Laws of Motion

In this chapter: Aristotle's fallacy and inertia, Newton's first, second and third laws, momentum and impulse, conservation of momentum, equilibrium of a particle, common forces, static, kinetic and rolling friction, circular motion on level and banked roads, and a method for solving mechanics problems.

Inertia and the first law

Aristotle held that a force is needed to keep a body moving. Galileo showed, with balls rolling on inclined planes, that a body moving on a smooth horizontal surface keeps moving; it is friction that stops it. The property of a body to resist a change in its state of rest or uniform motion is inertia, and mass is its measure.

Newton's first law: every body continues in its state of rest or of uniform motion in a straight line unless compelled by some external force to act otherwise. If the net external force is zero, the acceleration is zero. A passenger jerked backward when a bus starts suddenly is showing inertia of rest; one thrown forward when it brakes is showing inertia of motion.

Momentum and the second law

Momentum is p = mv, a vector, with SI unit kg m s⁻¹. Newton's second law: the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force.

• It is a vector equation, so it holds separately for each component: F_x = ma_x, F_y = ma_y.
• F is the net external force. Internal forces within a system cancel out.
• It is a local relation: the acceleration at an instant depends on the force at that instant, not on the history of the motion.

Impulse

When a large force acts for a very short time (a bat hitting a ball), we use impulse: J = F × Δt = Δp. Impulse has the same unit as momentum (N s = kg m s⁻¹). A cricketer draws the hands back while catching a fast ball to increase Δt, which reduces the average force for the same change in momentum.

The third law

Newton's third law: to every action there is always an equal and opposite reaction. If body A exerts a force F_AB on B, then B exerts F_BA = −F_AB on A. Action and reaction act at the same instant and on different bodies, so they never cancel each other. A book on a table is in equilibrium because its weight and the normal force on it add to zero; these two are not an action-reaction pair, because both act on the book.

Conservation of momentum

For an isolated system (no net external force), the total momentum stays constant. This follows from the second and third laws together. In a collision, the forces between the bodies are internal and equal and opposite, so whatever momentum one gains, the other loses.

Worked example: A bullet of mass 0.02 kg is fired at 400 m s⁻¹ from a gun of mass 4 kg. Find the recoil speed of the gun.
Solution: Initial momentum = 0. So 0.02 × 400 + 4 × V = 0, which gives V = −8/4 = −2 m s⁻¹. The gun recoils at 2 m s⁻¹ opposite to the bullet. Because the gun is much heavier, its speed is much smaller.

Equilibrium and common forces

A particle is in equilibrium when the net external force on it is zero: ∑F_x = 0, ∑F_y = 0, ∑F_z = 0. Three forces in equilibrium can be drawn head to tail to form a closed triangle.

Apart from gravity (a non-contact force), the forces in mechanics problems are contact forces: the normal reaction N perpendicular to the surface, friction parallel to it, tension in a string (the same all along a light, inextensible string over a smooth pulley), and the spring force F = −kx.

Apparent weight in a lift

For a person of mass m standing on a weighing scale in a lift with acceleration a:

Worked example: A 60 kg student stands on a scale in a lift that accelerates upward at 2 m s⁻². What does the scale read? (g = 9.8 m s⁻²)
Solution: Draw the free-body diagram: N upward, mg downward, net force upward. N − mg = ma, so N = 60(9.8 + 2) = 708 N. If the lift accelerated downward at the same rate, N = 60 × 7.8 = 468 N.

Friction

Friction opposes relative motion (or impending relative motion) between surfaces in contact.

• Static friction f_s is self-adjusting. It equals the applied force until the body is about to move, up to a limit: f_s ≤ μ_sN.
• Kinetic friction acts once sliding begins: f_k = μ_kN, roughly independent of speed and of the area of contact. Usually μ_k < μ_s.
• Rolling friction is much smaller than sliding friction, which is why wheels and ball bearings are used.
• Friction is not always a villain. Walking, a car's grip and the braking of a cycle all depend on it. Lubricants reduce it where it wastes energy.

On an incline, a block just begins to slide when tan θ = μ_s. This angle is called the angle of repose.

Worked example: A 4 kg block rests on a board. The board is slowly tilted, and the block just begins to slide when the angle reaches 15°. Find μ_s.
Solution: At the point of sliding, mg sin θ = μ_s mg cos θ, so μ_s = tan 15° ≈ 0.27. The mass does not matter.

Circular motion

A body moving on a circle of radius R at speed v needs a net force mv²/R towards the centre. This centripetal force is not a new kind of force; it is provided by tension, gravity, friction or a component of the normal force.

Car on a level road

Only static friction provides the centripetal force: mv²/R ≤ μ_smg, so the maximum safe speed is v_max = √μ_sRg, independent of the car's mass.

Car on a banked road

Tilting the road by an angle θ lets a component of the normal force, N sin θ, supply part of the centripetal force.

Worked example: A circular race track of radius 300 m is banked at 15°. The coefficient of friction between the tyres and the road is 0.2. Find the optimum speed and the maximum permissible speed. (g = 9.8 m s⁻², tan 15° = 0.268)
Solution: v_o = √(300 × 9.8 × 0.268) = √788 ≈ 28.1 m s⁻¹. v_max = √[300 × 9.8 × (0.2 + 0.268)/(1 − 0.2 × 0.268)] = √(2940 × 0.468/0.946) ≈ √1454 ≈ 38.1 m s⁻¹.

Solving problems in mechanics

Draw the free-body diagram before writing a single equation. Isolate one body at a time, show every force acting on it (not the forces it exerts), choose axes, often along and perpendicular to the acceleration, and apply F = ma along each axis. For connected bodies over a pulley with masses m₁ > m₂ (Atwood machine):

Common mistakes: (1) Treating weight and normal force as an action-reaction pair. (2) Writing static friction as μ_sN when the body is not about to slide; below that limit static friction just equals the applied force along the surface. (3) Adding "centripetal force" as an extra arrow on the free-body diagram. (4) Assuming N = mg on an incline or in an accelerating lift. (5) Forgetting that impulse is a vector: a ball bouncing straight back with the same speed has Δp = 2mv, not zero.

JEE and NEET focus

• Free-body diagrams for blocks, pulleys, inclines and connected bodies.
• Static versus kinetic friction, including the case where the applied force is less than the limiting friction.
• Apparent weight in a lift and conservation of momentum (recoil, explosions).
• Impulse from a force-time graph (area under the curve).
• Maximum speed on level and banked roads, and the optimum banking speed.

Practice questions