Newton's second law — F = ma (net force), impulse-momentum form: Worked examples

This page is the full case-map for the parent note Newton's Second Law. We will not learn new theory here — we will take the two formulas you already met and hammer them against every possible kind of input: forces pointing every which way, a moment of zero net force, a sign-flip collision, a limit where time shrinks to nothing, a full word problem, and a nasty exam twist.

The scenario matrix

Everything Newton's Second Law can throw at you sorts into these cells. The two tools are:

Force form: $F_{net} = m a$ (mass constant)
Impulse form: $J = F Δ t = Δ p = m v_{f} - m v_{i}$

#	Cell class	What is unusual about it	Example that hits it
A	All forces same direction (simple sum)	just add magnitudes	Ex 1
B	Opposing forces, positive net	subtraction, sign of net	Ex 2
C	Zero net force (degenerate)	$a = 0$ but forces exist	Ex 3
D	Two-dimensional forces (all quadrants)	vectors, not just $\pm$	Ex 4 (figure)
E	Sign-flip collision (velocity reverses)	$Δ p$ bigger than you'd guess	Ex 5
F	Same $Δ p$ , stretch the time (limiting behaviour)	force $\to \infty$ as $Δ t \to 0$	Ex 6 (figure)
G	Real-world word problem	you must build the numbers	Ex 7
H	Exam twist: variable mass	$F = ma$ lies; use $\frac{d p}{d t}$	Ex 8

Below, each worked example is tagged with the cell it fills.

Building blocks (so no symbol is unearned)

Worked Examples

Worked example Ex 3 — Cell C: zero net force (degenerate)

A car cruises at a constant $25 m/s$ on a straight road. The engine drives it forward with $600 N$ . What is the total resistive force, and what is $a$ ?

Forecast: "It's moving, so there must be a forward net force," right? Watch this trap spring.

Constant velocity means $a = 0$ . Why this step? Acceleration is the rate of change of velocity; unchanging velocity ⇒ rate $= 0$ .
Then $F_{net} = ma = m \cdot 0 = 0$ for any mass $m$ . Why this step? $F = ma$ with $a = 0$ forces the net to vanish — no matter the mass, since anything times zero is zero.
So resistance balances drive: resistive force $= 600 N$ backward. Why this step? If net is zero and drive is $+ 600$ , the other force must be $- 600$ .

Verify: $600 + (- 600) = 0 = F_{net}$ . ✓ This is the First Law living inside the Second: zero net force ⇒ velocity frozen. Speed is irrelevant.

Worked example Ex 4 — Cell D: two-dimensional forces in a

different quadrant A $5 kg$ puck on frictionless ice feels two forces at once: $F_{x} = - 30 N$ (i.e. west) and $F_{y} = + 40 N$ (north). Find the impulse delivered in $2 s$ , the magnitude of the acceleration, and its direction. (This lands in quadrant II: negative $x$ , positive $y$ — so we must watch the $arctan$ sign trap.)

Forecast: The answer is not $\frac{- 30 + 40}{5} = 2$ . Forces at right angles do not just add, and a west-north pull points up-and-left, not up-and-right. See the figure.

Draw the two forces tip-to-tail (figure): a west arrow, then a north arrow. Their sum is the diagonal (amber) reaching into the upper-left. Why this step? Forces are vectors; the net is the tip-to-tail resultant, whose length comes from the right triangle.
Magnitude of net force by Pythagoras: $F_{net} = (- 30)^{2} + 4 0^{2} = 900 + 1600 = 2500 = 50 N$ . Why this step? Squaring kills the sign, so the leg lengths are what set the hypotenuse; the resultant magnitude is the same $50 N$ as if both were positive.
Impulse over $2 s$ : $J = F_{net} Δ t$ , so $J_{x} = (- 30) (2) = - 60 N\cdots$ , $J_{y} = (40) (2) = + 80 N\cdots$ , and $∣ J ∣ = (- 60)^{2} + 8 0^{2} = 3600 + 6400 = 10000 = 100 N\cdots$ . Why this step? This is the impulse form $J = F Δ t$ from the header, used for real: a constant force for $2 s$ stamps $100 N\cdots$ of momentum change into the puck.
Acceleration magnitude $a = \frac{F _{net}}{m} = \frac{50}{5} = 10 m/s^{2}$ . Why this step? $F = ma$ works on the magnitude of the net vector.
Direction — the quadrant fix: the raw $arctan \frac{F _{y}}{F _{x}} = arctan \frac{40}{- 30} \approx - 53.1 3^{\circ}$ , which points into quadrant IV — wrong, because $tan$ repeats every $18 0^{\circ}$ and cannot tell quadrant II from IV. Since $F_{x} < 0, F_{y} > 0$ we are truly in quadrant II, so add $18 0^{\circ}$ : $θ = - 53.1 3^{\circ} + 18 0^{\circ} = 126.8 7^{\circ}$ measured counter-clockwise from east. Why this step? $arctan$ only ever returns answers between $- 9 0^{\circ}$ and $+ 9 0^{\circ}$ ; a west-pointing $x$ -component needs the $+ 18 0^{\circ}$ correction to land the angle in the correct (upper-left) quadrant.

Verify: The acceleration points the same way as the net force (they differ only by the scalar $1/ m$ ), so $a$ also lies at $126.8 7^{\circ}$ — up and to the left, exactly matching the amber arrow. Component check: $a_{x} = - 30/5 = - 6$ , $a_{y} = 40/5 = 8$ , and $(- 6)^{2} + 8^{2} = 36 + 64 = 100 = 10$ ✓. Impulse cross-check: $∣ J ∣ = m a Δ t = 5 \cdot 10 \cdot 2 = 100 N\cdots$ ✓.

Worked example Ex 5 — Cell E: the sign-flip collision

A $0.15 kg$ ball hits a wall at $20 m/s$ and rebounds at $30 m/s$ the opposite way in $0.01 s$ . Find the impulse and the average force.

Forecast: Careful — is the speed change $30 - 20 = 10$ , or something bigger? The sign decides.

Choose "+" = the ball's incoming direction. Then $v_{i} = + 20$ , $v_{f} = - 30$ (it now goes the other way). Why this step? A reversal must carry a minus sign, or the physics is wrong.
Impulse = change in momentum $J = Δ p = m (v_{f} - v_{i}) = 0.15 (- 30 - 20) = 0.15 (- 50) = - 7.5 N\cdots$ . Why this step? Impulse–momentum: $J = Δ p = m v_{f} - m v_{i}$ . The $- 50$ (not $- 10$ ) is the whole point of this cell.
Average force from $J = F Δ t$ : $F = \frac{J}{Δ t} = \frac{- 7.5}{0.01} = - 750 N$ . Why this step? For a constant force, impulse is just $F Δ t$ , so $F = J /Δ t$ .

Verify: The minus sign says the force points back against the ball's approach — exactly what a wall does. Magnitude $750 N$ . If you had wrongly used $Δ v = 10$ , you'd get $J = - 1.5$ and $F = - 150 N$ — five times too small. See Collisions and Elasticity for where the rebound speed comes from. ✓

Worked example Ex 6 — Cell F: same

$Δ p$ , stretch the time (limiting behaviour) A $70 kg$ driver at $15 m/s$ must stop. Compare the force for a hard stop ( $0.02 s$ ) versus an airbag stop ( $0.20 s$ ). Then ask: what happens as $Δ t \to 0$ ?

Forecast: Both stops kill the same momentum. Guess the ratio of the two forces before computing.

Impulse needed $J = Δ p = m Δ v = 70 \times 15 = 1050 N\cdots$ . Why this step? The person goes from $15$ to $0$ , so the impulse the seatbelt/airbag must deliver is $∣Δ p ∣ = m v_{i}$ .
Hard dashboard: $F = \frac{J}{Δ t} = \frac{1050}{0.02} = 52, 500 N$ . Why this step? $F = J /Δ t$ with the small time.
Airbag: $F = \frac{J}{Δ t} = \frac{1050}{0.20} = 5, 250 N$ . Why this step? Same impulse, ten times the time.
Limiting behaviour: as $Δ t \to 0$ , $F = \frac{J}{Δ t} \to \infty$ . Why this step? A finite impulse delivered in zero time needs infinite force — the mathematical reason instantaneous collisions are so violent. The figure's curve $F = 1050/Δ t$ shoots up as $Δ t$ shrinks (red asymptote).

Verify: Ratio $\frac{52500}{5250} = 10$ — exactly the ratio of the times ( $0.20/0.02 = 10$ ), confirming "10× the time ⇒ 1/10 the force". ✓ Units: $\frac{N\cdots}{s} = N$ ✓.

Worked example Ex 7 — Cell G: real-world word problem

A $1200 kg$ car speeds up from $0$ to $27 m/s$ (about $97 km/h$ ) in $9 s$ on a level road. (a) What constant net force does this need? (b) If air+rolling drag is $400 N$ , what forward drive force must the tyres supply?

Forecast: Two forces will be at play — the net that accelerates the car, and the extra to overcome drag. Guess which is bigger.

Acceleration $a = \frac{Δ v}{Δ t} = \frac{27 - 0}{9} = 3 m/s^{2}$ . Why this step? Constant acceleration = change in velocity over the time it took.
(a) Net force $F_{net} = ma = 1200 \times 3 = 3600 N$ . Why this step? This is the leftover force that actually accelerates the mass.
(b) Drive force must both accelerate and beat drag: $F_{drive} = F_{net} + F_{drag} = 3600 + 400 = 4000 N$ . Why this step? Net = drive − drag, so drive = net + drag. See Friction for the drag term.

Verify: Check by re-computing net from drive: $4000 - 400 = 3600 = ma$ ✓. Impulse cross-check: $J = F_{net} Δ t = 3600 \times 9 = 32400 N\cdots$ , and $Δ p = m Δ v = 1200 \times 27 = 32400 kg\cdotm/s$ — equal, as impulse–momentum demands. ✓

Worked example Ex 8 — Cell H: exam twist where

$F = ma$ lies (variable mass) An open railcar of mass $M = 500 kg$ rolls at $v = 4 m/s$ . Rain falls straight down and collects at $μ = \frac{d m}{d t} = 5 kg/s$ . There is no horizontal force. Find the car's speed after $10 s$ .

Forecast: No horizontal force ⇒ " $a = 0$ ⇒ speed stays $4$ "? That is the trap. Watch.

Horizontal Newton: $F_{net} = \frac{d ( m v )}{d t} = 0$ , since nothing pushes horizontally. Why this step? The momentum form never fails, even when mass changes. The rain adds mass but no horizontal momentum (it falls straight down).
Conserved horizontal momentum: $m v = constant$ , so $m_{i} v_{i} = m_{f} v_{f}$ . Why this step? If the time-derivative of $m v$ is zero, $m v$ never changes — this is Conservation of Linear Momentum.
New mass $m_{f} = M + μ t = 500 + 5 (10) = 550 kg$ . Why this step? Mass grows linearly at $5 kg/s$ .
Solve for $v_{f} = \frac{m _{i} v _{i}}{m _{f}} = \frac{500 \times 4}{550} = \frac{2000}{550} \approx 3.636 m/s$ . Why this step? As $m$ rises with $m v$ fixed, $v$ must drop — the exact opposite of the naive "constant speed" answer.

Verify: Momentum before $= 500 \times 4 = 2000$ ; after $= 550 \times 3.636 = 2000$ ✓. The $v \frac{d m}{d t}$ term the parent warned about is precisely what slows the car — see Variable Mass Systems & Rocket Equation. ✓

Recall Quick self-test across the whole matrix

Which cell does each phrase belong to? Constant velocity but forces present ::: Cell C — zero net force, $a = 0$ . Ball reverses direction on a wall ::: Cell E — sign-flip, $Δ p = m (v_{f} - v_{i})$ with $v_{f}$ negative. Two forces at $9 0^{\circ}$ with a west component ::: Cell D — Pythagorean net magnitude, $arctan$ plus $18 0^{\circ}$ for quadrant II. Airbag versus dashboard ::: Cell F — same $Δ p$ , larger $Δ t$ shrinks $F$ . Rain-filling railcar ::: Cell H — variable mass, use $\frac{d ( m v )}{d t}$ , not $ma$ .

Connections

Parent: Newton's Second Law — this page is its case-map.
Newton's First Law (Inertia) — Cell C is the First Law inside the Second.
Conservation of Linear Momentum — the engine behind Ex 8.
Collisions and Elasticity — where rebound speeds (Ex 5) come from.
Friction — the resistive term in Ex 2, 3, 7.
Variable Mass Systems & Rocket Equation — Ex 8's full generalisation.
Newton's Third Law — supplies the wall's reaction force in Ex 5.