4.2 Change in Momentum and Impulse

Impulse is the net force on an object multiplied by the time it acts, and it equals the object's change in momentum. This is the impulse momentum theorem ($\vec{J} = \vec{F} {avg}\Delta t = \Delta\vec{p}$), and it connects collision safety, force-time graphs, and momentum changes.

Why This Matters for the AP Physics 1 Exam

Impulse and change in momentum show up across multiple-choice and free-response questions in Unit 4, which carries about 10 to 15 percent of the exam. You will be expected to connect force, time, and momentum using equations, graphs, and written reasoning.

Common ways this topic appears:

• Reading impulse as the area under a force vs. time graph
• Finding net force from the slope of a momentum vs. time graph
• Explaining why extending impact time lowers the average force (safety design reasoning)
• Solving for an unknown force, velocity, or time using the impulse-momentum theorem
• Showing how Newton's second law comes from the impulse-momentum theorem for constant mass

Because Unit 4 connects to the Experimental Design and Analysis free-response question, you may also be asked to design or analyze data from impulse experiments, including linearizing graphs.

Key Takeaways

• Net force equals the rate of change of momentum: $\vec{F}_{net} = \frac{\Delta\vec{p}}{\Delta t}$.
• Impulse is average force times the time interval: $\vec{J} = \vec{F}_{avg}\Delta t$, and it is a vector pointing in the direction of the net force.
• The impulse-momentum theorem says impulse equals change in momentum: $\vec{J} = \Delta\vec{p} = \vec{p} - \vec{p}_0$.
• Area under a force vs. time graph equals impulse; slope of a momentum vs. time graph equals net force.
• Impulse units are N·s, which equal kg·m/s.
• Newton's second law ($\vec{F}_{net} = m\vec{a}$) comes directly from the impulse-momentum theorem when mass is constant.

Impulse Delivered to an Object or System

Force and the Rate of Momentum Change

The net external force on an object sets how fast its momentum changes.

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

What this tells you:

• Only the net force (sum of all external forces) changes momentum.
• If forces balance and the net force is zero, momentum stays constant.
• Doubling the net force doubles the rate of momentum change.
• This is a vector relationship, so you can analyze it component by component in 2D.

What Impulse Means

Impulse is the total effect of a force applied over a time interval. It is how much "push" an object gets during an interaction.

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

Key features:

• Measured in newton-seconds (N·s), which equal kg·m/s.
• You can increase impulse by using more force or by applying force for longer.
• Catching a ball, hitting a nail, and launching a rocket all involve impulse.
• Safety designs like airbags and cushioned shoes extend the time to reduce the average force.

Direction of Impulse

Impulse is a vector and points in the same direction as the net force.

• Find the net force first, then determine the impulse direction.
• Impulse can be split into x and y components for 2D problems.
• The impulse vector shows which way the object's motion will change.

Area Under a Force vs. Time Graph

A force vs. time graph shows how force changes during an interaction. The area between the curve and the time axis equals the impulse.

When reading these graphs:

• For a constant force, the area is a rectangle: force × time.
• For varying forces, use triangles, trapezoids, or estimate curved areas.
• Total impulse is the net area under the curve.
• Areas below the axis (force in the opposite direction) subtract from the total.

Slope of a Momentum vs. Time Graph

A momentum vs. time graph shows how momentum changes over time. The slope at any point equals the net force at that moment.

When reading these graphs:

• Steep slopes mean large net forces.
• Flat regions (zero slope) mean zero net force.
• Positive slopes mean force in the positive direction; negative slopes mean the opposite.
• The shape of the graph reveals how the net force varies.

Impulse and Change in Momentum

Calculating Change in Momentum

Change in momentum measures how much an object's motion was altered. It is the final momentum minus the initial momentum.

$\Delta \vec{p} = \vec{p} - \vec{p}_0 = m(\vec{v} - \vec{v}_0)$

When calculating it:

• The direction of $\Delta\vec{p}$ matches the direction of the net force.
• For constant mass, change in momentum depends only on the velocity change.
• Larger momentum changes require larger impulses.
• You can work component by component in 2D problems.

The Impulse-Momentum Theorem

The impulse-momentum theorem says the impulse on an object equals its change in momentum.

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

Why it is useful:

• Explains why extending impact time reduces force (catching an egg gently).
• Shows why very short impacts deliver large forces (hammering a nail).
• Lets engineers design safety systems that stretch out collision times.
• Lets you solve for an unknown force when you know the momentum change and time.

Deriving Newton's Second Law

Newton's second law follows directly from the impulse-momentum theorem when mass stays constant.

Starting from the theorem:

1. $\vec{F}_{net} \Delta t = \Delta \vec{p}$
2. For constant mass: $\vec{F}_{net} \Delta t = m \Delta \vec{v}$
3. Dividing both sides by $\Delta t$: $\vec{F}_{net} = m \frac{\Delta \vec{v}}{\Delta t} = m\vec{a}$

This shows that:

• Newton's second law is a special case of the impulse-momentum theorem.
• The impulse-momentum approach still works when acceleration is not constant.
• For varying forces, working with impulse is often simpler than tracking acceleration.

How to Use This on the AP Physics 1 Exam

Problem Solving

• Decide what you are solving for, then pick the matching form: $\vec{F}_{net} = \frac{\Delta\vec{p}}{\Delta t}$, $\vec{J} = \vec{F}_{avg}\Delta t$, or $\vec{J} = \Delta\vec{p}$.
• Always assign a positive direction first so your signs stay consistent.
• For rebound problems, remember the velocity reverses sign, so the change in momentum is larger than you might expect.
• Check units: impulse should come out in N·s or kg·m/s.

Graphs

• Force vs. time graph: find impulse by computing the area under the curve.
• Momentum vs. time graph: find net force by computing the slope.
• Do not mix these up. Area and slope answer different questions.

Free Response

• When a question asks you to explain a safety device, connect a longer time interval to a smaller average force using $\vec{J} = \vec{F}_{avg}\Delta t$ with the same impulse.
• Justify claims with the equation and the direction, not just a verbal answer.
• For experiments, you may be asked to linearize data so that a slope or area gives the quantity you want.

Practice Problem 1: Impulse Calculation

Solution

(a) To find the impulse, calculate the change in momentum: Initial momentum: $p_i = mv_i = (0.145 \text{ kg})(35 \text{ m/s}) = 5.075 \text{ kg}\cdot\text{m/s}$ Final momentum: $p_f = mv_f = (0.145 \text{ kg})(0 \text{ m/s}) = 0 \text{ kg}\cdot\text{m/s}$ Impulse: $J = \Delta p = p_f - p_i = 0 - 5.075 = -5.075 \text{ kg}\cdot\text{m/s} = -5.075 \text{ N}\cdot\text{s}$

The negative sign shows the impulse points opposite to the ball's initial velocity.

(b) The average force is: $F_{avg} = \frac{J}{\Delta t} = \frac{-5.075 \text{ N·s}}{0.050 \text{ s}} = -101.5 \text{ N}$

(c) During a real catch, the force is not constant. It starts at zero as the glove first touches the ball, rises to a peak as the glove compresses, then drops back to zero as the ball stops. A realistic force vs. time graph rises from zero to a peak and falls back to zero.

Practice Problem 2: Force from Momentum-Time Graph

Solution

The net force is the slope of the momentum vs. time graph:

Slope = $\frac{\Delta p}{\Delta t} = \frac{p_f - p_i}{t_f - t_i} = \frac{14 \text{ kg}\cdot\text{m/s} - 4 \text{ kg}\cdot\text{m/s}}{5 \text{ s} - 0 \text{ s}} = \frac{10 \text{ kg}\cdot\text{m/s}}{5 \text{ s}} = 2 \text{ N}$

Since $F_{net} = \frac{\Delta p}{\Delta t}$, the net force is 2 N in the positive direction.

Practice Problem 3: Impulse-Momentum Theorem

Solution

(a) The impulse delivered to the cart is: $J = F_{avg} \Delta t = (15 \, \text{N})(2.0 \, \text{s}) = 30 \, \text{N·s}$

(b) Using the impulse-momentum theorem: $J = \Delta p = m\Delta v = m(v_f - v_i)$

Since $v_i = 0$ (starts from rest): $30 \, \text{N·s} = (0.5 \, \text{kg})(v_f - 0)$

$v_f = \frac{30 \, \text{N·s}}{0.5 \, \text{kg}} = 60 \, \text{m/s}$

The cart's velocity is 60 m/s in the direction of the net force.

Common Misconceptions

• Impulse and force are not the same thing. Force acts at an instant; impulse is force applied over a time interval, and it equals the change in momentum.
• A larger force does not always mean a larger impulse. A small force over a long time can deliver the same impulse as a large force over a short time.
• Area and slope are not interchangeable. Area under a force vs. time graph gives impulse, while slope of a momentum vs. time graph gives net force.
• For rebound or bounce-back problems, do not forget the sign change. Reversing direction makes the change in momentum bigger than if the object simply stopped.
• Newton's second law is not a separate idea from impulse. $\vec{F}_{net} = m\vec{a}$ comes from the impulse-momentum theorem when mass is constant.
• Impulse depends on net external force, not internal forces. Forces inside the chosen system do not change the system's total momentum.

Related AP Physics 1 Guides

• 4.1 Linear Momentum
• 4.3 Conservation of Linear Momentum
• 4.4 Elastic and Inelastic Collisions