: every term in an equation must have the same physical dimensions. You cannot add metres to seconds.
We write dimensions of fundamental quantities as for length, for time, for mass. Then:
- Velocity: — ratio of a length to a time interval.
- Acceleration: — change in velocity over time.
- Force (Newton's 2nd law): .
- Work/Energy: .
Dimensional analysis cannot determine dimensionless constants (like ), but it is the first check to perform on any formula: if the dimensions do not match, the formula is certainly wrong.
To describe the motion of a point particle we first need to know where it is and how it moves. The position vector is the fundamental quantity: it locates the point in space relative to a fixed origin O, and it depends on time because the point moves.
When the point moves from to , we define the displacement . Crucially, displacement depends only on the initial and final points, not on the path taken. If you hike a winding mountain trail and return to the starting point, even if you have walked kilometres.
Average velocity is the ratio of displacement to time interval: . It tells us, on average, how fast the position is changing.
Instantaneous velocity is the limit of the average velocity as the time interval tends to zero: The derivative of the position vector with respect to time. Geometrically, is always tangent to the trajectory: imagine a car on a curved road — its velocity at every instant points in the exact direction the car is heading at that moment.
Speed (magnitude of velocity) is , where is the arc length travelled. The car's speedometer measures exactly .
When the point moves from to , we define the displacement . Crucially, displacement depends only on the initial and final points, not on the path taken. If you hike a winding mountain trail and return to the starting point, even if you have walked kilometres.
Average velocity is the ratio of displacement to time interval: . It tells us, on average, how fast the position is changing.
Instantaneous velocity is the limit of the average velocity as the time interval tends to zero: The derivative of the position vector with respect to time. Geometrically, is always tangent to the trajectory: imagine a car on a curved road — its velocity at every instant points in the exact direction the car is heading at that moment.
Speed (magnitude of velocity) is , where is the arc length travelled. The car's speedometer measures exactly .
Acceleration measures how velocity changes over time. Just as velocity is the derivative of position, acceleration is the derivative of velocity:
Note: a body can accelerate even if its speed is constant — it suffices that the direction of motion changes. This leads to the decomposition of acceleration into two components:
Physical example: a car driving around a curve at a constant 50 km/h has (constant speed) but (changing direction). The total acceleration is purely centripetal. Conversely, a car accelerating in a straight line has (straight trajectory) and .
For rectilinear motion, and therefore : the acceleration is purely tangential.
Note: a body can accelerate even if its speed is constant — it suffices that the direction of motion changes. This leads to the decomposition of acceleration into two components:
- Tangential acceleration : measures how the magnitude of velocity changes. If you press the accelerator, is positive and the car goes faster; if you brake, is negative. It is parallel to the velocity (same or opposite direction).
- Centripetal acceleration : measures how the direction of velocity changes. It is always perpendicular to the velocity and points toward the centre of curvature of the trajectory. On a curve, even at constant speed, you feel pushed outward — that is the effect of centripetal acceleration (which actually points inward).
Physical example: a car driving around a curve at a constant 50 km/h has (constant speed) but (changing direction). The total acceleration is purely centripetal. Conversely, a car accelerating in a straight line has (straight trajectory) and .
For rectilinear motion, and therefore : the acceleration is purely tangential.
Uniform linear motion (ULM) is the simplest kind of motion: velocity is constant in both magnitude and direction. Consequently, acceleration is zero ().
The law of motion is linear: where is the initial position (at ) and is the constant velocity. The graph is a straight line whose slope equals the velocity. The graph is a horizontal line (velocity unchanged). The graph is zero everywhere.
Physical meaning: in ULM the net force on the body is zero (Newton's first law). A car travelling at 130 km/h on a straight highway, neither accelerating nor braking, is in ULM — the engine provides just enough power to overcome friction and air resistance, but the resultant force is zero, so the velocity does not change.
Displacement over an interval is simply : the constant velocity multiplied by the elapsed time. If a train travels at 80 m/s for 30 seconds, it covers exactly metres.
ULM is also the motion of an inertial reference frame: an observer in ULM cannot distinguish their own motion from being at rest (Galilean principle of relativity).
The law of motion is linear: where is the initial position (at ) and is the constant velocity. The graph is a straight line whose slope equals the velocity. The graph is a horizontal line (velocity unchanged). The graph is zero everywhere.
Physical meaning: in ULM the net force on the body is zero (Newton's first law). A car travelling at 130 km/h on a straight highway, neither accelerating nor braking, is in ULM — the engine provides just enough power to overcome friction and air resistance, but the resultant force is zero, so the velocity does not change.
Displacement over an interval is simply : the constant velocity multiplied by the elapsed time. If a train travels at 80 m/s for 30 seconds, it covers exactly metres.
ULM is also the motion of an inertial reference frame: an observer in ULM cannot distinguish their own motion from being at rest (Galilean principle of relativity).
In UAM the acceleration is constant in both magnitude and direction, so the velocity changes linearly in time and the position changes quadratically. The three kinematic equations are obtained by integrating once (for ) and twice (for ), or by eliminating between them.
Free fall is UAM with constant downward acceleration (taking upward as positive). Air resistance is neglected. All objects fall with the same acceleration regardless of mass — Galileo's result.
Free fall is UAM with constant downward acceleration (taking upward as positive). Air resistance is neglected. All objects fall with the same acceleration regardless of mass — Galileo's result.
Projectile motion is the superposition of two independent motions: horizontal ULM (constant , zero horizontal force) and vertical UAM (constant downward , zero initial horizontal force). The trajectory is a parabola.
Key results: the time of flight is determined solely by the vertical motion; the range depends on . Since is maximised at , the optimal launch angle for maximum range on flat ground is . At the apex the vertical velocity is zero, so only the horizontal component remains.
Key results: the time of flight is determined solely by the vertical motion; the range depends on . Since is maximised at , the optimal launch angle for maximum range on flat ground is . At the apex the vertical velocity is zero, so only the horizontal component remains.
Uniform circular motion (UCM) is the motion of a point travelling along a circle at constant speed. Note: even though is constant, the direction of velocity changes continuously, so there is acceleration.
Angular quantities: rather than describing motion with Cartesian coordinates, it is convenient to use angular coordinates:
Non-uniform circular motion (NUCM): when the angular velocity is not constant, we introduce the angular acceleration [rad/s²], analogous to linear acceleration. The kinematic equations are formally identical to those of UAM:
Real-world examples:
Angular quantities: rather than describing motion with Cartesian coordinates, it is convenient to use angular coordinates:
- Angular position [rad] — the angle swept by the radius vector.
- Angular velocity [rad/s] — constant in UCM.
- Period [s] — time for one complete revolution.
- Frequency [Hz] — revolutions per second.
Non-uniform circular motion (NUCM): when the angular velocity is not constant, we introduce the angular acceleration [rad/s²], analogous to linear acceleration. The kinematic equations are formally identical to those of UAM:
Real-world examples:
- A geostationary satellite completes one revolution in 24 hours ().
- A rotating rigid disk: all points have the same , but linear speed is larger for points farther from the centre.
- A laboratory centrifuge uses centripetal acceleration to separate components of different densities.
Worked Examples
3Example 1Projectile with angle — range and maximum height
Given
initial speed
launch angle above horizontal
acceleration of gravity
Find
Range — horizontal distance to impact
Maximum height attained
Speed at the apex of the trajectory
Total time of flight
Step-by-step solution
1Problem setup: we launch a projectile with initial speed at an angle above the horizontal. The trajectory is parabolic because the motion is the composition of a horizontal ULM (no force in x) and a vertical UAM (only gravity). Air resistance is neglected.
2Decomposition of initial velocity: we resolve into Cartesian components using trigonometry: (constant throughout the flight), (subject to ). The component is the projection of onto the x-axis, the projection onto the y-axis.
3Time of flight: the total time is determined by the vertical motion. By symmetry, the projectile takes the same time to rise (until ) and to fall (until ). The ascent time is . Hence . Alternatively, solve , which gives .
4Range: during flight the projectile moves horizontally at constant speed . The range is , the horizontal distance between launch and impact.
5Maximum height: at the highest point the vertical velocity vanishes (). Using with gives . At the apex, the velocity is purely horizontal: .
6Energy check: in the absence of friction, mechanical energy is conserved. Initial energy: . At apex: . Check: , (up to rounding).
✓ Final result: , , speed at apex ,
Example 2Emergency braking
Given
initial speed (108 km/h)
constant deceleration (braking)
the vehicle comes to a complete stop
Find
Braking time
Stopping distance
Step-by-step solution
1Problem: a car is travelling at 30 m/s (about 108 km/h) when the driver slams on the brakes, applying a constant deceleration of . We want to know how long it takes to stop and how many metres it travels during braking. The motion is rectilinear uniformly decelerated (UAM with ).
2Braking time: use the velocity law . When the car stops, . Solving: → . The sign is consistent: the speed decreases by 6 m/s every second, so in 5 seconds it goes from 30 m/s to 0.
3Stopping distance: use the relation, which does not require time. With : → → . Alternatively, from with we get .
4Comment: the stopping distance grows with the square of the initial speed (). If the speed were 60 m/s (216 km/h), the distance would quadruple to 300 m. This is why speed limits are fundamental for road safety: doubling the speed quadruples the stopping distance.
✓ Final result: , stopping distance
Example 3Rotor at 1800 rpm
Given
rotation speed
rotor radius
Find
Frequency and period of the motion
Linear speed at the rim
Centripetal acceleration at the rim
Step-by-step solution
1Problem: an industrial rotor spins at 1800 revolutions per minute. Its radius is 12 cm. We want to characterise the uniform circular motion of a point on the outer rim.
2Frequency and period: frequency is the number of revolutions per second: (30 revolutions per second). The period is the time for one revolution: — about 33 milliseconds per complete rotation.
3Angular and linear velocity: the angular velocity is . The linear speed of a point at the rim is (about 81 km/h). Although the rotor is small, the rim moves very fast because it rotates at high frequency.
4Centripetal acceleration: a point on the rim continuously changes direction (circular motion), so it experiences centripetal acceleration: . For comparison, , so — over 400 times the acceleration of gravity! These enormous accelerations explain why rotors must be carefully balanced and built with strong materials: a mass of 100 grams on the rim "weighs" as much as 43 kg under centripetal acceleration.
✓ Final result: , , ,
Exercises with Solutions
3Exercise 1Projectile from heightHard
Problem to solve
A cannon is placed on top of a cliff above sea level. The cannon fires a projectile with initial speed at an angle above the horizontal. Determine: (a) the total time of flight, (b) the range (horizontal distance from the base of the cliff), (c) the velocity (magnitude and direction) at impact with the water.
Given data
h = 80 m (cliff height)v_0 = 50 m/s (initial speed)\theta = 30° (launch angle)g = 9.81 m/s²
Step-by-step solution
1Setup: choose the origin of axes at the base of the cliff, with y positive upward. The cannon is at . Impact with the water occurs at . The motion is parabolic: ULM in x and UAM in y.
2Velocity decomposition: (constant). (positive, upward).
3Time of flight: from the vertical law , set : . Solve the quadratic: . Solutions are . Discard the negative: .
4Range: the horizontal distance covered is . The projectile lands 317 metres from the base of the cliff.
5Velocity at impact: the velocity components at time are (constant) and (negative, downward). The magnitude is . The angle below the horizontal is .
✓ Final answer: , , at 47.3° below horizontal
Exercise 2Pursuit problemVery Hard
Problem to solve
Two cars A and B start on the same straight line. Car A starts from rest () with constant acceleration . Car B already has speed and accelerates with . Both start from the same point () at the same time (). Determine: (a) after how long A catches B, (b) at what distance from the start the overtaking occurs, (c) the relative velocity of A with respect to B at the overtaking instant.
Given data
a_A = 3\,m/s^2 (acceleration of A)v_{A0} = 0 (A starts from rest)a_B = 1.5\,m/s^2 (acceleration of B)v_{B0} = 10\,m/s (initial speed of B)
Step-by-step solution
1Setup: both cars move with UAM. Write the laws of motion: , . A catches B when .
2Overtaking time: → → . The solution corresponds to the start. The desired solution is .
3Overtaking position: substitute into (or ): .
4Relative velocity: at the overtaking instant, , . The relative velocity is : A pulls away from B at 10 m/s after overtaking.
5Interpretation: A has a larger acceleration ( vs ), so it progressively overcomes B's initial speed advantage () until it catches B after 13.3 seconds and 267 metres. After overtaking, A continues to pull away because its acceleration is double.
✓ Final answer: , ,
Exercise 3Boat in a riverHard
Problem to solve
A boat crosses a river wide. The boat speed relative to the water is , while the river current flows parallel to the bank at . (a) If the boat points perpendicular to the bank, how long does it take to cross and how far downstream is it carried? (b) At what angle relative to the perpendicular must the boat point to arrive exactly opposite the starting point (zero drift)?
Given data
v_b = 4 m/s (boat speed relative to water)v_f = 3 m/s (current speed)d = 120 m (river width)
Step-by-step solution
1Setup: this is a classic velocity composition problem (Relative Motion). Choose S = riverbank as the fixed frame, S' = water as the moving frame. The boat has relative velocity with respect to the water, and the water has transport velocity parallel to the bank.
2(a) Point perpendicular: the boat heads straight for the opposite bank, so all its relative speed is used for crossing: . Crossing time is . During these 30 seconds, the current carries the boat downstream by . The absolute velocity relative to the bank is (magnitude 5 m/s, direction relative to the current).
3(b) Perpendicular crossing (zero drift): to arrive exactly opposite, the boat must compensate for the current by pointing upstream. The horizontal component of the relative velocity must cancel the current: → → from the perpendicular (i.e. 48.6° upstream). The effective crossing speed is , so .
4Comparison: in case (a) you cross faster (30 s) but end up 90 m downstream. In case (b) it takes longer (45.3 s) but you arrive exactly opposite. The choice depends on the goal: if a precise destination awaits on the opposite bank, use the correct angle; if drift is acceptable, pointing perpendicular is quicker.
✓ Final answer: (a) 30 s, drift 90 m downstream. (b) 48.6° upstream, crossing in 45.3 s, zero drift.
Keep studying
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Integrative Problems
Problems combining all chapters — exam levelProblem 1Tower, Ballistic Pendulum, and Keplerian OrbitEXTREME
A cannon is placed on top of a tower tall and fires a projectile of horizontally at .
The projectile strikes and embeds in a wooden block hanging from a rope of length (ballistic pendulum), at ground level.
The Earth-Moon system is then used as a reference for Kepler's third law.
The projectile strikes and embeds in a wooden block hanging from a rope of length (ballistic pendulum), at ground level.
The Earth-Moon system is then used as a reference for Kepler's third law.
📌 Problem data
(a)Uniformly Accelerated Motion(b)Inelastic Collision(c)Potential Energy + Pendulum(d)Moment of Inertia — Rigid Body(e)Gravitation — Kepler's Third Law
Problem 2Spring, Rolling Disk, Inclined Plane Collision, and ConservationEXTREME
A spring (, compressed ) launches a solid disk (, ) up an inclined plane (, , ) that rolls without slipping.
At the top the disk is launched horizontally and strikes a pendulum (, ) — perfectly inelastic collision. What is asked (solved below, a→e): (a) the disk's speed at the top of the plane; (b) the range and impact speed of the horizontal launch; (c) the speed after the inelastic collision with the pendulum and the energy lost; (d) the pendulum's maximum angle, the maximum tension, and whether it completes the loop; (e) the full energy balance (from spring to maximum angle).
At the top the disk is launched horizontally and strikes a pendulum (, ) — perfectly inelastic collision. What is asked (solved below, a→e): (a) the disk's speed at the top of the plane; (b) the range and impact speed of the horizontal launch; (c) the speed after the inelastic collision with the pendulum and the energy lost; (d) the pendulum's maximum angle, the maximum tension, and whether it completes the loop; (e) the full energy balance (from spring to maximum angle).
📌 Problem data
(a)Energy + Rigid Body (rolling)(b)Kinematics — Projectile(c)Inelastic Collision + CM(d)Pendulum Dynamics + Forces(e)Conservation Laws — Complete Energy Balance