Foundations — Impulse-momentum theorem — derivation

By the end of this page you will be able to read this line and know why every mark on it is there: $\vec{J}={\int }_{t_1}^{t_2}\vec{F}dt=\Delta \vec{p}$ But do not try to read it yet — none of its symbols are defined. We build them from nothing, in order, each one anchored to a picture. Nothing here assumes you've seen an arrow-over-a-letter or an $\int$ sign before. We come back and assemble that line, symbol by symbol, in Section 7.

1. The arrow on top: what a vector means

Look at Figure 1 below. Both pushes are "5 units strong," but one shoves right and one shoves left. A plain number like $5$ cannot tell them apart — it forgot the direction.

• Plain words: "an amount that points somewhere."
• The picture: Figure 1 — a longer arrow means bigger; rotate it and you change direction.
• Why the topic needs it: a ball hitting a wall and bouncing back has the same speed but the opposite direction. If we only used plain numbers we'd think "nothing changed," which is wrong. The arrow is what lets us say the motion reversed.

2. $m$, $\vec{v}$, and the product $m\vec{v}$: momentum

Now multiply them. A heavy slow truck and a light fast pebble can both be "hard to stop." What captures that combined stubbornness?

• The picture: Figure 2 — the velocity arrow, stretched (or shrunk) by the mass. Same direction, new length.
• Why the topic needs it: the whole theorem is about changing momentum. You can't talk about a change in something you haven't named. $\vec{p}$ is the "motion-currency" the theorem spends.

3. $\Delta$: the "change in" symbol

• The picture: two arrows, "before" and "after." $\Delta \vec{p}$ is the arrow you'd draw from the tip of the old one to the tip of the new one — the correction needed to turn "before" into "after."
• Why the topic needs it: the theorem's whole right-hand side is $\Delta \vec{p}$. When a ball bounces, ${\vec{p}}_i=+mv$ and ${\vec{p}}_f=-mv$, so $\Delta \vec{p}=-mv-(+mv)=-2mv$ — that famous "doubling" is just subtraction done carefully with signs.

4. Force $\vec{F}$, net force ${\vec{F}}_{\text{net}}$, and the rate-of-change idea $\frac{d\vec{p}}{dt}$

Real objects rarely feel a single force. A thrown ball feels gravity pulling down, air pushing back, and a bat slamming it forward — all at once. Which one changes its momentum? The answer is: their combined effect.

• Why the topic needs it: only the net push changes momentum. Writing ${\vec{F}}_{\text{net}}$ (rather than a lone $\vec{F}$) is our promise to always add up everything acting on the object before we say how its motion changes.

Newton's deepest statement is not $F=ma$ but this: the net force is whatever changes momentum, and the bigger it is, the faster momentum changes. To write "how fast $\vec{p}$ changes" we need one more symbol.

• Why this tool and not just $\Delta p/\Delta t$? $\Delta p/\Delta t$ gives the average change over a whole interval. But force during a collision changes moment-to-moment. The derivative $d\vec{p}/dt$ is the instantaneous version — the average taken over a slice so short the force has no time to change inside it. It answers "what is the force at this exact instant?"
• The picture: Figure 3 — on a momentum-vs-time graph, $d\vec{p}/dt$ is the steepness (slope) of the curve at one point — steep means momentum is changing fast, so the net force is large.

So Newton's law, in its original honest form: ${\vec{F}}_{\text{net}}=\frac{d\vec{p}}{dt}$ "The net force equals how fast momentum is changing." (See Newton's Second Law for why this is more fundamental than $F=ma$.)

5. The integral $\int$: adding up infinitely many tiny pushes

The derivative broke time into slices. To get the total effect we must glue the slices back together. That gluing operation is the integral.

• The picture: Figure 4 — on a force-vs-time graph, each $\vec{F}dt$ is the area of one thin rectangle. Summing them all gives the total area under the force curve. That area is the impulse.
• Why this tool and not simple multiplication? If the force were constant you could just do $F\times \Delta t$ (one big rectangle). But real collision forces spike and fade — you can't multiply by a single number when the number keeps changing. The integral is "multiplication for quantities that vary." (See Force–Time Graphs.)

6. Assembling the theorem: integrating Newton's law

Now every symbol is earned, so we can watch them combine. We start from Newton's law (Section 4) and integrate both sides over the collision time — nothing new, just using the tools.

Step 1 — Start from Newton's second law. ${\vec{F}}_{\text{net}}=\frac{d\vec{p}}{dt}$ WHAT: the net force equals the instantaneous rate momentum changes. WHY: it's the most honest form of the law (Section 4). LOOKS LIKE: Figure 3 — force is the slope of the $p$–$t$ curve at each instant.

Step 2 — Multiply both sides by the tiny time slice $dt$. ${\vec{F}}_{\text{net}}dt=d\vec{p}$ WHAT: we isolated one sliver's worth of push on the left and one sliver's worth of momentum change on the right. WHY: we want a total over an interval, so we prepare to add slivers up. LOOKS LIKE: one thin rectangle in Figure 4 on the left; one tiny nudge of the arrow-tip in Figure 3 on the right.

Step 3 — Sum every sliver from $t_1$ to $t_2$ (integrate both sides). ${\int }_{t_1}^{t_2}{\vec{F}}_{\text{net}}dt={\int }_{t_1}^{t_2}\frac{d\vec{p}}{dt}dt={\int }_{{\vec{p}}_1}^{{\vec{p}}_2}d\vec{p}$ WHAT: the left side sums all the force rectangles; the right side sums all the tiny momentum changes. WHY: summing (integrating) is exactly how we glue instantaneous facts into a total (Section 5). LOOKS LIKE: the whole shaded area of Figure 4 on the left.

Step 4 — Evaluate the right-hand sum (the Fundamental Theorem of Calculus). Adding up every tiny change $d\vec{p}$ from the start value ${\vec{p}}_1$ to the end value ${\vec{p}}_2$ just leaves the overall change — the little steps telescope: ${\int }_{{\vec{p}}_1}^{{\vec{p}}_2}d\vec{p}={\vec{p}}_2-{\vec{p}}_1=\Delta \vec{p}$ WHAT: summing all the tiny momentum changes gives final-minus-initial. WHY: that's the meaning of $\Delta$ (Section 3); integrating a change gives back the total change. LOOKS LIKE: the single "before-to-after" arrow of Section 3's picture.

Step 5 — Recognise the left side as impulse. The left integral is exactly our definition of impulse $\vec{J}$ (Section 5). Putting the two sides together:

7. Putting the symbols in order

Each symbol above needed the one before it. The prerequisite map below shows the build order, not the physics flow. How to read it: start at the top boxes (the raw ideas you meet first) and follow the arrows downward — each arrow means "this idea is needed to build the next one." All streams converge at the bottom on the theorem itself.

Read top to bottom: arrows and mass build momentum; force gathered into a net force plus the rate-of-change idea build Newton's law; delta plus momentum build $\Delta \vec{p}$; the integral plus net force build impulse; and the theorem is where all three streams meet.

Equipment checklist

Test yourself — cover the right side. If any answer is fuzzy, re-read its section before the derivation.

Connections

• इसी को Hinglish में पढ़ो → — the parent topic, in Hinglish.
• Newton's Second Law — where ${\vec{F}}_{\text{net}}=d\vec{p}/dt$ comes from.
• Force–Time Graphs — the picture behind the integral (impulse = area).
• Conservation of Momentum — what $\Delta \vec{p}=0$ leads to.
• Collisions — Elastic and Inelastic — where bouncing-vs-sticking sign work pays off.
• Work–Energy Theorem — the same ideas but force over distance instead of time.
• Center of Mass Motion — momentum bookkeeping for whole systems.