Classical-Mechanics · Unit 12 · Video 1 · Interactive Practice

Impulse Equals Change in Momentum: Newton's Second Law Written as an Integral

IKey Formulas

Formula	Name	What you need
$\vec{p} = m\vec{v}$	Momentum — the quantity of motion	Mass and velocity in a named frame
$\vec{F} = \dfrac{d\vec{p}}{dt}$	Second Law, differential form	Constant mass
$\vec{I} = \int_{t}^{t+\Delta t}\vec{F}(t')\,dt' = \Delta\vec{p}$	Second Law, integral form	The force history across $\Delta t$
$\vec{F}_{\mathrm{ave}} = \dfrac{1}{\Delta t}\int_{t}^{t+\Delta t}\vec{F}(t')\,dt' \;\Longrightarrow\; \vec{F}_{\mathrm{ave}}\Delta t = \Delta\vec{p}$	Average force	Impulse and elapsed time

Key Insight: Only the area counts. Two force histories that enclose the same area under $F$ versus $t$ deliver the same $\Delta\vec{p}$ — and that area is measured in $\mathrm{N\,s} = \mathrm{kg\,m\,s^{-1}}$ , momentum's units, never newtons.

IIImpulse Accumulates Into Momentum

The shaded area on the left is not merely an area — it is the momentum the block has gained.

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IIIThe Constant Force That Does the Same Job

One constant force acting across the whole interval delivers the same impulse — how tall must it be?

Force history Your constant force 8.0 N

💡 Only the average force is pinned down by the impulse. Squeeze the same area into a tenth of the time and the peak force grows tenfold — which is precisely what a crumple zone exists to undo.

IVImpulse Adds as a Vector

The momentum a force delivers points along the force and adds head to tail to what was already there.

VQuiz Questions

Problem 1 · From Impulse to Speed

Given: a constant force of $12\ \mathrm{N}$ acts for $0.25\ \mathrm{s}$ on a $3.0\ \mathrm{kg}$ block at rest on a frictionless surface — find the block's final speed.

✅ Correct!

I = (12)(0.25) = 3.0\ \mathrm{N\,s}

, so

\Delta p = 3.0\ \mathrm{kg\,m\,s^{-1}}

and

v = \Delta p / m = 1.0\ \mathrm{m\,s^{-1}}

❌ That is the acceleration.

F/m = 4.0\ \mathrm{m\,s^{-2}}

is a rate of change of velocity; it still has to act for

0.25\ \mathrm{s}

❌ That is the impulse, not the speed.

F\Delta t = 3.0\ \mathrm{N\,s} = 3.0\ \mathrm{kg\,m\,s^{-1}}

is a momentum; divide by the mass to get a velocity.

❌ Not quite. For a constant force the integral is a rectangle:

I = F\Delta t

, and that impulse equals

\Delta p = mv - 0

Show solution

Step 1 — the impulse. The force is constant, so the area under $F$ versus $t$ is a rectangle:

I = F\,\Delta t = (12\ \mathrm{N})(0.25\ \mathrm{s}) = 3.0\ \mathrm{N\,s}

Step 2 — impulse is the change in momentum. The block starts at rest, so $\Delta p = p_f$ :

p_f = \Delta p = I = 3.0\ \mathrm{kg\,m\,s^{-1}}

Step 3 — momentum to velocity.

v = \frac{p_f}{m} = \frac{3.0\ \mathrm{kg\,m\,s^{-1}}}{3.0\ \mathrm{kg}} = 1.0\ \mathrm{m\,s^{-1}}

Check with $F = ma$ : $a = 12/3.0 = 4.0\ \mathrm{m\,s^{-2}}$ , and $v = a\Delta t = (4.0)(0.25) = 1.0\ \mathrm{m\,s^{-1}}$ ✓. The impulse route reaches the same answer without ever naming the acceleration — and it keeps working when the force is not constant.

Problem 2 · The Sign of a Reversal

Given: a $0.145\ \mathrm{kg}$ baseball arrives at $40\ \mathrm{m\,s^{-1}}$ , leaves along the same line at $30\ \mathrm{m\,s^{-1}}$ in the opposite direction, and is in contact with the bat for $1.5\ \mathrm{ms}$ — find the magnitude of the average force.

✅ Correct! The reversal makes

\Delta v = -70\ \mathrm{m\,s^{-1}}

, so

|\Delta p| = 10.15\ \mathrm{N\,s}

and

F_{\mathrm{ave}} = 10.15 / 0.0015 = 6.8\times10^{3}\ \mathrm{N}

❌ You subtracted the speeds. With

+x

along the incoming velocity,

v_i = +40

and

v_f = -30

, so

\Delta v = -70

, not

-10\ \mathrm{m\,s^{-1}}

. The ball must be stopped and thrown back.

❌ Those are the wrong units.

10.15

is the impulse in

\mathrm{N\,s}

— a momentum. A force is that impulse divided by the

1.5\ \mathrm{ms}

of contact.

❌ You used only the final momentum.

m|v_f| / \Delta t

ignores the

5.80\ \mathrm{kg\,m\,s^{-1}}

the ball arrived with; impulse is the change.

❌ Not quite. Work in one dimension with signs:

F_{\mathrm{ave}} = m(v_f - v_i)/\Delta t

, with

v_f

and

v_i

of opposite sign.

Show solution

Take $+x$ along the ball's incoming velocity. The reversal is the whole point — $v_f$ is negative:

p_i = (0.145)(+40) = +5.80\ \mathrm{kg\,m\,s^{-1}}, \qquad p_f = (0.145)(-30) = -4.35\ \mathrm{kg\,m\,s^{-1}}

\Delta p = p_f - p_i = -4.35 - 5.80 = -10.15\ \mathrm{kg\,m\,s^{-1}}

By the integral form of the Second Law that $\Delta p$ is the impulse the bat delivered: $10.15\ \mathrm{N\,s}$ directed back along $-x$ . Now use the average force:

F_{\mathrm{ave}} = \frac{|\Delta p|}{\Delta t} = \frac{10.15\ \mathrm{N\,s}}{1.5\times10^{-3}\ \mathrm{s}} = 6.77\times10^{3}\ \mathrm{N} \approx 6.8\times10^{3}\ \mathrm{N}

Common mistakes:

Using $40 - 30 = 10\ \mathrm{m\,s^{-1}}$ , which describes a ball that merely slowed down, and gives $9.7\times10^{2}\ \mathrm{N}$ — seven times too small.
Reporting $10.15$ as the answer: those are $\mathrm{N\,s}$ , not $\mathrm{N}$ .

Note the trade the average force records: $6.8\ \mathrm{kN}$ for $1.5\ \mathrm{ms}$ changes the momentum by exactly as much as $10.15\ \mathrm{N}$ would in a full second.

Problem 3 · A Force That Rises and Falls

Given: a push along $+x$ lasting $\Delta t = 0.80\ \mathrm{s}$ , with $\vec{F}(t) = bt\,\hat{i}$ on the first half of the interval and $\vec{F}(t) = (d - bt)\,\hat{i}$ on the second, where $b = 30\ \mathrm{N\,s^{-1}}$ and $d = 24\ \mathrm{N}$ — find the peak force and the impulse delivered.

What is the peak force?

What is the impulse?

✅ Correct! The graph is a triangle of base

\Delta t

and height

b\Delta t/2 = 12\ \mathrm{N}

, so

\vec{I} = \tfrac{1}{4}b(\Delta t)^2\,\hat{i} = 4.8\ \mathrm{N\,s}

along

+x

— and that is the momentum gained.

❌ Check where the peak is. The two branches meet at the midpoint

t = \Delta t/2 = 0.40\ \mathrm{s}

, where

F = b\Delta t/2 = (30)(0.40)

. The constants

b = 30

and

d = 24

are not forces the push ever reaches.

❌ Check the area. Split the triangle down the midline: each half is

\tfrac{1}{2}(0.40)(12) = 2.4\ \mathrm{N\,s}

, and there are two of them. A full

12\ \mathrm{N}

rectangle over

0.80\ \mathrm{s}

would be

9.6\ \mathrm{N\,s}

— twice too much, because the force is only briefly at its peak.

Show solution

Step 1 — the peak. The rising branch ends at the midpoint:

F_{\max} = b\,\frac{\Delta t}{2} = (30\ \mathrm{N\,s^{-1}})(0.40\ \mathrm{s}) = 12\ \mathrm{N}

Step 2 — check that the pieces join. The falling branch at the midpoint gives $d - b(0.40) = 24 - 12 = 12\ \mathrm{N}$ — continuous ✓ — and at the end $d - b(0.80) = 24 - 24 = 0$ ✓.

Step 3 — the impulse is the area. The triangle splits into two right triangles, each of base $\Delta t/2$ and height $b\Delta t/2$ :

\vec{I} = \left[\tfrac{1}{2}\!\left(\frac{b\Delta t}{2}\right)\!\left(\frac{\Delta t}{2}\right) + \tfrac{1}{2}\!\left(\frac{b\Delta t}{2}\right)\!\left(\frac{\Delta t}{2}\right)\right]\hat{i} = \tfrac{1}{4}b(\Delta t)^2\,\hat{i}

\vec{I} = \tfrac{1}{4}(30\ \mathrm{N\,s^{-1}})(0.80\ \mathrm{s})^2\,\hat{i} = 4.8\ \mathrm{N\,s}\,\hat{i}

Each half contributes $\tfrac{1}{2}(12)(0.40) = 2.4\ \mathrm{N\,s}$ , and $2.4 + 2.4 = 4.8$ ✓.

Step 4 — read it as physics. By $\vec{I} = \Delta\vec{p}$ the object's momentum is larger by $4.8\ \mathrm{kg\,m\,s^{-1}}$ along $+x$ than before the push, whatever it was to begin with.

Problem 4 · Transfer — Why a Crumple Zone Works

Given: a $1200\ \mathrm{kg}$ car travelling at $15\ \mathrm{m\,s^{-1}}$ is brought to rest. A rigid barrier stops it in $0.10\ \mathrm{s}$ ; a crumple zone stretches the same stop out to $0.60\ \mathrm{s}$ — find the average force in the crumpling case, and compare the two impulses.

Average force with the crumple zone?

How does the impulse compare with the rigid stop?

✅ Correct! Both stops remove the same

1.8\times10^{4}\ \mathrm{kg\,m\,s^{-1}}

, so both deliver the same impulse; spreading it over six times as long divides the average force by six.

❌ That is the rigid barrier.

1.8\times10^{5}\ \mathrm{N}

comes from

\Delta t = 0.10\ \mathrm{s}

. The crumple zone gives the same impulse

0.60\ \mathrm{s}

to arrive.

❌ That is the impulse itself.

1.8\times10^{4}

|\Delta p|

\mathrm{N\,s}

; still divide by

\Delta t = 0.60\ \mathrm{s}

to get a force.

❌ You averaged the velocity. Using

7.5\ \mathrm{m\,s^{-1}}

gives

1.5\times10^{4}\ \mathrm{N}

, but the impulse uses the change in momentum

m\Delta v

, with the full

15\ \mathrm{m\,s^{-1}}

❌ Check the two steps. First

|\Delta p| = m|\Delta v| = (1200)(15)

, then

F_{\mathrm{ave}} = |\Delta p| / \Delta t

with

\Delta t = 0.60\ \mathrm{s}

❌ Compare the momenta, not the forces. The car goes from

15\ \mathrm{m\,s^{-1}}

to rest either way, so

\Delta p

— and therefore the impulse — is identical. Only its distribution in time differs.

Show solution

Step 1 — the impulse, which is set by the momenta alone.

|\vec{I}| = |\Delta \vec{p}| = m|\Delta v| = (1200\ \mathrm{kg})(15\ \mathrm{m\,s^{-1}}) = 1.8\times10^{4}\ \mathrm{N\,s}

Nothing in that line mentions time, so both collisions deliver the same impulse.

Step 2 — the average force, which does depend on the time.

F_{\mathrm{rigid}} = \frac{1.8\times10^{4}}{0.10} = 1.8\times10^{5}\ \mathrm{N}, \qquad F_{\mathrm{crumple}} = \frac{1.8\times10^{4}}{0.60} = 3.0\times10^{4}\ \mathrm{N}

Step 3 — read the design. Stretching the collision by a factor of six divides the average force by six. The area under $F$ versus $t$ is fixed at $1.8\times10^{4}\ \mathrm{N\,s}$ ; a crumple zone simply re-shapes that area from a tall narrow spike into a low broad plateau, and it is the height — the force — that injures people.

The same reasoning explains airbags, landing with bent knees, and catching a ball by drawing your hand back.

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