Intuition The "amount of motion" ideaImagine getting hit by a ping-pong ball vs. a bowling ball, both moving at the same speed. The bowling ball hurts way more . Now imagine the same bowling ball rolling slowly vs. hurled fast — the fast one hurts more. So how hard something is to stop depends on two things together : how much stuff it has (mass) and how fast it's going (velocity). Momentum is exactly this combined "quantity of motion."
Definition Linear momentumThe linear momentum p ⃗ \vec{p} p mass and its velocity :
p ⃗ = m v ⃗ \vec{p} = m\vec{v} p = m v
It is a vector — it points in the same direction as the velocity.
SI unit: ==kg⋅m/s \text{kg·m/s} kg⋅m/s N⋅s \text{N·s} N⋅s
It is not the same as kinetic energy (a scalar). Same word "motion" in plain English, different physics.
WHY does this specific combination m v mv m v m v 2 mv^2 m v 2 m + v m+v m + v
Because Newton's second law, in its original form, is written in terms of p ⃗ \vec{p} p rate of change equals force. That makes it the natural bookkeeping tool for "how motion gets passed around" — especially in collisions.
Newton actually stated his second law as: force is the rate of change of momentum.
F ⃗ = d p ⃗ d t \vec{F} = \frac{d\vec{p}}{dt} F = d t d p
Let's unpack this to recover the familiar F = m a F=ma F = ma p = m v p=mv p = m v
Intuition Why momentum is "conserved-able"Rearrange: F ⃗ d t = d p ⃗ \vec{F}\,dt = d\vec{p} F d t = d p is the change in momentum (this is impulse , see Impulse–Momentum Theorem ). If two objects only push on each other (internal forces), Newton's third law makes those pushes equal and opposite, so the total p ⃗ \vec{p} p
Because p ⃗ = m v ⃗ \vec{p}=m\vec{v} p = m v m m m
p x = m v x , p y = m v y , ∣ p ⃗ ∣ = m ∣ v ⃗ ∣ = p x 2 + p y 2 p_x = m v_x, \qquad p_y = m v_y, \qquad |\vec{p}| = m|\vec{v}| = \sqrt{p_x^2 + p_y^2} p x = m v x , p y = m v y , ∣ p ∣ = m ∣ v ∣ = p x 2 + p y 2
In 1D, momentum can be negative . A ball moving left at 3 m/s 3\text{ m/s} 3 m/s opposite momentum to one moving right. When they collide, these signs do the cancelling/adding for you — that's the whole point of treating it as a vector.
Kinetic energy is K E = 1 2 m v 2 KE = \tfrac{1}{2}mv^2 K E = 2 1 m v 2
K E = 1 2 m v 2 = ( m v ) 2 2 m = p 2 2 m KE = \frac{1}{2}mv^2 = \frac{(mv)^2}{2m} = \frac{p^2}{2m} K E = 2 1 m v 2 = 2 m ( m v ) 2 = 2 m p 2
Why this step? Multiply and divide by m m m 1 2 m v 2 = 1 2 m 2 v 2 m = ( m v ) 2 2 m \tfrac12 mv^2 = \tfrac{1}{2}\frac{m^2v^2}{m} = \frac{(mv)^2}{2m} 2 1 m v 2 = 2 1 m m 2 v 2 = 2 m ( m v ) 2 m v = p mv=p m v = p
Intuition Two objects can share momentum but not energyMomentum scales linearly with v v v v 2 v^2 v 2 same momentum but very different kinetic energies. This is why momentum is conserved in all collisions but KE only in elastic ones.
Worked example 1) Basic magnitudeA 0.50 kg 0.50\text{ kg} 0.50 kg 20 m/s 20\text{ m/s} 20 m/s p = m v = ( 0.50 ) ( 20 ) = 10 kg⋅m/s, east p = mv = (0.50)(20) = 10\ \text{kg·m/s, east} p = m v = ( 0.50 ) ( 20 ) = 10 kg⋅m/s, east Why "east"? Momentum is a vector — direction is part of the answer.
Worked example 2) Same momentum, different massesA 2 kg 2\text{ kg} 2 kg 3 m/s 3\text{ m/s} 3 m/s 6 kg 6\text{ kg} 6 kg 1 m/s 1\text{ m/s} 1 m/s p 1 = 2 × 3 = 6 kg⋅m/s , p 2 = 6 × 1 = 6 kg⋅m/s p_1 = 2\times3 = 6\ \text{kg·m/s}, \quad p_2 = 6\times1 = 6\ \text{kg·m/s} p 1 = 2 × 3 = 6 kg⋅m/s , p 2 = 6 × 1 = 6 kg⋅m/s K E 1 = 1 2 ( 2 ) ( 3 2 ) = 9 J , K E 2 = 1 2 ( 6 ) ( 1 2 ) = 3 J KE_1 = \tfrac12(2)(3^2)=9\text{ J}, \quad KE_2=\tfrac12(6)(1^2)=3\text{ J} K E 1 = 2 1 ( 2 ) ( 3 2 ) = 9 J , K E 2 = 2 1 ( 6 ) ( 1 2 ) = 3 J Why this matters: equal momentum ≠ equal energy. The fast light cart carries 3× the energy.
Worked example 3) Negative momentum (1D)Cart A: 2 kg 2\text{ kg} 2 kg + 4 m/s +4\text{ m/s} + 4 m/s 3 kg 3\text{ kg} 3 kg − 2 m/s -2\text{ m/s} − 2 m/s p A = 2 ( + 4 ) = + 8 , p B = 3 ( − 2 ) = − 6 p_A = 2(+4)=+8,\quad p_B=3(-2)=-6 p A = 2 ( + 4 ) = + 8 , p B = 3 ( − 2 ) = − 6 p total = + 8 + ( − 6 ) = + 2 kg⋅m/s p_{\text{total}} = +8 + (-6) = +2\ \text{kg·m/s} p total = + 8 + ( − 6 ) = + 2 kg⋅m/s Why add with signs? Because it's a vector sum; opposite directions partially cancel.
Worked example 4) 2D momentumA 0.2 kg 0.2\text{ kg} 0.2 kg v x = 3 m/s v_x = 3\text{ m/s} v x = 3 m/s v y = 4 m/s v_y = 4\text{ m/s} v y = 4 m/s p x = 0.2 ( 3 ) = 0.6 , p y = 0.2 ( 4 ) = 0.8 p_x = 0.2(3)=0.6,\quad p_y=0.2(4)=0.8 p x = 0.2 ( 3 ) = 0.6 , p y = 0.2 ( 4 ) = 0.8 ∣ p ⃗ ∣ = 0.6 2 + 0.8 2 = 0.36 + 0.64 = 1.0 kg⋅m/s |\vec{p}|=\sqrt{0.6^2+0.8^2}=\sqrt{0.36+0.64}=1.0\ \text{kg·m/s} ∣ p ∣ = 0. 6 2 + 0. 8 2 = 0.36 + 0.64 = 1.0 kg⋅m/s Why Pythagoras? Perpendicular components add as a right triangle.
Common mistake "Momentum and kinetic energy are basically the same thing."Why it feels right: both grow when you go faster, both are zero at rest, both involve m m m v v v feel interchangeable.
The fix: p ∝ v p \propto v p ∝ v K E ∝ v 2 KE \propto v^2 K E ∝ v 2 ≥ 0 \ge 0 ≥ 0 p p p quadruples K E KE K E p p p K E KE K E
Common mistake "I'll just use the speeds and add the numbers."Why it feels right: in everyday talk we ignore direction.
The fix: Momentum is a vector . In 1D you MUST keep + / − +/- + / −
F = m a F = ma F = ma F = d p / d t F = dp/dt F = d p / d t Why it feels right: F = m a F=ma F = ma The fix: F = d p / d t F=dp/dt F = d p / d t more general . When mass changes (rockets, raindrops growing, conveyor belts), F = m a F=ma F = ma F = d p / d t F=dp/dt F = d p / d t
Recall Feynman: explain it to a 12-year-oldMomentum is "how much oomph a moving thing has." A thing has more oomph if it's heavy or fast — multiply those two and you get its momentum. A truck and a bicycle can both have the same oomph if the bicycle is going super fast and the truck is crawling. When things bump into each other, oomph never disappears — it just gets passed around, like sharing marbles. That sharing rule is what makes momentum so useful: count the total oomph before a crash, and you'll find the same total after.
"Mass Velocity = Punch you'll receive." p = m v p = mv p = m v
What is the definition of linear momentum? p ⃗ = m v ⃗ \vec{p} = m\vec{v} p = m v , a vector pointing along the velocity.
What are the SI units of momentum? kg⋅m/s \text{kg·m/s} kg⋅m/s (also written
N⋅s \text{N·s} N⋅s ).
Is momentum a scalar or a vector? A vector — it has direction (signs in 1D, components in 2D).
State Newton's 2nd law in momentum form. F ⃗ = d p ⃗ d t \vec{F} = \dfrac{d\vec{p}}{dt} F = d t d p .
How do you get F = m a F=ma F = ma F = d p / d t F=dp/dt F = d p / d t Substitute
p = m v p=mv p = m v ; if
m m m is constant,
F = m d v / d t = m a F=m\,dv/dt=ma F = m d v / d t = ma .
When does F = m a F=ma F = ma F = d p / d t F=dp/dt F = d p / d t When mass changes (rockets, falling-and-growing raindrops, conveyor belts).
Express kinetic energy in terms of momentum. K E = p 2 2 m KE = \dfrac{p^2}{2m} K E = 2 m p 2 .
Two objects have equal momentum but different mass — equal KE? No. The lighter (faster) one has more KE, since
K E = p 2 / 2 m KE=p^2/2m K E = p 2 /2 m .
If speed doubles, what happens to p p p K E KE K E p p p doubles (linear in
v v v );
K E KE K E quadruples (∝
v 2 v^2 v 2 ).
How do you find total momentum of objects moving in opposite directions (1D)? Add with signs: opposite directions get opposite signs and partially cancel.
Intuition Hinglish mein samjhoDekho, momentum ka matlab hai kisi cheez ke "motion ki quantity" — yani woh kitni mushkil se rukegi. Yeh do cheezon par depend karta hai: mass (kitna bhaari hai) aur velocity (kitna tej ja raha hai). Inko multiply karo: p = m v p = mv p = m v vector hai, matlab direction bhi matter karti hai — left wali cheez ka momentum negative, right wali ka positive.
Ab yeh important kyun hai? Kyunki Newton ka asli second law F = d p / d t F = dp/dt F = d p / d t F = m a F=ma F = ma
Ek cheez ka khaas dhyaan rakho: momentum aur kinetic energy ek jaise lagte hain par alag hain. Momentum v v v v 2 v^2 v 2 K E = p 2 / 2 m KE = p^2/2m K E = p 2 /2 m 1 2 m v 2 \tfrac12 mv^2 2 1 m v 2 m m m
