Visual walkthrough — Impulse-momentum theorem — derivation

This is a deep-dive under Impulse–Momentum Theorem — Derivation. If you want the same story in Hinglish, read the Hinglish version →.

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Step 1 — What is "momentum"? Draw it as an arrow with weight

WHAT. We start with a single moving block. It has a mass — call it $m$, "how much stuff is in it" — and a velocity $\vec{v}$, "how fast and which way it moves." We glue them into one quantity: $\vec{p}=m\vec{v}$

Read this equation left to right:

• $\vec{p}$ — the little arrow on top means "this points somewhere"; $p$ stands for momentum, the object's quantity of motion.
• $m$ — a plain number (no arrow), the mass in kilograms. It only scales the arrow, it can't turn it.
• $\vec{v}$ — the velocity arrow: its length is the speed, its direction is where the block heads.

WHY. Because "hard to stop" isn't just about speed. A slow truck and a fast pebble can be equally hard to stop. Momentum bundles both mass and velocity into the one number that actually resists change.

PICTURE. Two blocks below. The small block moves fast (long velocity arrow) but is light, so its momentum arrow (blue) is short. The heavy block moves slowly but its momentum arrow is long. Same idea — "how hard to stop" — captured by the blue arrow's length.

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Step 2 — What is "force"? The thing that changes the arrow

WHAT. Now push the block. Newton's deepest statement of a force is not $F=ma$ — it is: ${\vec{F}}_{\text{net}}=\frac{d\vec{p}}{dt}$

Term by term:

• ${\vec{F}}_{\text{net}}$ — the net (total) force arrow, the sum of all pushes and pulls.
• $\frac{d\vec{p}}{dt}$ — this fraction reads "the rate at which the momentum arrow changes." The little $d$'s mean "an infinitely tiny slice": $d\vec{p}$ is a tiny change in momentum, $dt$ a tiny slice of time. Their ratio is how fast $\vec{p}$ is changing right now.

WHY this tool — the derivative. We use $d\vec{p}/dt$ because force is instantaneous: it can change from moment to moment. We need a tool that reports the change at an instant, not over a long stretch. That tool is the derivative — the slope of the momentum-versus-time curve at a single point. Nothing else answers "how fast is it changing right now?"

Why this form, not $F=m\vec{a}$? Because $F=m\vec{a}$ quietly assumes $m$ never changes. The momentum form survives rockets that burn fuel and lose mass. We build on the sturdier foundation.

PICTURE. The momentum arrow at time $t$, and a slightly longer one a tiny instant $dt$ later. The green arrow is the tiny difference $d\vec{p}$ — and the force points exactly along it.

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Step 3 — Multiply by $dt$: freeze one tiny push

WHAT. Take Step 2 and multiply both sides by the tiny time slice $dt$. The $dt$ under the fraction cancels: ${\vec{F}}_{\text{net}}dt=d\vec{p}$

Term by term:

• ${\vec{F}}_{\text{net}}dt$ — force times a tiny time. A tiny "push-parcel." Its size is the strength of the push $\times$ the sliver of time it lasted.
• $d\vec{p}$ — the tiny change in momentum that this one push-parcel caused.

WHY. A collision is not one clean push — it is thousands of tiny push-parcels happening back to back. To add them up later, we first need to know what one parcel does. This equation says: one tiny push-parcel produces exactly one tiny momentum change of the same size and direction. They are two names for one thing.

PICTURE. A thin vertical sliver under a force curve — a rectangle of width $dt$ and height $F$. Its area is the push-parcel $Fdt$. Beside it, the matching tiny green nudge to the momentum arrow.

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Step 4 — Add up every parcel: the integral appears

WHAT. Real pushes last a finite time, from a start $t_1$ to an end $t_2$. Sum all the tiny push-parcels across that stretch. "Sum of infinitely many infinitely thin slices" has a name and a symbol — the integral $\int$: ${\int }_{t_1}^{t_2}{\vec{F}}_{\text{net}}dt={\int }_{{\vec{p}}_1}^{{\vec{p}}_2}d\vec{p}$

Term by term:

• ${\int }_{t_1}^{t_2}$ — the stretched-S means "add up," and the little $t_1,t_2$ say "from start time to end time."
• ${\vec{F}}_{\text{net}}dt$ inside — each tiny push-parcel we are summing.
• Right side: ${\int }_{{\vec{p}}_1}^{{\vec{p}}_2}d\vec{p}$ — add up all the tiny momentum nudges, from the momentum at the start ${\vec{p}}_1$ to the momentum at the end ${\vec{p}}_2$.

WHY the integral, not multiplication? If the force were constant we could just do $F\times \Delta t$. But collision forces spike wildly — they are not one height. The integral is the only tool that adds up a changing height correctly: it is the area under the force–time curve. Multiplication is the special case of a flat-top rectangle.

PICTURE. The whole messy force curve filled in with many thin slivers — the total shaded area is the left-hand integral. Every sliver's green nudge, stacked, grows the momentum arrow from ${\vec{p}}_1$ to ${\vec{p}}_2$.

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Step 5 — Evaluate the right side: the theorem is born

WHAT. Adding up tiny changes in momentum just gives the total change — final minus initial: ${\int }_{{\vec{p}}_1}^{{\vec{p}}_2}d\vec{p}={\vec{p}}_2-{\vec{p}}_1$ So the whole equation collapses to: $\boxed{\vec{J}\equiv {\int }_{t_1}^{t_2}{\vec{F}}_{\text{net}}dt={\vec{p}}_2-{\vec{p}}_1=\Delta \vec{p}}$

Term by term:

• $\vec{J}$ — we give the left-hand area its own name, impulse. The $\equiv$ means "is defined as."
• ${\vec{p}}_2-{\vec{p}}_1$ — final momentum minus initial momentum.
• $\Delta \vec{p}$ — the triangle $\Delta$ is shorthand for "change in": $\Delta \vec{p}={\vec{p}}_2-{\vec{p}}_1$.

WHY this is the payoff. We never needed to know the messy force in detail. The left side (total push, hard to know) is forced to equal the right side (change in motion, easy to measure). That is the entire usefulness of the theorem.

PICTURE. On the left, the shaded area of the force curve labelled $\vec{J}$. On the right, the initial arrow ${\vec{p}}_1$ and the final arrow ${\vec{p}}_2$, with the red arrow $\Delta \vec{p}$ bridging their tips. The area equals the red arrow.

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Step 6 — The average-force shortcut (flatten the spike)

WHAT. The theorem holds for any force shape. So replace the ugly spike with a flat rectangle of the same area — a made-up constant force $\bar{F}$ (the bar means "average"): $\overline{\stackrel{⃗}{F}}\Delta t=\Delta \vec{p}⟹\overline{\stackrel{⃗}{F}}=\frac{\Delta \vec{p}}{\Delta t}$

Term by term:

• $\overline{\stackrel{⃗}{F}}$ — the average force: the single height that, over time $\Delta t$, gives the same area.
• $\Delta t=t_2-t_1$ — the total duration of the push.
• $\Delta \vec{p}/\Delta t$ — total momentum change divided by total time.

WHY. In real problems we rarely have $\vec{F}(t)$; we have "how much did the velocity change, and over how long." This rearranged form answers that directly and solves most collision/catch questions.

PICTURE. The tall thin spike and a short wide rectangle side by side — identical shaded areas. That's the trade-off: same impulse, either a big force for a short time OR a small force for a long time.

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Step 7 — Every case: stop, bounce, and zero

Impulse is a vector, so signs decide everything. Pick rightward $=+$.

WHAT — three scenarios for a ball of mass $m$ arriving at speed $v$:

1. Stops (glove catches it): $v_i=+v\to v_f=0$, so $\Delta p=m(0)-mv=-mv$.
2. Bounces back at the same speed: $v_i=+v\to v_f=-v$, so $\Delta p=m(-v)-mv=-2mv$.
3. Passes through unchanged (degenerate / no interaction): $v_f=v_i$, so $\Delta p=0$, hence $\vec{J}=0$.

WHY these cases matter. They show the sign is not decoration:

• A bounce demands twice the impulse of a mere stop — reversing motion is "harder" than killing it. That's why a bouncing hailstone dents a car more than a sticking snowflake.
• The zero case is the sanity check: no change in momentum ⇒ no impulse ⇒ (if $\Delta t$ is finite) no net force. It links straight to Conservation of Momentum.

PICTURE. Three rows. Top: incoming red arrow shrinks to nothing (stop, $\Delta p=-mv$). Middle: incoming red arrow flips to a full-length leftward arrow (bounce, $\Delta p=-2mv$). Bottom: arrow unchanged (pass-through, $\Delta p=0$). The length of each $\Delta p$ arrow shows the impulse the wall had to supply.

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The one-picture summary

WHAT it compresses. The whole chain, one image: a tiny push-parcel $\vec{F}dt$ → summed into the area $\int \vec{F}dt$ → equal to the red change-arrow $\Delta \vec{p}={\vec{p}}_2-{\vec{p}}_1$ → flattenable into $\bar{F}\Delta t$. Left side lives in the force–time world; right side lives in the momentum-arrow world; the theorem is the bridge between them.

Recall Feynman: retell the whole walkthrough to a friend

Picture a moving block. Its "moving-ness" is an arrow — long if it's heavy or fast. That arrow is momentum. A force is anything that changes that arrow, and Newton says the rate it changes equals the force. Now imagine a punch that lasts only a blink: chop that blink into thousands of even tinier blinks. In each one, "force times that sliver of time" makes a tiny nudge to the momentum arrow — and the nudge is exactly as big as the push. Stack every sliver up (that's the area under the force–time graph, which we call impulse) and the arrow has grown from its starting length to its ending length. So total push = total change in the momentum arrow. That's the whole theorem. And since only the area matters, you can trade a huge spike for a gentle, long push of the same area — which is exactly what airbags, crumple zones, and bending your knees do: same change in motion, spread over more time, so the force you feel is small.

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Connections

• Newton's Second Law — Step 2 is literally this law in momentum form.
• Force–Time Graphs — Step 4's shaded area is the impulse.
• Conservation of Momentum — Step 7's zero case: no net impulse ⇒ momentum unchanged.
• Collisions — Elastic and Inelastic — the stop/bounce split of Step 7.
• Work–Energy Theorem — force over distance instead of over time; the spatial twin of this walkthrough.
• Center of Mass Motion — external impulse shifts the whole system's momentum arrow.