Impulse-Momentum Theorem

Equivalent forms

\vec{J} = \int \vec{F}\,dt = \Delta\vec{p}

F_{\text{avg}} = \frac{\Delta p}{\Delta t}

Force times time = mass times velocity change. The hidden physics of every safety device.

Symbol	Name	SI unit	Dimension	Range
$J$	Impulseoutput Impulse delivered to the object	N·s	MLT^-1	0 – 1000
$F$	Average Force Average force during the impulse interval	N	MLT^-2	0 – 5000
$Δt$	Time Interval Duration over which the force acts	s	T	0.01 – 5

Unit systems

Symbol	SI	CGS	Imperial
$J$	N·s	dyn·s	lbf·s
$F$	N	dyn	lbf
$Δt$	s	s	s

Where it holds

Valid in any inertial frame for non-relativistic speeds. Variable-mass systems (rockets) require thrust equation derived from same principle.

Dimensional analysis

J \to [MLT^{-1}]

F\cdot \Delta t \to [MLT^{-2}]\cdot [T] = [MLT^{-1}]

Both sides are

kg\cdot m/s

.

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Discovery

Isaac Newton · 1687

Direct integral form of Newton's second law: integrate F = dp/dt over time to get impulse equals change in momentum.

Try this

Why does bending your knees when landing from a jump save your ankles?

A 70 kg person lands at 5 m/s. Compare the average force when stopping in 0.05 s (stiff legs) vs 0.5 s (bent knees). Apply F·Δt = Δp.

Research status: stable

Real-world applications

Airbags and seatbelts in cars
Crumple zones in vehicle frames
Boxing gloves and gymnastic mats
Catching a ball by pulling hands back

Common misconceptions

Impulse is not energy — it has units of momentum $(N\cdot s = kg\cdot m/s)$
A soft landing reduces force, not the change in momentum
Impulse is a vector quantity along the force direction

Experimental verification

Force plates, high-speed video of collisions, and accelerometer data on car crash dummies all confirm

J = \Delta p

to within instrumentation precision.

Derivation

Start from

F = dp/dt

.

Integrate both sides over

\Delta t

:

\int F dt = \int dp = p_f - p_i = \Delta p

.

The left side is the impulse J.

For constant F:

J = F\cdot \Delta t = \Delta p

.

Limiting cases

\Delta t \to 0

⟶

F \to \infty

Instantaneous stops require infinite force (idealized point collisions).

\Delta t \to \infty

⟶

F \to 0

Spread the same momentum change over infinite time and the force vanishes.

F = 0

⟶

\Delta p = 0

No net force → momentum conserved (Newton's first law).

What if…

What

if \Delta t

halves?

Average force doubles for the same momentum change. Halving stop-time on a 70 kg, 5 m/s landing pushes force from 700 N to 1400 N.

What if the object bounces back instead of stopping?

$\Delta p$ doubles (from mv to 2mv), so impulse and average force both double for the same $\Delta t$ .

What if force varies in time?

Use the integral form $J = \int F dt$ . The area under the F-t curve equals the impulse.

1

Stiff vs bent-knee landing

Given ·

m:: 70
v:: 5
Δt stiff:: 0.05
Δt bent:: 0.5

Find · F

Steps

$\Delta p = m\Delta v = 70 \times 5 = 350 kg\cdot m/s$
Stiff landing: $F = \Delta p/\Delta t = 350/0.05 = 7000\,\mathrm{N} (\approx 10\times$ body weight, ankles break)
Bent knees: $F = 350/0.5 = 700\,\mathrm{N} (\approx 1\times$ body weight, safe)

Result · F_stiff

= 7000\,\mathrm{N}

, F_bent

= 700\,\mathrm{N}

—

10\times

reduction

2

Tennis serve impulse

Given ·

m:: 0.058
v i:: 0
v f:: 60
Δt:: 0.004

Find · F

Steps

$\Delta p = 0.058 \times 60 = 3.48 kg\cdot m/s$
$F = \Delta p/\Delta t = 3.48/0.004$
$F \approx 870\,\mathrm{N}$ — explains the racket compression and ball deformation

Result ·

F = 870\,\mathrm{N}

average force on the ball during contact

Linear Momentum→Newton's Second Law→Work-Energy Theorem→