Newton's third law of motion
Third law: to every action there is always an equal and opposite reaction. In simple words: if body A puts a force on body B, then B puts an equal force on A in the opposite direction.
FAB = − FBA
- The two forces act on different bodies, so they never cancel each other. (They cancel only when you treat A and B together as one system.)
- They act at the same instant. Neither is "first".
- They are of the same type: both contact forces, or both gravitational, and so on.
A book on a table: the Earth pulls the book down (weight) and the book pulls the Earth up; that is one pair. The book presses the table and the table pushes the book up (normal force); that is another pair. Weight and normal force on the book are equal here, but they are not a third-law pair, because both act on the same body.
Examples: walking (foot pushes the ground back, ground pushes you forward), swimming, rowing, a rocket, a bouncing ball.
Conservation of linear momentum
Law: if no net external force acts on a system, its total linear momentum stays constant.
Proof from the second and third laws
Let two balls A and B collide. During the short contact time Δt, A pushes B with FBA and B pushes A with FAB. By the second law, change in momentum of A: ΔpA = FAB Δt. Of B: ΔpB = FBA Δt. By the third law FAB = − FBA, so ΔpA + ΔpB = 0. Total momentum does not change:
m1u1 + m2u2 = m1v1 + m2v2
Uses
- Recoil of a gun: 0 = M V + m v, so V = − m v / M. The gun moves back slowly because it is heavy.
- Rocket: gas is thrown backward fast, the rocket gains forward momentum.
- Collisions and explosions: total momentum just before = total momentum just after, even if kinetic energy is lost as heat and sound.
- A shell that bursts in mid-air: the pieces' momenta add up to the momentum of the shell before bursting.
Equilibrium of concurrent forces
Concurrent forces are forces whose lines of action pass through one point. A particle is in equilibrium when the net force on it is zero: it stays at rest or moves with constant velocity.
F1 + F2 + … + Fn = 0, which means ΣFx = 0, ΣFy = 0 and ΣFz = 0.
- Two forces are in equilibrium only if they are equal and opposite.
- Three forces in equilibrium, drawn head to tail, make a closed triangle (triangle law). Any one force is equal and opposite to the resultant of the other two.
- Lami's theorem (for three forces): each force is proportional to the sine of the angle between the other two: F1/sin α = F2/sin β = F3/sin γ.
How to solve
Draw a free-body diagram of the point where the forces meet (a knot, a ring). Resolve each force into x and y parts. Set the sum along x and along y to zero and solve.
Try it at home
Balloon rocket: thread a straw on a long string tied across a room. Tape a blown-up balloon to the straw and let go of its mouth. Air rushes back, the balloon rushes forward (third law and momentum).
Spring balances: hook two spring balances together and pull. Both always show the same reading, whoever pulls harder.
In the 3D: on the last step set m₁ = m₂ and u₁ = 4 m/s. Predict v before you watch (answer: 2 m/s).
Key formulas and definitions
- F_AB = − F_BA (third law)
- m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
- Recoil: V = − m v / M
- Stick together: v = (m₁u₁ + m₂u₂)/(m₁ + m₂)
- Equilibrium: ΣFx = 0, ΣFy = 0
- Lami: F₁/sin α = F₂/sin β = F₃/sin γ
Worked examples
1. A 4 kg gun fires a 20 g bullet at 400 m/s. Find the recoil speed of the gun.
Momentum before = 0. After: 4 V + 0.02 × 400 = 0 → V = −8/4 = −2 m/s. The gun moves back at 2 m/s.
2. A 2 kg cart at 3 m/s hits a 1 kg cart at rest and they stick. Find their common speed.
2 × 3 + 1 × 0 = (2 + 1) v → v = 6/3 = 2 m/s.
3. A 50 kg boy jumps off a 100 kg boat at 2 m/s towards the shore. The boat was at rest. Find the boat's speed.
0 = 50 × 2 + 100 × V → V = −1 m/s. The boat moves away from the shore at 1 m/s.
4. A 3 kg shell at rest explodes into two pieces of 1 kg and 2 kg. The 1 kg piece flies at 20 m/s. Find the speed of the other piece.
0 = 1 × 20 + 2 × v → v = −10 m/s: 10 m/s in the opposite direction.
5. Two forces of 6 N (east) and 8 N (north) act on a ring. What third force keeps the ring in equilibrium?
Resultant of the two = √(6² + 8²) = 10 N, towards north-east at tan⁻¹(8/6) ≈ 53° from east. The third force is 10 N in exactly the opposite direction.
6. A 10 kg mass hangs from two strings, each at 45° to the vertical. Find the tension in each (g = 9.8 m/s²).
Vertical: 2T cos 45° = 98 → T = 98/(2 × 0.707) ≈ 69.3 N. Horizontal parts T sin 45° cancel.
7. A 6 kg mass hangs from a string. A horizontal force pulls it until the string makes 30° with the vertical. Find the force and the tension (g = 10 m/s²).
Vertical: T cos 30° = 60 → T = 60/0.866 ≈ 69.3 N. Horizontal: F = T sin 30° = 69.3 × 0.5 ≈ 34.6 N (= 60 tan 30°).
Common mistakes
- Thinking action and reaction cancel. They act on different bodies, so each body can still accelerate.
- Calling weight and normal force a third-law pair. Both act on the same body; the true pairs are book–Earth and book–table.
- Forgetting the minus sign for velocities in the opposite direction in momentum problems.
- Using conservation of momentum when a big outside force acts for a long time (like friction over many seconds). It holds when the net external force is zero or the event is very short.