Classical-Mechanics · Unit 6 · Video 3 · Interactive Practice

Newton's Second and Third Laws: Force as the Rate of Change of Momentum

IKey Formulas

Formula	Name	Meaning
$\vec{p} = m\vec{v}$	Momentum	Mass $\times$ velocity — a vector along $\vec v$
$\vec{F} = \dfrac{d\vec{p}}{dt} = m\vec{a}$	Second Law	Force $=$ rate of change of momentum; reduces to $m\vec a$ when $m$ is constant
$\vec{I} = \vec{F}\,\Delta t = \Delta\vec{p}$	Impulse–momentum	A constant force over $\Delta t$ delivers a momentum kick
$\vec{F}_{1,2} = -\vec{F}_{2,1}$	Third Law	Interaction pair: equal magnitude, opposite direction

Key Insight: A force does not set a velocity — it sets the rate of change of momentum. The familiar $\vec F = m\vec a$ is only the special case of $\vec F = d\vec p/dt$ in which the mass stays constant.

IIVisualization 1 — Impulse Is the Area, and It Equals $\Delta p$

A constant force $F$ acting for a time $\Delta t$ delivers an impulse $F\,\Delta t$ — the shaded area — and that impulse equals the change in momentum $\Delta p = p_f - p_i$ .

Force

F

(N) 2 Duration

\Delta t

(s) 3

💡 Same $\Delta p$ , different force: equal areas mean a large force over a short time delivers the same momentum change as a small force over a long time. An airbag stretches $\Delta t$ to shrink the force $F$ your body feels for a fixed $\Delta p$ .

IIIVisualization 2 — From Average Rate to Instantaneous Force

Shrink the interval $\Delta t$ and the average rate $\Delta p/\Delta t$ (a secant slope) approaches the instantaneous force $dp/dt$ (the tangent slope).

Interval

\Delta t

(s) 3

💡 Where $ma$ comes from: with constant mass, $p = mv$ , so $\dfrac{dp}{dt} = m\dfrac{dv}{dt} = ma$ . The tangent slope on this curve is the net force.

IVVisualization 3 — Equal Forces, Unequal Accelerations

The interaction forces obey $\vec F_{1,2} = -\vec F_{2,1}$ always, but $a = F/m$ — so unequal masses receive unequal accelerations from the very same force.

Force

F

(N) 6 Mass

m_1

(kg) 2 Mass

m_2

(kg) 4

💡 Why the pair never cancels: $\vec F_{1,2}$ and $\vec F_{2,1}$ are equal and opposite, but they act on different bodies. Forces only cancel when they act on the same body — so a Third-Law pair can never add to zero on one object.

VQuiz Questions

Problem 1 · Compute a Momentum

Given: A cart of mass $m = 3$ kg moves at velocity $v = 4$ m/s — find its momentum $p$ .

✅ Correct!

p = mv = 3 \times 4 = 12

kg·m/s, pointing the same way as

\vec v

❌ Not quite. Momentum is a product, not a sum — you added

m + v = 3 + 4

. Multiply instead.

❌ Not quite. Use

p = mv

, mass times velocity.

Show solution

Momentum is defined as $\vec p = m\vec v$ :

p = mv = (3\ \text{kg})(4\ \text{m/s}) = 12\ \text{kg}\cdot\text{m/s}

The vector $\vec p$ points in the same direction as the velocity.

Problem 2 · Impulse on a Bouncing Ball (Watch the Vector!)

Given: A $0.5$ kg ball strikes a wall at $6$ m/s and rebounds straight back at $6$ m/s. Find the magnitude of the impulse the wall delivers. ⚠️ Velocity is a vector.

✅ Correct! Taking the outgoing direction as positive,

\Delta v = 6 - (-6) = 12

m/s, so

I = m\,\Delta v = 0.5 \times 12 = 6

N·s.

❌ The speed is unchanged, but the velocity is not. It reverses direction, so

\Delta v \neq 0

❌ You used only one leg of the trip. The ball both stops and then reverses; the total change is

|{-6} - 6| = 12

m/s.

❌ That is $\Delta v = 12$ m/s. Impulse is

m\,\Delta v

— multiply by the mass

0.5

kg.

❌ Not quite. Use

I = \Delta p = m(v_f - v_i)

with opposite signs on the velocities.

Show solution

Choose the rebound direction as positive: $v_i = -6$ m/s (toward the wall), $v_f = +6$ m/s (away).

I = \Delta p = m(v_f - v_i) = 0.5\big(6 - (-6)\big) = 0.5(12) = 6\ \text{N}\cdot\text{s}

The key is that momentum is a vector: reversing direction doubles the change compared with simply stopping.

Problem 3 · Impulse and Force Together

Given: A constant force acts on a $2$ kg object for $\Delta t = 4$ s, changing its velocity from $3$ m/s to $11$ m/s. Find the impulse and the force.

What is the impulse $I$ ?

What is the force $F$ ?

✅ Correct!

I = m\,\Delta v = 2(8) = 16

N·s, and

F = I/\Delta t = 16/4 = 4

❌ Check the impulse.

I = \Delta p = m\,\Delta v

, and

\Delta v = 11 - 3 = 8

m/s — don't forget the mass, and use the change in velocity.

❌ Check the force. A constant force means

F = \Delta p / \Delta t

. Divide the impulse by the time; don't confuse

F

with the acceleration

a = 2

m/s².

Show solution

Impulse (change in momentum):

I = \Delta p = m\,\Delta v = 2\,(11 - 3) = 2(8) = 16\ \text{N}\cdot\text{s}

Force (constant, so the impulse relation applies directly):

F = \frac{\Delta p}{\Delta t} = \frac{16}{4} = 4\ \text{N}

Check via $F = ma$ : $a = \dfrac{\Delta v}{\Delta t} = \dfrac{8}{4} = 2$ m/s², so $F = ma = 2 \times 2 = 4$ N. ✓

Problem 4 · Third Law: Car Pushes Truck (Transfer)

Given: A $1000$ kg car in contact with a $2000$ kg truck pushes the truck with a force of $3000$ N at one instant. Find the reaction force and compare the accelerations from this interaction.

What force does the truck exert on the car?

Which vehicle gets the larger acceleration from this interaction?

✅ Excellent! By the Third Law the truck pushes back with

3000

N. Equal forces on unequal masses give

a_{\text{car}} = 3

m/s²

>

a_{\text{truck}} = 1.5

m/s².

❌ The Third Law equates the forces, not force-per-mass.

\vec F_{1,2} = -\vec F_{2,1}

makes the magnitudes identical, regardless of the masses.

❌ The heavier body still pushes back. Contact forces always come in equal, opposite pairs — the truck cannot feel a force without exerting one.

❌ Equal forces, but not equal accelerations.

a = F/m

, and the masses differ, so the accelerations must differ.

❌ Check the reaction force. The Third Law says it is equal in magnitude and opposite in direction:

3000

❌ Check the accelerations. Same force,

a = F/m

— the smaller mass accelerates more.

Show solution

Reaction force (Third Law):

\vec F_{\text{truck→car}} = -\vec F_{\text{car→truck}} \quad\Rightarrow\quad |F| = 3000\ \text{N (opposite direction)}

Accelerations from this force alone:

a_{\text{car}} = \frac{3000}{1000} = 3\ \text{m/s}^2, \qquad a_{\text{truck}} = \frac{3000}{2000} = 1.5\ \text{m/s}^2

The forces are the equal pair; the motions are not. The lighter car accelerates twice as hard as the truck.

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Newton's Second and Third Laws: Force as the Rate of Change of Momentum

IKey Formulas

IIVisualization 1 — Impulse Is the Area, and It Equals Δp\Delta pΔp

IIIVisualization 2 — From Average Rate to Instantaneous Force

IVVisualization 3 — Equal Forces, Unequal Accelerations

VQuiz Questions

IIVisualization 1 — Impulse Is the Area, and It Equals $\Delta p$