In short: Kinematics describes motion, Newton’s laws explain what changes it, torque explains rotational effects, and kinetic energy measures the energy of motion. This guide connects all five, with solved problems (g = 10 m/s² unless stated).
What are Newton’s laws of motion?
Newton’s laws of motion are three laws that relate force to the motion of a body: (1) a body keeps its state of rest or uniform motion unless a net external force acts; (2) net force equals the rate of change of momentum; (3) forces occur in equal and opposite pairs. They form the base of classical mechanics.
A force is a push or pull that can change a body’s velocity or shape. Mechanics is the study of motion and its causes. Inertia is a body’s resistance to any change in its state of motion; mass measures it.
Newton’s first law (law of inertia)
A body stays at rest or in uniform motion in a straight line unless a net external force acts on it. Seat belts work because your body tends to keep moving forward when the car stops suddenly.
| Type | Meaning | Example |
|---|---|---|
| Inertia of rest | Body resists starting to move | Passengers jerk backward when a bus starts |
| Inertia of motion | Body resists stopping or slowing | Passengers lurch forward on braking |
| Inertia of direction | Body resists change of direction | Passengers slide outward on a sharp turn |
Newton’s second law (F = ma)
The net external force on a body equals the rate of change of its momentum. For constant mass this gives F = ma: acceleration is directly proportional to net force and inversely proportional to mass, and points along the net force.
F = dp/dt = d(mv)/dt = m(dv/dt) = ma (m constant)SI unit: newton (N) = kg·m/s². Dimensional formula: [M L T⁻²]. With mass fixed, doubling the force doubles the acceleration. With force fixed, doubling the mass halves the acceleration. Acceleration is along the net force because force is the vector cause of the change in velocity; a free-body diagram (one body, all forces on it drawn as arrows) lets you add them to find that net force.
Newton’s third law
When body A exerts a force on body B, B exerts an equal and opposite force on A. The two forces act simultaneously, on different bodies, and are of the same type. Because they act on different bodies, they never cancel.
Examples: walking (foot pushes ground back, ground pushes you forward), swimming, gun recoil, rocket propulsion (exhaust pushed down, rocket pushed up) and jumping.
| Action–reaction pair | Balanced forces |
|---|---|
| Act on two different bodies | Act on the same body |
| Always equal, opposite and simultaneous | Equal and opposite, net force zero |
| Cannot cancel | Cancel, so no acceleration |
Applications: tension, normal reaction, lifts and pseudo force
Tension is the pull along a string; normal reaction is the contact force perpendicular to a surface; friction acts along the surface, opposing relative motion. In a lift accelerating upward with a, apparent weight is N = m(g + a); downward, N = m(g − a). In a non-inertial frame with acceleration a, a pseudo force of magnitude ma acts opposite to a.
Given m = 5 kg, F = 20 N. Formula a = F/m = 20/5 = 4 m/s².
N − mg = ma ⇒ N = m(g + a) = 60 × 12 = 720 N.
a = (5 − 3)g/(5 + 3) = 2 × 10/8 = 2.5 m/s². For the 3 kg mass, T − 30 = 3 × 2.5 ⇒ T = 37.5 N.
What are the equations of motion?
The equations of motion (kinematics equations) relate initial velocity u, final velocity v, acceleration a, time t and displacement s for motion in a straight line with constant acceleration: v = u + at, s = ut + ½at², v² = u² + 2as.
Distance is the total path length (scalar); displacement is the shortest straight-line change in position (vector). Speed is distance/time; velocity is displacement/time; acceleration is the rate of change of velocity.
v = u + at | s = ut + ½at² | v² = u² + 2asDerivation. Since a = (v − u)/t, rearranging gives the first equation. The v–t graph is a straight line, so the area under it (a trapezoid) is s = ½(u + v)t; substituting v = u + at gives the second. Eliminating t between the first two gives the third. Symbols: u, v in m/s; a in m/s²; t in s; s in m.
| Equation | Missing variable | Use when |
|---|---|---|
| v = u + at | s | Time and velocity are asked; distance is not involved |
| s = ut + ½at² | v | Displacement and time are involved |
| v² = u² + 2as | t | Time is neither given nor asked |
v = 0 + 2 × 5 = 10 m/s; s = ½ × 2 × 25 = 25 m.
Take a = −5: 0 = 400 − 10s ⇒ s = 40 m.
Take up as positive, a = −10. H = u²/2g = 400/20 = 20 m. Time to top = 2 s, so total time = 4 s.
a = (0 − 20)/10 = −2 m/s²; s = 20 × 10 + ½(−2)(100) = 100 m.
s₃ = u + a(2n − 1)/2 = 2 + 4 × 5/2 = 12 m (check: s(3) − s(2) = 24 − 12).
45 = ½ × 10 × t² ⇒ t = 3 s; v = 10 × 3 = 30 m/s.
Limits and signs. These equations fail if acceleration changes with time; then you need calculus (v = ∫a dt) or graph areas. Pick one positive direction and keep it: a quantity opposite to it is negative, including g in upward throws.
What is the average velocity formula?
Average velocity is total displacement divided by total time: v_avg = Δx/Δt = (x₂ − x₁)/(t₂ − t₁). It is a vector, so direction matters, and it can be zero even when the body has travelled a long distance.
For uniformly accelerated motion only, v_avg = (u + v)/2. Do not use this when acceleration varies, because the velocity–time graph is then not a straight line. Instantaneous velocity is the limit of Δx/Δt as Δt → 0.
| Property | Average speed | Average velocity |
|---|---|---|
| Formula | Total distance / time | Displacement / time |
| Type | Scalar | Vector |
| Possible values | Always positive | Positive, negative or zero |
| Magnitude | ≥ |average velocity| | ≤ average speed |
What is moment of force (torque)?
Moment of force (torque) is the turning effect of a force about an axis. Its magnitude is τ = F × perpendicular distance from the axis = rF sinθ, where θ is the angle between the position vector r and force F. SI unit: newton-metre (N·m).
τ = r × F ⇒ |τ| = rF sinθ [M L² T⁻²]Perpendicular distance matters because only the force component perpendicular to r (F sinθ) turns the body. Torque is largest at θ = 90° and zero when the force passes through the axis (θ = 0° or 180°). Torque is a vector along the axis, given by the right-hand rule (curl fingers from r toward F; the thumb gives the direction). Convention: anticlockwise positive, clockwise negative. Examples: pushing a door far from the hinges, a longer spanner, a seesaw, a steering wheel.
Principle of moments. For rotational equilibrium, total clockwise moment = total anticlockwise moment (net torque zero). Full equilibrium also needs net force zero.
What is kinetic energy?
Kinetic energy is the energy a body has because of its motion: KE = ½mv². It is a scalar, measured in joules (J = kg·m²/s²), with dimensional formula [M L² T⁻²].
Derivation. A constant net force F accelerates a body of mass m from rest to speed v over distance s. Work W = Fs = (ma)s. From v² = 2as, as = v²/2, so W = ½mv², which is stored as KE. It is a scalar because it depends on v² (the dot product v·v), which has no direction.
| Change | Effect on KE | Example (m = 2 kg, v = 3 m/s, KE = 9 J) |
|---|---|---|
| Mass × 2 | × 2 | 18 J |
| Velocity × 2 | × 4 | 36 J |
| Velocity × 3 | × 9 | 81 J |
Velocity is squared because work depends on distance, and the distance needed to reach a speed grows as v². This is why stopping distance quadruples when speed doubles.
Work–energy theorem. The net work done on a body equals its change in kinetic energy: W_net = ½mv² − ½mu². Positive work raises KE, negative work lowers it, and zero work (e.g. a force perpendicular to motion) leaves it unchanged.
How are the five topics connected?
Kinematics (and average velocity) describe motion; Newton’s laws give the cause; torque is the rotational counterpart of force; work done by force changes kinetic energy. Torque links back to Newton’s laws through τ = Iα in rotation.
Formula sheet for Class 11 Mechanics
| Formula | Quantity | SI unit | Condition |
|---|---|---|---|
| F = ma | Net force | N | Constant mass |
| F = dp/dt | Force (momentum) | N | General |
| v = u + at | Final velocity | m/s | Constant a |
| s = ut + ½at² | Displacement | m | Constant a |
| v² = u² + 2as | Velocity–displacement | m²/s² | Constant a |
| v_avg = Δx/Δt | Average velocity | m/s | Always |
| v_avg = (u + v)/2 | Average velocity | m/s | Constant a only |
| τ = rF sinθ | Torque | N·m | About a fixed axis |
| ΣCW = ΣACW | Principle of moments | N·m | Rotational equilibrium |
| KE = ½mv² | Kinetic energy | J | Speed ≪ speed of light |
| W_net = ΔKE | Work–energy theorem | J | All forces included |
What are the common conceptual mistakes?
15 misconceptions to avoid
- Action and reaction cancel. They act on different bodies.
- No force means no motion. No net force means constant velocity.
- Zero velocity means zero acceleration. At the top of a throw, a = g.
- Acceleration always points along velocity. It points along net force.
- Average speed equals the magnitude of average velocity. Only for straight-line motion without reversal.
- Using (u + v)/2 for any motion. It needs constant acceleration.
- Distance equals displacement. Equal only in straight-line motion without reversal.
- Forgetting signs. Choose one positive direction; g is negative if up is positive.
- Using equations of motion when a varies.
- Mass equals weight. Mass is invariant (kg); weight mg changes with g (N).
- Torque equals force. Torque depends on lever arm and angle too.
- Using r instead of the perpendicular distance r sinθ.
- KE is proportional to v. It is proportional to v².
- KE can be negative. It cannot; work can.
- Apparent weight is true weight. N = m(g ± a) in accelerating lifts.