Impulse and momentum — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: Linear momentum definition for point masses, the impulse-momentum theorem, average and variable force impulse calculation, impulse as area under force-time curves, and momentum change for single particle systems.

You should already know: Newton's second law for point masses. Definite integration of functions over time. Interpreting force vs. time graphs.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: Mechanics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Impulse and momentum?

Impulse and linear momentum are core concepts in the study of particle and system motion, forming the foundation for all collision and system momentum problems on the AP Physics C: Mechanics exam. Per the College Board CED, Unit 4 (Systems of Particles and Linear Momentum) makes up 14-18% of the total exam score, and the impulse and momentum subtopic accounts for roughly one-third of that unit’s content, appearing regularly in both multiple-choice (MCQ) and free-response (FRQ) sections. Linear momentum (defined formally as $\vec{p}=m\vec{v}$) is a vector quantity that describes the inertia of motion of a moving object, with units of kilogram-meters per second ($\text{kg}\cdot \text{m/s}$), equivalent to newton-seconds ($\text{N}\cdot \text{s}$). Impulse (denoted $\vec{J}$) is the change in momentum caused by a net force acting over a finite time interval; it captures how the cumulative effect of a force over time changes an object’s motion. Unlike work (which relates force over displacement to kinetic energy change), impulse relates force over time to momentum change, making it ideal for analyzing short-duration interactions like collisions, where forces vary rapidly and cannot be described easily with constant acceleration kinematics.

2. The Impulse-Momentum Theorem for Single Particles

The impulse-momentum theorem is derived directly from Newton’s second law, rewritten in terms of momentum: ${\vec{F}}_{net}=\frac{d\vec{p}}{dt}$. Rearranging to separate variables gives $d\vec{p}={\vec{F}}_{net}dt$. Integrate both sides from the initial time $t_i$ to final time $t_f$: $${\int }_{p_i}^{p_f}d\vec{p}={\int }_{t_i}^{t_f}{\vec{F}}_{net}dt$$ The left-hand side simplifies to $\Delta \vec{p}={\vec{p}}_f-{\vec{p}}_i$, and the right-hand side is defined as the total impulse $\vec{J}$ from the net force. This gives the core impulse-momentum theorem: $$\vec{J}=\Delta \vec{p}$$ For any interaction, impulse equals the change in momentum of the object. For a variable force, impulse is equal to the signed area under the $F(t)$ vs. $t$ graph between $t_i$ and $t_f$, one of the most common question formats on the AP exam. Because impulse and momentum are vectors, direction matters: forces acting opposite to motion decrease momentum, while forces in the direction of motion increase momentum.

Worked Example

A 2.0 kg object moving along the x-axis has an initial velocity $v_i=+3.0\text{ m/s}$. A net force acts on the object, varying with time to form three segments: a linear increase from 0 N to 4 N between $t=0$ and $t=2$ s, constant 4 N between $t=2$ s and $t=3$ s, and a linear decrease from 4 N back to 0 N between $t=3$ s and $t=5$ s. Find the final velocity of the object.

1. Total impulse equals the signed area under the $F(t)$ graph; all force values are positive here, so we just sum the area of each segment.
2. Area 1 (0–2 s, triangle): $A_1=\frac{1}{2}\times 2\times 4=4\text{ Ncdotps}$. Area 2 (2–3 s, rectangle): $A_2=1\times 4=4\text{ Ncdotps}$. Area 3 (3–5 s, triangle): $A_3=\frac{1}{2}\times 2\times 4=4\text{ Ncdotps}$.
3. Total impulse $J=4+4+4=12\text{ Ncdotps}$. By the impulse-momentum theorem, $J=\Delta p=m(v_f-v_i)$.
4. Solve for $v_f$: $v_f=v_i+\frac{J}{m}=3.0+\frac{12}{2.0}=9.0\text{ m/s}$.

Exam tip: Always mark the sign of force on your F(t) graph before calculating area; any segment of the graph below the time axis contributes negative impulse, which subtracts from total momentum change.

3. Average Force and Impulse

For short-duration interactions like collisions, we rarely need the full time dependence of the force. Instead, we use the concept of average force: the constant force that produces the same total impulse (and thus same momentum change) as the actual variable force over the same time interval $\Delta t$. By definition: $$\vec{J}={\vec{F}}_{avg}\Delta t=\Delta \vec{p}$$ Rearranged gives the common form used for collision problems: ${\vec{F}}_{avg}=\frac{\Delta \vec{p}}{\Delta t}$. This formula is frequently tested in FRQs, and it aligns with Newton’s third law: for two colliding objects, the average force on object 1 from object 2 is equal and opposite to the force on object 2 from object 1, so their impulses are also equal and opposite. The most common mistake here is mishandling signs for objects that reverse direction during a collision, so always set a clear coordinate system first.

Worked Example

A 0.15 kg baseball travels horizontally toward a bat at 35 m/s. After being hit, it travels at 45 m/s away from the bat along the same line. If the collision lasts 0.002 s, what is the magnitude of the average force exerted on the ball by the bat?

1. Set a coordinate system where positive x is away from the bat. Initial velocity is $v_i=-35\text{ m/s}$, final velocity is $v_f=+45\text{ m/s}$.
2. Calculate initial and final momentum: $p_i=m{v}_i=(0.15)(-35)=-5.25\text{ kgcdotpm/s}$, $p_f=m{v}_f=(0.15)(45)=6.75\text{ kgcdotpm/s}$.
3. Find momentum change: $\Delta p=p_f-p_i=6.75-(-5.25)=12\text{ kgcdotpm/s}$.
4. Solve for average force: $F_{avg}=\frac{\Delta p}{\Delta t}=\frac{12}{0.002}=6000\text{ N}$. The magnitude of the average force is 6000 N.

Exam tip: When an object reverses direction, do not drop the negative sign on initial velocity. Failing to do this cuts your calculated momentum change (and thus average force) in half, which is a very common exam error.

4. Impulse for Calculus-Based Variable Force Problems

AP Physics C: Mechanics regularly tests your ability to calculate impulse for a force given as an explicit function of time, using the definition of impulse as a definite integral. For a 1D net force $F_{net}(t)$, impulse is: $$J={\int }_{t_1}^{t_2}{F}_{net}(t)dt=\Delta p$$ For 2D motion, the impulse-momentum theorem holds independently for each vector component: you calculate $J_x=\Delta p_x$ and $J_y=\Delta p_y$ separately, since x and y components of motion do not mix. This is one of the core calculus applications of the unit, and it appears at least once on almost every AP exam.

Worked Example

A variable force $F(t)=10t^2-5t$ (in N, t in s) acts on a 4.0 kg stationary object along the x-axis from $t=0$ s to $t=2.0$ s. No other forces act on the object. Find the speed of the object at $t=2.0$ s.

1. The object starts from rest, so initial momentum $p_i=0$, so $J=\Delta p=p_f=m{v}_f$, which gives $v_f=\frac{J}{m}$.
2. Set up the definite integral for impulse: $$J={\int }_0^2(10t^2-5t)dt$$
3. Evaluate the integral: the antiderivative is $\frac{10}{3}{t}^3-\frac{5}{2}{t}^2$. Substitute bounds: $$J=\left(\frac{10}{3}(2{)}^3-\frac{5}{2}(2{)}^2\right)-0=\frac{80}{3}-10=\frac{50}{3}\approx 16.67\text{ Ncdotps}$$
4. Solve for final speed: $v_f=\frac{50/3}{4.0}\approx 4.17\text{ m/s}$.

Exam tip: If the force function changes form at different time boundaries, split your integral into separate intervals matching each function, then sum the results from each integral to get total impulse.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Calculating momentum change for a bouncing object as $\Delta p=m(v_f-v_i)$ where $v_i$ and $v_f$ are both positive magnitudes. Why: Students forget momentum is a vector, so reversing direction changes the sign of velocity, leading to a momentum change half the correct size. Correct move: Always set a coordinate system before starting, assign positive/negative signs to velocity based on direction, then calculate $\Delta p=m{v}_f-m{v}_i$ with signed velocities.
• Wrong move: Treating impulse as force times distance instead of force times time. Why: Students confuse impulse with work, which is force times displacement. Correct move: Label each quantity by its units on your paper: impulse has units $\text{Ncdotps}$, work has units $\text{Ncdotpm}$, so mismatched units will immediately flag a mistake.
• Wrong move: Counting negative area (force below the time axis) as positive impulse when calculating area under F(t). Why: Students calculate total absolute area regardless of force direction, especially for graphs that cross the time axis. Correct move: Split the area into positive regions (above the axis) and negative regions (below the axis), add positive areas and subtract negative areas to get total impulse.
• Wrong move: When asked for the impulse exerted on object A by object B, use the change in momentum of object B instead of object A. Why: Students mix up which object the impulse acts on when applying Newton's third law. Correct move: Explicitly write that impulse on X equals the change in momentum of X, then match your calculation to the correct object.
• Wrong move: Calculate impulse using only the applied force, ignoring other forces like gravity or friction. Why: Problems often only mention the applied force, leading students to forget the impulse-momentum theorem uses net force. Correct move: Always check for other forces acting on the object, and add them to get net force before calculating impulse.

6. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

A 0.50 kg cart moves to the right along a horizontal track at an initial speed of 2.0 m/s. A net force varying with time acts on the cart, where positive force corresponds to the right direction:

• From $t=0$ to $t=2.0$ s: The force forms a triangle rising from 0 N at $t=0$ to +1.0 N at $t=1$ s, then falling back to 0 N at $t=2$ s.
• From $t=2$ s to $t=4$ s: The force forms a triangle falling from 0 N at $t=2$ s to -1.0 N at $t=3$ s, then rising back to 0 N at $t=4$ s.

What is the speed of the cart at $t=4.0$ s? A) 0 m/s B) 2.0 m/s C) 3.0 m/s D) 4.0 m/s

Worked Solution: Total impulse equals the sum of the signed areas of the two triangles. The first triangle (0–2 s) has area $\frac{1}{2}\times 2\times 1=+1\text{ Ncdotps}$. The second triangle (2–4 s) has area $\frac{1}{2}\times 2\times (-1)=-1\text{ Ncdotps}$. Total impulse $J=1-1=0\text{ Ncdotps}$. By the impulse-momentum theorem, $J=\Delta p=m(v_f-v_i)$, so $v_f=v_i=2.0\text{ m/s}$. The most common mistake is adding absolute areas, which incorrectly gives 4 m/s. The correct answer is B.

Question 2 (Free Response)

A 3.0 kg block is pulled along a frictionless horizontal surface by a time-varying force $F(t)=6t-2t^2$, where $F$ is in newtons and $t$ is in seconds. The block starts from rest at $t=0$. (a) Calculate the total impulse exerted on the block from $t=0$ to $t=2$ s. (b) What is the speed of the block at $t=2$ s? (c) At what time $t>2$ s does the block have the same momentum it has at $t=2$ s?

Worked Solution: (a) Impulse is the integral of $F(t)$ over the interval: $$J={\int }_0^2(6t-2t^2)dt={\left[3t^2-\frac{2}{3}{t}^3\right]}_0^2=3(4)-\frac{2}{3}(8)=\frac{20}{3}\approx 6.67\text{ Ncdotps}$$

(b) The block starts from rest, so $J=\Delta p=m{v}_f$, so: $$v_f=\frac{J}{m}=\frac{20/3}{3}=\frac{20}{9}\approx 2.22\text{ m/s}$$

(c) We set the impulse from $0$ to $t$ equal to $\frac{20}{3}$: $$3t^2-\frac{2}{3}{t}^3=\frac{20}{3}⟹2t^3-9t^2+20=0$$ We know $t=2$ is a root, so factor out $(t-2)$ to get $2t^2-5t-10=0$. Solving the quadratic gives the positive root greater than 2 as $t=\frac{5+\sqrt{105}}{4}\approx 3.81\text{ s}$.

Question 3 (Application / Real-World Style)

A 60 kg stunt person jumps from a 10 m height onto an air bag, which stops them over a time interval of 0.80 seconds. Estimate the magnitude of the average force exerted by the air bag on the stunt person, accounting for gravity during the impact. Assess if the landing is safe (human bodies tolerate impacts up to ~3x body weight).

Worked Solution: First, find the speed of the stunt person just before impact with kinematics: $v=\sqrt{2gh}=\sqrt{2(9.8)(10)}\approx 14\text{ m/s}$ downward. Set upward as positive: $p_i=m(-v)=60(-14)=-840\text{ kgcdotpm/s}$, $p_f=0$, so $\Delta p=840\text{ kgcdotpm/s}$. Net average force is $F_{net,avg}=\frac{\Delta p}{\Delta t}=\frac{840}{0.8}=1050\text{ N}$. Net force is $F_{bag}-mg=F_{net,avg}$, so $F_{bag}=1050+(60)(9.8)\approx 1640\text{ N}$. The stunt person's weight is 588 N, so the average force from the bag is ~2.8x body weight. This is within the safe tolerance for a human landing.

7. Quick Reference Cheatsheet

8. What's Next

Mastering impulse and momentum for single particles is an absolute prerequisite for the rest of Unit 4, starting with conservation of linear momentum for multi-particle systems. This chapter gives you the foundation to prove that for an isolated system (no net external force), total momentum is conserved, the core tool for solving all collision and interaction problems. Without being able to calculate impulse and momentum change correctly, you will not be able to analyze elastic and inelastic collisions, center of mass motion, or variable-mass systems like rockets, all of which are heavily tested on the AP exam. Impulse and momentum also connect to energy conservation, as you will combine momentum and energy analysis to solve collision problems that cannot be solved with force or energy alone.

Conservation of Linear Momentum Elastic and Inelastic Collisions Center of Mass Rocket Propulsion