2.4.2.2. Impulse-Momentum Theorem

Table of Contents

Core Idea

The impulse-momentum theorem connects force, time, and change in motion. It says that when a force acts on an object for some time, the object's momentum changes.

Momentum is defined earlier as the product of mass and velocity, $ \vec{p} = m\vec{v} $. The impulse-momentum theorem tells us exactly how force changes that momentum.

The impulse-momentum theorem is
$$
\vec{J} = \Delta \vec{p}
$$
where $ \vec{J} $ is impulse and $ \Delta \vec{p} = \vec{p}_f - \vec{p}_i $ is the change in momentum.

This theorem is one of the most useful ideas in mechanics because many real interactions happen over a short time, such as a kick, a collision, or a bounce.

From Newton's Second Law to the Theorem

Start with Newton's second law in momentum form,

$$
\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}
$$

If the net force acts during a time interval from $t_i$ to $t_f$, we multiply by $dt$ and integrate:

$$
\vec{F}_{\text{net}}\,dt = d\vec{p}
$$

$$
\int_{t_i}^{t_f} \vec{F}_{\text{net}} \, dt = \int_{\vec{p}_i}^{\vec{p}_f} d\vec{p}
$$

Therefore,

$$
\vec{J} = \int_{t_i}^{t_f} \vec{F}_{\text{net}} \, dt = \vec{p}_f - \vec{p}_i = \Delta \vec{p}
$$

This is the general form of the theorem.

If the net force is constant during the time interval, the integral becomes simpler:

$$
\vec{J} = \vec{F}_{\text{net}} \Delta t
$$

and so

$$
\vec{F}_{\text{net}} \Delta t = \Delta \vec{p}
$$

For a constant net force,
$$
\vec{F}_{\text{net}} \Delta t = \Delta \vec{p}
$$
For a varying net force,
$$
\int \vec{F}_{\text{net}}\,dt = \Delta \vec{p}
$$

Meaning of Impulse

Impulse is the effect of a force acting over a time interval. A large force acting briefly can produce the same impulse as a smaller force acting for a longer time.

This is why catching a ball by moving your hands backward is easier. The ball's momentum must still change, but increasing the time of contact reduces the average force needed.

Impulse is a vector quantity, so direction matters. If the momentum changes direction, the impulse points in the direction of that change.

Average Force

Often the exact force changes during contact, but we can still use an average force.

If $ \vec{F}_{\text{avg}} $ is the average net force over the interval $ \Delta t $, then

$$
\vec{J} = \vec{F}_{\text{avg}} \Delta t
$$

and therefore

$$
\vec{F}_{\text{avg}} \Delta t = \Delta \vec{p}
$$

So the average force is

$$
\vec{F}_{\text{avg}} = \frac{\Delta \vec{p}}{\Delta t}
$$

This form is especially useful in collisions.

Average net force during an interaction is
$$
\vec{F}_{\text{avg}} = \frac{\Delta \vec{p}}{\Delta t}
$$
A longer interaction time gives a smaller average force for the same momentum change.

Direction and Sign

In one dimension, signs help track direction. Suppose an object moves to the right, and we choose right as positive.

If its velocity changes from $v_i$ to $v_f$, then

$$
\Delta p = mv_f - mv_i = m(v_f - v_i)
$$

If the object slows down while moving right, then $v_f < v_i$, so $ \Delta p $ is negative. That means the impulse and the net force are directed to the left.

If the object reverses direction, the change in momentum can be even larger because the initial and final momenta point opposite ways.

For example, if a ball with mass $m$ moves right at speed $v$ and then rebounds left at speed $v$, then

$$
p_i = +mv, \qquad p_f = -mv
$$

$$
\Delta p = -mv - (+mv) = -2mv
$$

The magnitude of the impulse is $2mv$.

Units

Impulse has the same units as momentum.

Quantity	SI unit	Equivalent form
Momentum	$\mathrm{kg \cdot m/s}$	$\mathrm{N \cdot s}$
Impulse	$\mathrm{N \cdot s}$	$\mathrm{kg \cdot m/s}$

Since

$$
1\,\mathrm{N} = 1\,\mathrm{kg \cdot m/s^2}
$$

then

$$
1\,\mathrm{N \cdot s} = 1\,\mathrm{kg \cdot m/s}
$$

Common Physical Situations

In many short interactions, such as hitting, catching, or crashing, the force may be very large but act only briefly. The theorem is useful because we do not need every detail of the motion during the interaction. We only need the change in momentum.

A tennis racket changes the momentum of a ball. A car airbag increases the collision time, reducing average force on the passenger. A hammer delivers a large impulse to a nail in a short time.

These examples all follow the same rule: momentum changes because impulse is applied.

Simple Example

Suppose a $0.20\,\mathrm{kg}$ ball is initially at rest and is struck so that it moves to the right at $15\,\mathrm{m/s}$.

The initial momentum is

$$
p_i = 0
$$

The final momentum is

$$
p_f = mv = (0.20)(15) = 3.0\,\mathrm{kg \cdot m/s}
$$

So the change in momentum is

$$
\Delta p = 3.0\,\mathrm{kg \cdot m/s}
$$

Therefore the impulse is

$$
J = 3.0\,\mathrm{N \cdot s}
$$

directed to the right.

If the contact time was $0.010\,\mathrm{s}$, then the average force would be

$$
F_{\text{avg}} = \frac{\Delta p}{\Delta t}
= \frac{3.0}{0.010}
= 300\,\mathrm{N}
$$

Rebound Example

Consider a $0.50\,\mathrm{kg}$ ball moving right at $8.0\,\mathrm{m/s}$ that bounces back left at $6.0\,\mathrm{m/s}$.

Take right as positive. Then

$$
p_i = (0.50)(8.0) = 4.0\,\mathrm{kg \cdot m/s}
$$

$$
p_f = (0.50)(-6.0) = -3.0\,\mathrm{kg \cdot m/s}
$$

$$
\Delta p = p_f - p_i = -3.0 - 4.0 = -7.0\,\mathrm{kg \cdot m/s}
$$

Thus the impulse is

$$
J = -7.0\,\mathrm{N \cdot s}
$$

The negative sign means the impulse is to the left.

If the bounce lasted $0.020\,\mathrm{s}$, then

$$
F_{\text{avg}} = \frac{-7.0}{0.020} = -350\,\mathrm{N}
$$

So the average force on the ball is $350\,\mathrm{N}$ to the left.

Visualizing the Interaction

A short collision often changes momentum quickly.

Rendering scientific sketch…

Impulse changing momentum

The ball begins with momentum to the right, then leaves with momentum to the left. The impulse points left because that is the direction of the momentum change.

Relation to Force-Time Graphs

A force can vary during contact. In that case, impulse is the area under the force-time graph.

If force is plotted on the vertical axis and time on the horizontal axis, then

$$
J = \int F\,dt
$$

In one dimension, positive area gives positive impulse and negative area gives negative impulse.

This graphical interpretation is very useful when force is not constant.

Rendering scientific sketch…

Force-time graph and impulse

What the Theorem Helps You Find

The impulse-momentum theorem can be used in several directions. If you know the force and time, you can find the momentum change. If you know the momentum change and contact time, you can estimate average force. If you know the impulse, you can determine how velocity changed.

Known	Can find
$F_{\text{avg}}$ and $\Delta t$	$\Delta p$
$\Delta p$ and $\Delta t$	$F_{\text{avg}}$
$m$, $v_i$, $v_f$	impulse
impulse and mass	change in velocity

For constant mass in one dimension,

$$
J = \Delta p = m(v_f - v_i)
$$

Important Distinction

Impulse is not the same as force, and it is not the same as momentum. Impulse measures the effect of force over time. Momentum measures motion.

A large force does not always mean a large impulse. If the force acts for only an extremely tiny time, the impulse may still be small. Likewise, a modest force acting for a long time can produce a large impulse.

Do not confuse these quantities:
$$
\text{Force} \neq \text{Impulse} \neq \text{Momentum}
$$
Force has units of $\mathrm{N}$.
Impulse has units of $\mathrm{N \cdot s}$.
Momentum has units of $\mathrm{kg \cdot m/s}$.

Summary

The impulse-momentum theorem states that the impulse delivered to an object equals its change in momentum. In symbols,

$$
\vec{J} = \Delta \vec{p}
$$

For constant force,

$$
\vec{F}_{\text{net}} \Delta t = \Delta \vec{p}
$$

For variable force,

$$
\int \vec{F}_{\text{net}}\,dt = \Delta \vec{p}
$$

This theorem is especially useful for collisions and short interactions, where forces may be large and time intervals small. It explains why changing the time of impact changes the force, even when the overall change in momentum stays the same.

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