Newton's Laws (Calculus-based) — AP Physics C: Mechanics Phys C Mech Study Guide

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

Covers: Calculus-based formulation of Newton's 2nd Law, variable force problem-solving methods, drag force and terminal velocity calculations, and free-body diagrams for non-inertial reference frames, with exam-specific tips and practice.

You should already know: Strong calculus (concurrent OK), AP Physics 1 helpful.

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 papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Newton's Laws (Calculus-based)?

Calculus-based Newton's Laws extend the algebra-level dynamics framework from AP Physics 1 to solve for motion with non-constant forces and accelerations, using differentiation and integration to model real-world dynamic systems. This topic accounts for 18-22% of your total AP Physics C: Mechanics exam score per the official CED, and is the foundational link between kinematics and all later units including work, energy, momentum, and rotational dynamics. It is sometimes referred to as "variable force dynamics" or "calculus Newtonian mechanics" in exam materials.

2. Newton's 2nd law as $\vec{F}=m\frac{d\vec{v}}{dt}$

The algebra-level form of Newton's 2nd Law ($\vec{F}=m\vec{a}$) is a special case of the general calculus-based formulation, where acceleration is defined explicitly as the first time derivative of velocity. This vector form applies to all non-relativistic motion, even when acceleration is not constant.

Key Definitions

• $\vec{F}$: Net external vector force acting on the system (sum of all real interaction forces, excluding internal forces between components of the system)
• $m$: Inertial mass of the system, constant for all testable non-relativistic problems on the exam
• $\frac{d\vec{v}}{dt}$: Instantaneous vector acceleration, equal to the rate of change of velocity over time For systems with changing mass (e.g., rockets burning fuel), you will use the full momentum form $\vec{F}=\frac{d\vec{p}}{dt}$, where $\vec{p}=m\vec{v}$ is linear momentum, but constant-mass problems make up 90% of test questions for this unit.

Worked Example

A 2.0kg block moves along the x-axis with velocity given by $\vec{v}(t)=(3t^2+2t)\hat{i}$ m/s. Find the net force on the block at $t=1.5$s.

1. Differentiate velocity to get acceleration: $$\frac{d\vec{v}}{dt}=(6t+2)\hat{i}{\text{ m/s}}^2$$
2. Substitute $t=1.5$s: $a=6(1.5)+2=11{\text{ m/s}}^2\hat{i}$
3. Apply Newton's 2nd Law: $\vec{F}=m\vec{a}=2.0\ast 11=22\text{ N }\hat{i}$ Exam tip: Examiners often deduct 1 point for missing unit vectors or direction labels on force answers, so always align your result with the defined coordinate system.

3. Variable force problems

Most real-world forces are not constant: spring force depends on displacement, impulsive collision forces depend on time, and drag force depends on velocity. When force varies as a function of time, position, or velocity, you will rearrange Newton's 2nd Law to solve for velocity or position using integration, rather than relying on constant-acceleration kinematics.

Core Integration Rules

1. Force as a function of time ($F(t)$): Rearrange $\frac{dv}{dt}=\frac{F(t)}{m}$ and integrate over time: $${\int }_{v_0}^{v_f}dv=\frac{1}{m}{\int }_{t_0}^{t_f}F(t)dt$$
2. Force as a function of position ($F(x)$): Use the chain rule to rewrite acceleration: $\frac{dv}{dt}=v\frac{dv}{dx}$, then rearrange and integrate over position: $${\int }_{v_0}^{v_f}vdv=\frac{1}{m}{\int }_{x_0}^{x_f}F(x)dx$$

Worked Example

A 3.0kg object is at rest at $x=0$, subject to a position-dependent force $F(x)=4x+2$ N along the x-axis. Find its speed at $x=5.0$m.

1. Apply the position-dependent force integration formula: $${\int }_0^{v_f}vdv=\frac{1}{3}{\int }_0^5(4x+2)dx$$
2. Evaluate both integrals: Left side: $\frac{1}{2}{v}_f^2$ Right side: $\frac{1}{3}{\left[2x^2+2x\right]}_0^5=\frac{1}{3}(50+10)=20$
3. Solve for $v_f$: $\frac{1}{2}{v}_f^2=20⟹v_f=\sqrt{40}\approx 6.32$ m/s Common trap: Never use constant-acceleration kinematics for variable force problems, even if you get a numerically similar result: examiners explicitly award 0 points for this approach on FRQs.

4. Drag force and terminal velocity

Drag is a resistive force exerted by a fluid (air, water) on a moving object, always acting opposite to the direction of the object's velocity relative to the fluid. AP Physics C: Mechanics almost exclusively tests linear drag, proportional to velocity, for terminal velocity problems.

Key Formulas

• Linear drag force: ${\vec{F}}_d=-b\vec{v}$, where $b$ is a drag coefficient dependent on object shape and fluid viscosity, and the negative sign indicates the force opposes velocity.
• Terminal velocity: Occurs when net force on a falling object is zero, so drag equals gravitational force: $$mg=b{v}_t⟹v_t=\frac{mg}{b}$$
• Velocity as a function of time for a falling object starting from rest: Derived by solving the differential equation $mg-bv=m\frac{dv}{dt}$: $$v(t)=v_t\left(1-e^{-bt/m}\right)$$

Worked Example

A 0.10kg raindrop has a linear drag coefficient $b=0.050$ kg/s. Find (a) its terminal velocity, and (b) its speed 2.0s after being released from rest.

1. Part (a): Calculate terminal velocity: $v_t=\frac{mg}{b}=\frac{0.10\ast 9.8}{0.050}=19.6$ m/s
2. Part (b): Substitute $t=2.0$s into the time-dependent velocity formula: $$v(2.0)=19.6\left(1-e^{-(0.050\ast 2.0)/0.10}\right)=19.6(1-e^{-1})\approx 12.4\text{ m/s}$$ Exam note: If you forget the exponential velocity formula, you can derive it in 1-2 steps on the exam by separating variables and integrating Newton's 2nd Law for drag.

5. Free-body diagrams in non-inertial frames

Non-inertial reference frames are frames that are accelerating (e.g., a turning car, an elevator speeding up). Newton's 1st Law does not hold in these frames unless you introduce fictitious forces: apparent forces that are not real interaction forces, but allow you to apply $F=ma$ to the accelerating frame.

Core Rules for Non-Inertial FBDs

1. First draw all real forces: gravity, normal force, friction, tension, drag, etc.
2. Add the fictitious force: ${\vec{F}}_{fict}=-m{\vec{a}}_{frame}$, where ${\vec{a}}_{frame}$ is the acceleration of the non-inertial frame relative to an inertial frame (e.g., the ground). The fictitious force always points opposite to the frame's acceleration.
3. In the non-inertial frame, net force equals $m$ times the object's acceleration relative to the frame.

Worked Example

A 5.0kg box sits on the floor of an elevator accelerating upward at 2.0 m/s². Draw the FBD for the box in the elevator's non-inertial frame, and find the normal force acting on the box.

1. Real forces: Downward gravitational force $mg=49$ N, upward normal force $N$.
2. Fictitious force: The elevator accelerates upward, so fictitious force points downward: $F_{fict}=ma=5.0\ast 2.0=10$ N (label explicitly as fictitious on your FBD).
3. In the elevator frame, the box is stationary, so net force = 0: $N=mg+F_{fict}=49+10=59$ N. Exam warning: You will lose 1-2 points if you add fictitious forces to inertial frame FBDs, or fail to label fictitious forces explicitly on non-inertial FBDs.

6. Common Pitfalls (and how to avoid them)

• Pitfall 1: Using constant-acceleration kinematics (e.g., $v=u+at$) for variable force problems. Why it happens: Students default to memorized AP Physics 1 equations. Fix: Always check if force depends on time, position, or velocity first; if yes, use integration of Newton's 2nd Law.
• Pitfall 2: Ignoring the vector nature of $F=m\frac{dv}{dt}$ and mixing up sign conventions. Why it happens: Students treat all force and velocity values as scalars. Fix: Define a positive coordinate system at the start of every problem, and assign signs consistent with that system.
• Pitfall 3: Using quadratic drag for terminal velocity problems when linear drag is specified. Why it happens: Students confuse high-speed and low-speed drag cases. Fix: The exam will explicitly state if drag is proportional to $v$ or $v^2$; default to linear drag for terminal velocity unless told otherwise.
• Pitfall 4: Forgetting the chain rule for position-dependent force problems. Why it happens: Students try to integrate $F(x)$ over time instead of position. Fix: When $F$ is a function of $x$, always rearrange to $vdv=F(x)dx/m$ before integrating.
• Pitfall 5: Adding fictitious forces to inertial frame FBDs. Why it happens: Students do not confirm the reference frame type first. Fix: Start every FBD question by noting if the frame is inertial (constant velocity) or non-inertial (accelerating); only add fictitious forces for the latter.

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

Question 1 (3 points)

A 4.0kg particle moves along the y-axis, subject to a net force given by $F(t)=12t-6$ N, where $t$ is in seconds. If the particle has an initial velocity of 3.0 m/s upward at $t=0$, what is its velocity at $t=2.0$s?

Solution

1. Rearrange Newton's 2nd Law for time-dependent force: $dv=\frac{F(t)}{m}dt$
2. Integrate both sides with appropriate bounds: $${\int }_{3.0}^{v_f}dv=\frac{1}{4.0}{\int }_0^{2.0}(12t-6)dt$$
3. Evaluate integrals: Left side = $v_f-3.0$, Right side = $\frac{1}{4.0}{\left[6t^2-6t\right]}_0^{2.0}=\frac{1}{4.0}(24-12)=3.0$
4. Solve for $v_f$: $v_f=3.0+3.0=6.0$ m/s upward.

Question 2 (6 points)

A skydiver of mass 80kg is falling through air, with linear drag coefficient $b=10$ kg/s. (a) Calculate their terminal velocity. (b) How much time passes before they reach 90% of their terminal velocity?

Solution

Part (a): Terminal velocity occurs when net force = 0, so $mg=b{v}_t$: $$v_t=\frac{mg}{b}=\frac{80\ast 9.8}{10}=78.4\text{ m/s}$$ Part (b): Use the time-dependent velocity formula for linear drag, set $v=0.9v_t$: $$0.9v_t=v_t\left(1-e^{-bt/m}\right)$$ Cancel $v_t$, rearrange: $e^{-bt/m}=0.1$ Take natural log of both sides: $-\frac{bt}{m}=\ln (0.1)\approx -2.303$ Solve for $t$: $t=\frac{2.303\ast m}{b}=\frac{2.303\ast 80}{10}\approx 18.4$ s.

Question 3 (4 points)

A 2.0kg book rests on the dashboard of a car that is accelerating to the right at 3.0 m/s² relative to the road. Draw the free-body diagram for the book in the car's non-inertial reference frame, and calculate the minimum static friction coefficient required to keep the book stationary relative to the car.

Solution

FBD components:

1. Real forces: Downward gravitational force $mg=19.6$ N, upward normal force $N$, rightward static friction force $f_s$.
2. Fictitious force: Leftward $F_{fict}=ma=6.0$ N, labeled explicitly as fictitious. Calculations:

• Vertical equilibrium: $N=mg=19.6$ N
• Horizontal equilibrium: $f_s=F_{fict}=6.0$ N
• Minimum static friction coefficient: ${\mu }_s=\frac{f_s}{N}=\frac{6.0}{19.6}\approx 0.31$

8. Quick Reference Cheatsheet

9. What's Next

This topic is the foundational backbone of all remaining AP Physics C: Mechanics content. Your mastery of variable force integration directly prepares you for work-energy theorem calculations, where you integrate force over displacement to find work, as well as momentum and impulse problems, where you integrate force over time to find impulse. The non-inertial frame and drag concepts covered here will also appear in advanced rotation problems, including rolling without slipping in accelerating frames and air resistance effects on projectile motion, a common FRQ topic worth 10-15 points on the exam. If you have gaps in your understanding of any of the integration steps, force vector directions, or FBD rules covered in this guide, you can ask Ollie, our AI tutor, for personalized practice problems or step-by-step walkthroughs of more complex questions. Once you are confident with these concepts, move on to our study guide for Work, Energy, and Power, the next unit in the AP Physics C: Mechanics syllabus, to build on the dynamics skills you have learned here.