The complete lecture — every idea comes alive in the live 3D panel on the right as you read. Scroll down; the animation keeps pace, and you can push the trolley and slide the blocks yourself.
1 — Force & Newton's first law
- Force — an agent that changes (or tends to change) a body's state of rest or uniform motion. Vector; unit the newton (N).
- Inertia — a body's resistance to any change in its motion. Mass is the measure of inertia.
First law: a body stays at rest, or moves with uniform velocity in a straight line, unless an unbalanced external force acts on it.
Karachi moment: when the bus on University Road brakes hard, you lurch forward — your body keeps moving by inertia. That is why seat-belts exist.
2 — Newton's second law (F = ma)
Second law: an unbalanced force produces an acceleration in its own direction, proportional to the force and inversely proportional to the mass.
The equation of dynamicsF = m a · a = F/m
1 N = 1 kg × 1 m/s² · weight W = m g
Same engine, heavier load → smaller acceleration: an empty Suzuki pickup darts away from the signal; loaded with mango crates it crawls.
3 — Acceleration: putting F = ma to work
worked numerical · f = ma
A 200 N force acts on a 250 kg rickshaw starting from rest. Find a and the velocity after 5 s.
a = F/m = 200/250 = 0.8 m/s² · v = u + at = 0.8 × 5 = 4 m/s
worked numerical · braking force
A 1000 kg car at 20 m/s stops in 4 s. Find the braking force.
a = (0 − 20)/4 = −5 m/s² · F = ma = −5000 N (opposes motion)
4 — Newton's third law (action & reaction)
Third law: to every action there is an equal and opposite reaction. The pair acts on different bodies, so they never cancel.
- Walking — foot pushes road back; road pushes you forward.
- Gun & bullet — bullet forward, gun recoils backward.
- Rocket — gases pushed down, rocket pushed up; no air needed.
5 — Momentum & impulse
- Momentum — p = m v, the quantity of motion. Unit kg·m/s (= N·s).
- Impulse — J = F t = change in momentum.
Second law, momentum formF = (mv − mu)/t ⇒ F t = mv − mu
worked numerical · the fielder's catch
A 0.15 kg cricket ball at 30 m/s is stopped in 0.05 s. Find the force. And if stopped in 0.5 s?
F = (0 − 0.15×30)/0.05 = 90 N on the palms · with t = 0.5 s → only 9 N
6 — Law of conservation of momentum
When no external force acts, the total momentum of a system stays constant:
Two-body collisionm₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
worked numerical · recoil
A 4 kg rifle fires a 20 g bullet at 400 m/s. Find the recoil velocity.
0 = (0.02 × 400) + 4V ⇒ V = −2 m/s (backwards)
7 — Elastic & inelastic collisions
| Elastic | Inelastic |
|---|
| Momentum | conserved | conserved |
| Kinetic energy | conserved | lost (heat, sound, deformation) |
| Example | billiard / carrom strike | mud on a wall, coupling trolleys |
Classic result: equal masses in an elastic head-on collision exchange velocities — the carrom striker stops dead and the target shoots off.
8 — Friction: static & kinetic
Friction opposes relative motion between surfaces. Static friction is self-adjusting up to a limiting value; kinetic friction acts during sliding and is slightly smaller.
Coefficient of frictionμ = F / R · f = μ m g (level ground) · μ_k < μ_s
worked numerical · friction
A 30 kg crate just starts to slide under 117.6 N. Find μ_s. (g = 9.8)
R = mg = 294 N ⇒ μ_s = 117.6/294 = 0.4
9 — Inclined plane & tension in strings
Incline of angle θalong slope: mg sin θ · normal: R = mg cos θ
a = g(sin θ − μ cos θ)
Atwood machine (m₁ > m₂)a = (m₁ − m₂)g/(m₁ + m₂) · T = 2m₁m₂g/(m₁ + m₂)
worked numerical · atwood
5 kg and 3 kg hang over a frictionless pulley. Find a and T. (g = 9.8)
a = 2 × 9.8/8 = 2.45 m/s² · T = 294/8 = 36.75 N
10 — Exam recap
- 1st law: inertia — no unbalanced force, no change in velocity.
- 2nd law: F = ma; W = mg; 1 N = 1 kg·m/s².
- 3rd law: equal & opposite forces on two different bodies.
- Impulse F t = mv − mu; longer time, smaller force.
- Conservation: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
- Elastic collisions conserve KE; inelastic do not.
- Friction μ = F/R; μ_k < μ_s.
- Incline a = g(sin θ − μ cos θ); Atwood a = (m₁−m₂)g/(m₁+m₂).