Impulse-momentum theorem — derivation: Worked examples

This is the problem gym for the parent theorem — Impulse–Momentum. The parent proved why $J = Δ p$ . Here we hit every kind of situation that theorem can be thrown into, so no exam case surprises you. (Prefer Hindi-English? See the Hinglish version.)

Before anything else, one reminder of the two tools we will reuse constantly:

Everything below is these three relations, applied to a different "shape" of problem each time.

The scenario matrix

A momentum problem varies along a few independent knobs. Each cell below is one distinct thing that can go wrong or feel new. The worked examples are labelled with the cell(s) they cover, and together they fill the whole table.

Cell	What changes	Example(s)
A. Stop — object ends at rest	$v_{f} = 0$ , $Δ p = - m v_{i}$	Ex 1
B. Bounce — velocity reverses sign	$v_{f} = - v_{i}$ , $Δ p = - 2 m v_{i}$	Ex 2
C. Partial bounce — reverses but slower	$	v_f
D. Speed-up (same direction)	$v_{f}, v_{i}$ same sign, $	v_f
E. Varying force — must integrate	$F = F (t)$ , area under curve	Ex 5
F. Zero / degenerate — no net impulse	$Δ p = 0$ even though force acts	Ex 6
G. Limiting case — $Δ t \to 0$	force blows up as time shrinks	Ex 7
H. Real-world word problem — 2D	vector components, pick $+$ axes	Ex 8
I. Exam twist — invert a graph, back-solve velocity	given $J$ , run the theorem in reverse	Ex 9

The single trap running through all of them: impulse is a vector, so signs (or components) are the whole game. We fix a $+$ direction every single time.

Ex 1 — Cell A: the clean stop

Worked example Sandbag catch

A $0.40 kg$ beanbag flies in at $12 m/s$ and is caught (brought to rest) in $0.30 s$ . Find the impulse on it and the average force.

Forecast: Will the impulse be positive or negative if we call the incoming direction $+$ ? Guess the sign before reading on.

Step 1 — Fix a $+$ direction. Let the beanbag's incoming motion be $+$ . Then $v_{i} = + 12$ , $v_{f} = 0$ . Why this step? Impulse is a vector; without a chosen axis "the force" has no sign and we'd get nonsense.

Step 2 — Change in momentum. $Δ p = m v_{f} - m v_{i} = 0.40 (0) - 0.40 (12) = - 4.8 kg\cdotm/s$ Why this step? The theorem says $J = Δ p$ , so computing $Δ p$ is computing the impulse. The minus is physics telling us the force points backwards, opposing the incoming beanbag. (Written $kg\cdotm/s$ because we used mass × velocity; it equals $- 4.8 N\cdots$ .)

Step 3 — Average force. $\overset{ˉ}{F} = \frac{Δ p}{Δ t} = \frac{- 4.8}{0.30} = - 16 N$ Why this step? $\overset{ˉ}{F} Δ t = Δ p$ rearranged. The sign again says "backwards."

Verify: Units $\frac{kg\cdotm/s}{s} = kg\cdotm/s^{2} = N$ ✓. Sign is negative (opposes motion) as forecast. Magnitude $16 N$ ≈ the weight of a $1.6 kg$ object — plausible for a gentle catch.

Ex 2 — Cell B: the perfect bounce (doubling)

Worked example Ball off a wall

A $0.25 kg$ ball hits a wall at $8 m/s$ and rebounds at the same speed $8 m/s$ . Impulse from the wall?

Forecast: Is this impulse bigger, smaller, or equal to just stopping the ball? By how much?

Step 1 — Fix $+$ = into the wall. So $v_{i} = + 8$ , $v_{f} = - 8$ . Why this step? The ball leaves in the opposite direction, so its final velocity must carry a minus. This flip is the entire reason bouncing is special.

Step 2 — Change in momentum. $Δ p = 0.25 (- 8) - 0.25 (8) = - 2 - 2 = - 4 kg\cdotm/s = - 4 N\cdots$ Why this step? Because $v_{f}$ and $v_{i}$ have opposite signs, the two terms add in magnitude instead of partly cancelling. (Same quantity, either unit spelling.)

Step 3 — Compare to a stop. A pure stop would give $Δ p = 0.25 (0) - 0.25 (8) = - 2 kg\cdotm/s$ . Why this step? To see the "doubling" explicitly: $- 4$ vs $- 2$ .

Verify: $- 4 = 2 \times (- 2)$ ✓. Bouncing needs twice the impulse of stopping — which is why a bouncing hailstone dents a roof more than a splatting raindrop of equal mass and speed.

Ex 3 — Cell C: partial bounce (both signs, unequal)

Worked example Squishy ball

A $0.30 kg$ ball hits the floor moving down at $10 m/s$ and rebounds up at only $5 m/s$ (energy lost). Find impulse from the floor.

Forecast: Will the magnitude be closer to a stop ( $3 kg\cdotm/s$ ) or a full bounce ( $6 kg\cdotm/s$ )?

Step 1 — Fix $+$ = upward. Then the incoming velocity is downward: $v_{i} = - 10$ . The rebound is upward: $v_{f} = + 5$ . Why this step? Choosing up as $+$ is arbitrary but must be consistent; the incoming ball then carries the minus.

Step 2 — Change in momentum. $Δ p = 0.30 (+ 5) - 0.30 (- 10) = 1.5 + 3.0 = + 4.5 kg\cdotm/s$ Why this step? Subtracting a negative $v_{i}$ adds to the total — the floor both stopped the fall and launched it back up.

Step 3 — Interpret the $+$ sign. Positive means the floor's impulse points upward. Why this step? Sanity: the floor must push up to reverse a downward-falling ball.

Verify: $4.5$ lies between $3.0$ (stop) and $6.0$ (perfect bounce) ✓ — exactly as forecast, since the bounce is partial. Sign is up ✓.

Ex 4 — Cell D: speed-up in the same direction

Worked example Rocket sled kick

A $50 kg$ cart already rolling right at $4 m/s$ gets a booster that raises it to $9 m/s$ (still right) over $2.0 s$ . Find impulse and average force.

Forecast: Both velocities point the same way. Do the momentum terms add or subtract here?

Step 1 — Fix $+$ = right. $v_{i} = + 4$ , $v_{f} = + 9$ — same sign. Why this step? No reversal here, so this is the case where signs do not flip. We must not accidentally introduce a minus.

Step 2 — Change in momentum. $Δ p = 50 (9) - 50 (4) = 450 - 200 = + 250 kg\cdotm/s$ Why this step? Same-sign velocities mean we take a genuine difference, not a sum — contrast with the bounce cases.

Step 3 — Average force. $\overset{ˉ}{F} = \frac{250}{2.0} = + 125 N$ Why this step? $\overset{ˉ}{F} = Δ p /Δ t$ ; the $+$ says the push is in the direction of motion (it speeds it up).

Verify: Units → N ✓. Positive force on an already-forward object → speeds it up ✓, consistent with $v_{f} > v_{i}$ .

Ex 5 — Cell E: force that varies (must integrate)

Worked example Triangular push, from a graph

A block starts at rest. A force (pushing in the $+$ direction, which we take as the direction of motion) ramps linearly from $0 N$ up to $30 N$ over $t = 0$ to $t = 4 s$ , forming a triangle on the force–time graph (see figure). The block has mass $3 kg$ . Find its final speed.

The red shaded triangle in the figure is the impulse: its area equals

\int F d t

. Notice the force never stays at its peak — that is exactly why the naive

F Δ t

(dashed rectangle at

30 N

) overshoots.

Forecast: The force is not constant. Is $J = F Δ t$ with $F = 30$ legal here? (No — but guess the true answer's ballpark.)

Step 0 — Fix the $+$ direction. Take the direction the force pushes as $+$ . Then $F (t) > 0$ throughout, so the impulse comes out positive and the block speeds up in the $+$ direction. Why this step? Even a "1D" integration needs a sign convention, or the area's sign is meaningless.

Step 1 — Find the ramp's slope. A straight line from $(0, 0)$ to $(4 s, 30 N)$ has slope $slope = \frac{30 - 0}{4 - 0} = 7.5 N/s, so F (t) = 7.5 t .$ Why this step? We need the actual function $F (t)$ before we can integrate it; slope = rise over run gives it from the two endpoints.

Step 2 — Impulse = area under the curve. The red triangle has base $4 s$ and height $30 N$ : $J = \frac{1}{2} \times base \times height = \frac{1}{2} (4) (30) = 60 N\cdots$ Why this step? $J = \int F d t$ is the area under $F (t)$ . For a straight ramp the area is a triangle — no calculus needed, geometry does it.

Step 3 — Check with the integral. Using $F (t) = 7.5 t$ from Step 1: $J = \int_{0}^{4} 7.5 t d t = [3.75 t^{2}]_{0}^{4} = 3.75 (16) = 60 N\cdots$ Why this step? To prove the "area = integral" claim rather than trust the picture.

Step 4 — Theorem gives final speed. $J = m v_{f} - 0 \Rightarrow 60 = 3 v_{f} \Rightarrow v_{f} = 20 m/s$ Why this step? The area we found is the momentum change; divide by mass for velocity.

Verify: If we had wrongly used $F Δ t = 30 \times 4 = 120$ , we'd get $v_{f} = 40$ — double the truth, because the average force is only $15 N$ , not $30$ . $\overset{ˉ}{F} Δ t = 15 (4) = 60$ ✓ matches. That is exactly why we integrate.

Ex 6 — Cell F: zero net impulse (degenerate)

Worked example Full round trip, back to start

A $2 kg$ puck is moving east at $5 m/s$ . It is then pushed (some forces slow it, stop it, reverse it, then speed it up again) so that after a whole sequence it ends up once more moving east at $5 m/s$ , identical to how it began. What is the net impulse delivered over the entire sequence?

Forecast: Forces clearly acted (they turned the puck around and back!). Can the net impulse still be zero?

Step 1 — Fix $+$ = east. Identify only the start and end states. Start: $v_{i} = + 5$ . End of the full sequence: $v_{f} = + 5$ (same as start). Why this step? The theorem cares only about the endpoints, not the messy path between — so we ignore the intermediate stopping and reversing entirely.

Step 2 — Net impulse. $J_{net} = m v_{f} - m v_{i} = 2 (5) - 2 (5) = 0 N\cdots$ Why this step? Same start and end velocity → $Δ p = 0$ → zero net impulse, even though individual forces along the way were large.

Step 3 — Reconcile with intuition. Big forces happened, but they came in canceling pairs: a westward impulse that reversed the puck, then an eastward impulse that restored it. As vectors they sum to zero. Why this step? To show "no net impulse" ≠ "no forces" — it means the vector sum of the impulses cancels.

Verify: $Δ p = 0$ ✓. This is the seed idea of Conservation of Momentum: when net external impulse is zero, momentum is unchanged.

Ex 7 — Cell G: the limiting case ( $Δ t \to 0$ )

Worked example Why hard floors hurt

A $0.5 kg$ phone falls and hits at $6 m/s$ , then stops. Compute the average force for two stopping times: (a) onto foam, $Δ t = 0.10 s$ ; (b) onto concrete, $Δ t = 0.002 s$ . What happens as $Δ t \to 0$ ?

Forecast: The momentum change is the same both times. So what carries the difference in "damage"?

Step 1 — Same $Δ p$ for both. Take $+$ = downward direction of impact; $v_{i} = + 6$ , $v_{f} = 0$ : $Δ p = 0.5 (0) - 0.5 (6) = - 3 kg\cdotm/s$ Why this step? The floor material can't change the momentum the phone must lose — only how long it takes.

Step 2 — Foam. $\overset{ˉ}{F} = \frac{- 3}{0.10} = - 30 N$ . Step 3 — Concrete. $\overset{ˉ}{F} = \frac{- 3}{0.002} = - 1500 N$ . Why these steps? Same numerator $Δ p$ , tiny denominator $Δ t$ → the fraction explodes.

Step 4 — The limit. As $Δ t \to 0$ with $Δ p$ fixed: $\overset{ˉ}{F} = \frac{Δ p}{Δ t} ⟶ \infty$ Why this step? Shows the general law: an instantaneous stop implies an infinite force. Real materials save us by stretching $Δ t$ (crumple zones, airbags, bending knees).

Verify: Concrete force is $\frac{0.10}{0.002} = 50 \times$ the foam force ✓ ( $30 \times 50 = 1500$ ). Same $Δ p$ , $50 \times$ shorter time, $50 \times$ bigger force — a clean inverse relationship.

Ex 8 — Cell H: 2D real-world word problem (components)

Worked example Hockey puck deflected

A $0.16 kg$ puck slides east at $15 m/s$ . A stick strikes it so it leaves at $15 m/s$ but now moving due north. Find the impulse vector (magnitude and direction) from the stick.

In the figure, the two black arrows are the initial (east) and final (north) momenta; the red arrow joining the tip of

p_{i}

to the tip of

p_{f}

is the impulse

J = p_{f} - p_{i}

. Impulse literally closes the triangle of momenta.

Forecast: Speed didn't change, only direction. Is the impulse zero? (Watch out — it's a vector!)

Step 1 — Set axes. $+ x$ = east, $+ y$ = north. Initial: $p_{i} = (0.16 \times 15, 0) = (2.4, 0) kg\cdotm/s$ . Final: $p_{f} = (0, 0.16 \times 15) = (0, 2.4) kg\cdotm/s$ . Why this step? In 2D each axis is its own 1D problem; we handle $x$ and $y$ separately.

Step 2 — Impulse per component. $J_{x} = 0 - 2.4 = - 2.4 kg\cdotm/s, J_{y} = 2.4 - 0 = + 2.4 kg\cdotm/s$ Why this step? Apply $J = Δ p$ independently in each direction. $J_{x}$ is negative (stick killed the eastward motion); $J_{y}$ positive (stick created northward motion).

Step 3 — Magnitude. $∣ J ∣ = J_{x}^{2} + J_{y}^{2} = (- 2.4)^{2} + (2.4)^{2} = 2.4 2 \approx 3.39 N\cdots$ Why this step? The total impulse is the vector sum of its perpendicular parts (Pythagoras on the arrows in the figure).

Step 4 — Direction. It points northwest — at $4 5^{\circ}$ from north toward west (i.e. $13 5^{\circ}$ measured counter-clockwise from east), since $J_{x} < 0$ , $J_{y} > 0$ and they are equal in size. Why this step? Equal-and-opposite-sign components → the arrow bisects into the second quadrant.

Verify: Even with constant speed, $J \neq = 0$ — changing direction is a momentum change ✓. $2.4 2 \approx 3.394$ ✓, and the impulse points away from the incoming east arrow, toward the outgoing north arrow, as the figure shows.

Ex 9 — Cell I: exam twist (invert the theorem, back-solve velocity)

Worked example Read the theorem backwards

During a $0.020 s$ collision a sensor records that the impulse delivered was $J = 12 N\cdots$ . The struck object has mass $0.60 kg$ and was moving at $v_{i} = - 8 m/s$ (taking rightward as $+$ ). Find (a) the average force, (b) the final velocity.

Forecast: We're given impulse and asked to run the theorem in reverse to get velocity. Will the object end up moving right or left?

Step 1 — Fix $+$ = right (already given). $v_{i} = - 8$ means it starts moving left; the recorded $J = + 12$ is a rightward impulse. Why this step? Signs are the whole game; we lock the convention before inverting anything.

Step 2 — Average force from impulse. $\overset{ˉ}{F} = \frac{J}{Δ t} = \frac{12}{0.020} = 600 N$ Why this step? $J = \overset{ˉ}{F} Δ t$ solved for $\overset{ˉ}{F}$ ; this is the height of the equivalent rectangle whose area equals the recorded impulse (the inverse of Ex 5's "find the area").

Step 3 — Final velocity from the theorem. $J = m v_{f} - m v_{i} \Rightarrow 12 = 0.60 v_{f} - 0.60 (- 8)$ $12 = 0.60 v_{f} + 4.8 \Rightarrow 0.60 v_{f} = 7.2 \Rightarrow v_{f} = + 12 m/s$ Why this step? We know $J$ and the initial state, so the theorem pins down the final state — the inverse of every earlier example, where we knew velocities and found $J$ .

Step 4 — Interpret. The object came in at $- 8$ (leftward) and leaves at $+ 12$ (rightward): it was reversed and sped up by the large rightward impulse. Why this step? Sign check: a large $+$ impulse should push a leftward object into rightward motion, and it does.

Verify: Plug back: $Δ p = 0.60 (12) - 0.60 (- 8) = 7.2 + 4.8 = 12 = J$ ✓ (closes the loop). And $\overset{ˉ}{F} Δ t = 600 (0.020) = 12 = J$ ✓. Units: $\frac{N\cdots}{s} = N$ ✓.

Recall

Recall Which cell is which?

A stop makes $Δ p = ?$ ::: $- m v_{i}$ (final velocity zero). A perfect bounce makes $Δ p = ?$ ::: $- 2 m v_{i}$ — double the stop, because velocity reverses sign. When the force varies with time, impulse equals ::: the area under the force–time curve, $\int F d t$ . Can net impulse be zero even when large forces act? ::: Yes — if start and end velocities match, $Δ p = 0$ (impulses cancel). As stopping time $Δ t \to 0$ with fixed $Δ p$ , the force ::: blows up toward infinity, since $\overset{ˉ}{F} = Δ p /Δ t$ . In 2D, is impulse zero when only the direction changes? ::: No — direction change is still a momentum change; combine components with Pythagoras. Is impulse the same quantity as momentum? ::: No — impulse is the change in momentum, $J = Δ p$ ; they merely share a unit.

Connections

Newton's Second Law — every $\overset{ˉ}{F} = Δ p /Δ t$ step is $F = d p / d t$ averaged; the theorem's origin.
Hinglish exposition — same derivation in Hindi-English.
Force–Time Graphs — Ex 5 and Ex 9 are pure "area under the curve" problems.
Collisions — Elastic and Inelastic — bounce vs stop (Ex 2, 3) is the doorway to collision types.
Conservation of Momentum — Ex 6's zero-net-impulse case is the seed of conservation.
Work–Energy Theorem — the "force over distance" cousin; contrast with force over time here.
Center of Mass Motion — external impulse moves the whole system's total momentum.