Intuition The one-sentence soul of this lawA force is not what keeps things moving — it is what changes their motion. The net force on an object equals the rate at which its momentum changes. No net force ⇒ momentum stays constant (that's the First Law hiding inside the Second).
Definition Two equivalent statementsMomentum form (the fundamental one):
F ⃗ net = d p ⃗ d t , p ⃗ = m v ⃗ \vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}, \qquad \vec{p} = m\vec{v} F net = d t d p , p = m v Mass-times-acceleration form (special case, constant mass):
F ⃗ net = m a ⃗ \vec{F}_{\text{net}} = m\vec{a} F net = m a F ⃗ net \vec{F}_{\text{net}} F net vector sum of ALL forces acting on the object, m m m a ⃗ \vec{a} a p ⃗ \vec{p} p
WHY "net"? Because only the total push/pull matters. If you pull a box right with 10 N and friction pulls left with 10 N, the net force is zero and a ⃗ = 0 \vec{a}=0 a = 0
Start with the fundamental statement:
F ⃗ net = d p ⃗ d t = d ( m v ⃗ ) d t \vec{F}_{\text{net}} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt} F net = d t d p = d t d ( m v )
Apply the product rule (because both m m m v ⃗ \vec{v} v d ( m v ⃗ ) d t = m d v ⃗ d t + v ⃗ d m d t \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt} d t d ( m v ) = m d t d v + v d t d m
Why this step? Momentum is a product m v ⃗ m\vec{v} m v
Now assume mass is constant (a single rigid object, not a leaking rocket), so d m d t = 0 \dfrac{dm}{dt}=0 d t d m = 0 F ⃗ net = m d v ⃗ d t = m a ⃗ \vec{F}_{\text{net}} = m\frac{d\vec{v}}{dt} = m\vec{a} F net = m d t d v = m a
Why this step? d v ⃗ d t \dfrac{d\vec{v}}{dt} d t d v is the definition of acceleration. Done.
Intuition Why momentum form is "more correct"F = m a F=ma F = ma fails when mass changes (rockets burning fuel, raindrops growing, conveyor belts). The d p ⃗ d t \frac{d\vec p}{dt} d t d p never fails . So memorise the momentum form; treat m a ma ma
Take the fundamental form and multiply both sides by d t dt d t t 1 → t 2 t_1\to t_2 t 1 → t 2 ∫ t 1 t 2 F ⃗ net d t = ∫ t 1 t 2 d p ⃗ d t d t = p ⃗ 2 − p ⃗ 1 \int_{t_1}^{t_2}\vec{F}_{\text{net}}\,dt = \int_{t_1}^{t_2}\frac{d\vec{p}}{dt}\,dt = \vec{p}_2 - \vec{p}_1 ∫ t 1 t 2 F net d t = ∫ t 1 t 2 d t d p d t = p 2 − p 1
Why this step? Integrating a derivative over time just gives the net change (Fundamental Theorem of Calculus).
Define the left side as impulse J ⃗ \vec{J} J
If the force is constant , the integral becomes a simple multiplication:
J ⃗ = F ⃗ Δ t = Δ p ⃗ \vec{J} = \vec{F}\,\Delta t = \Delta \vec{p} J = F Δ t = Δ p
Intuition What impulse buys youTo change an object's momentum by a fixed amount Δ p \Delta p Δ p big force for a short time OR a small force for a long time — the product F Δ t F\,\Delta t F Δ t increase Δ t \Delta t Δ t to shrink the force F = Δ p / Δ t F=\Delta p/\Delta t F = Δ p /Δ t
F = m a F=ma F = ma a a a Step 1: Net force = 12 − 4 = 8 N = 12 - 4 = 8\text{ N} = 12 − 4 = 8 N Why? Forces are vectors along one line; add with signs.
Step 2: a = F net m = 8 2 = 4 m/s 2 a = \dfrac{F_{\text{net}}}{m} = \dfrac{8}{2} = 4\ \text{m/s}^2 a = m F net = 2 8 = 4 m/s 2 Why? Rearranged F = m a F=ma F = ma
Worked example 2 — Impulse from a constant forceA 0.15 kg ball moving at 20 m/s 20\ \text{m/s} 20 m/s 30 m/s 30\ \text{m/s} 30 m/s 0.01 s 0.01\ \text{s} 0.01 s Step 1: Pick a positive direction (initial direction). v i = + 20 v_i=+20 v i = + 20 v f = − 30 v_f=-30 v f = − 30 Why? Momentum is a vector; the reversal must carry a sign.
Step 2: Δ p = m ( v f − v i ) = 0.15 ( − 30 − 20 ) = − 7.5 kg⋅m/s \Delta p = m(v_f - v_i) = 0.15(-30 - 20) = -7.5\ \text{kg·m/s} Δ p = m ( v f − v i ) = 0.15 ( − 30 − 20 ) = − 7.5 kg⋅m/s Why? Impulse–momentum theorem.
Step 3: F = Δ p Δ t = − 7.5 0.01 = − 750 N F = \dfrac{\Delta p}{\Delta t} = \dfrac{-7.5}{0.01} = -750\ \text{N} F = Δ t Δ p = 0.01 − 7.5 = − 750 N Why? J = F Δ t J=F\Delta t J = F Δ t 750 750 750
Worked example 3 — Why airbags work (sameΔ p \Delta p Δ p Δ t \Delta t Δ t 15 m/s 15\ \text{m/s} 15 m/s Δ p = 70 × 15 = 1050 kg⋅m/s \Delta p = 70\times15 = 1050\ \text{kg·m/s} Δ p = 70 × 15 = 1050 kg⋅m/s
Hard dashboard, stop in 0.02 s 0.02\ \text{s} 0.02 s F = 1050 / 0.02 = 52,500 N F = 1050/0.02 = 52{,}500\ \text{N} F = 1050/0.02 = 52 , 500 N
Airbag, stop in 0.20 s 0.20\ \text{s} 0.20 s F = 1050 / 0.20 = 5,250 N F = 1050/0.20 = 5{,}250\ \text{N} F = 1050/0.20 = 5 , 250 N Why this matters: Same momentum change, but 10× the time ⇒ 1/10 the force . Survivable.
Worked example 4 — Variable mass (wherem a ma ma M M M v v v d m d t = μ \frac{dm}{dt}=\mu d t d m = μ F net = 0 F_{\text{net}}=0 F net = 0 d ( m v ) d t = 0 \frac{d(mv)}{dt}=0 d t d ( m v ) = 0 m m m v v v drop . Using only "F = m a F=ma F = ma a = 0 ⇒ v a=0 \Rightarrow v a = 0 ⇒ v v d m d t v\frac{dm}{dt} v d t d m
Common mistake "Moving objects must have a forward force on them."Why it feels right: Everyday experience — a car needs the engine running to keep moving. The truth: that's only because friction/air drag are removing momentum. Without any net force, an object keeps its velocity forever (First Law). Fix: Ask "what's the NET force?" Constant velocity ⇒ net force = 0, no matter how fast.
Common mistake "Heavier objects fall faster becauseF = m a F=ma F = ma Why it feels right: Bigger F F F a a a The truth: Weight F = m g F=mg F = m g a = F / m = m g / m = g a = F/m = mg/m = g a = F / m = m g / m = g Fix: A larger mass has both more force AND more inertia; they cancel.
Common mistake "Impulse is just force."Why it feels right: Both relate to pushing. The truth: Impulse = F Δ t =F\Delta t = F Δ t accumulated over time . A tiny force over hours can give a huge impulse. Fix: Always include the time.
Common mistake Forgetting momentum is a vector in collisions.Why it feels right: Speeds look like plain numbers. The truth: A ball bouncing back from 20 20 20 30 30 30 50 m 50m 50 m 10 m 10m 10 m Fix: Choose a positive direction and keep signs.
Recall Feynman: explain it to a 12-year-oldImagine pushing a shopping trolley. Push harder → it speeds up faster. Pack it heavier → harder to speed up. That's F = m a F=ma F = ma how hard you push, but how long . A gentle push for a long time can do the same job as a hard shove for a split second. That's why catching a fast ball, you pull your hands back — you stretch out the time so the sting (force) is smaller.
"FIND" → F orce = I mpulse over time, and impulse = change iN D elta-momentum.
And for airbags: "More Time = Less Crime (force)."
What is the most fundamental statement of Newton's 2nd law? F ⃗ net = d p ⃗ d t \vec{F}_{\text{net}} = \dfrac{d\vec{p}}{dt} F net = d t d p — net force equals the rate of change of momentum.
When does F = m a F=ma F = ma When mass changes; use
F ⃗ = d ( m v ⃗ ) d t = m a ⃗ + v ⃗ d m d t \vec F = \frac{d(m\vec v)}{dt} = m\vec a + \vec v\frac{dm}{dt} F = d t d ( m v ) = m a + v d t d m .
Derive F = m a F=ma F = ma F ⃗ = d ( m v ⃗ ) d t = m d v ⃗ d t + v ⃗ d m d t \vec F=\frac{d(m\vec v)}{dt}=m\frac{d\vec v}{dt}+\vec v\frac{dm}{dt} F = d t d ( m v ) = m d t d v + v d t d m ; constant mass ⇒ second term 0 ⇒
F ⃗ = m a ⃗ \vec F=m\vec a F = m a .
State the impulse–momentum theorem. J ⃗ = ∫ F d t = Δ p ⃗ = m v ⃗ f − m v ⃗ i \vec J=\int F\,dt=\Delta\vec p = m\vec v_f-m\vec v_i J = ∫ F d t = Δ p = m v f − m v i .
Units of impulse? N·s, equal to kg·m/s (same as momentum).
For a constant force, impulse = ? J ⃗ = F ⃗ Δ t \vec J=\vec F\,\Delta t J = F Δ t .
Why do airbags reduce injury? They increase the stopping time
Δ t \Delta t Δ t , so for the same
Δ p \Delta p Δ p the force
F = Δ p / Δ t F=\Delta p/\Delta t F = Δ p /Δ t is smaller.
Why do all objects fall with acceleration g g g a = F / m = m g / m = g a=F/m=mg/m=g a = F / m = m g / m = g ; mass cancels.
A 2 kg block: 10 N right, 6 N friction left. Acceleration? Net 4 N →
a = 2 m/s 2 a=2\ \text{m/s}^2 a = 2 m/s 2 right.
What does "net" mean in F n e t = m a F_{net}=ma F n e t = ma The vector sum of all forces acting on the object.
product rule + constant mass
net force = vector sum of all forces
Intuition Hinglish mein samjhoDekho, Newton ka second law ek hi baat keh raha hai: force koi cheez ko chalti rakhne ke liye nahi chahiye — force toh motion ko badalne ke liye chahiye. Asli fundamental form hai F ⃗ n e t = d p ⃗ d t \vec F_{net} = \frac{d\vec p}{dt} F n e t = d t d p F ⃗ = m a ⃗ \vec F = m\vec a F = m a
Impulse–momentum ka funda ekdum simple hai: F ⃗ \vec F F J ⃗ = Δ p ⃗ \vec J = \Delta \vec p J = Δ p time Δ t \Delta t Δ t F = Δ p / Δ t F=\Delta p/\Delta t F = Δ p /Δ t . Same Δ p \Delta p Δ p
Ek important warning: momentum ek vector hai. Agar ball 20 20 20 30 30 30 50 m 50m 50 m 10 m 10m 10 m a = m g / m = g a=mg/m=g a = m g / m = g
