AP Physics C: Mechanics common mistakes
AP Physics C: Mechanics · Understand the traps and how to avoid them.
When a motion reverses direction or has two stages, the student keeps the sign convention from the first stage, then substitutes a negative displacement while leaving acceleration positive, and accepts an unphysical negative time or speed.
The student memorized the constant-acceleration equations as symbol-matching recipes instead of assigning a positive axis and tracking the sign of every vector through the motion.
What to do instead
Before choosing an equation, draw the axis and list x0, v0, vf, a, and t with explicit signs. Solve the quadratic and discard roots that violate the physical situation, then check that the sign of v matches the direction of motion at that stage.
On an incline, the student writes N = mg and then uses that value in friction, even though the perpendicular acceleration is zero and the correct normal force is N = mg cos θ.
The flat-surface result N = mg is overlearned, and the student treats the normal force as a gravitational counterweight rather than as whatever force is needed to satisfy Newton’s second law perpendicular to the surface.
What to do instead
Write ΣF_perp = ma_perp first. For a straight incline, set a_perp = 0 and solve for N = mg cos θ. On a vertical curve or loop, include the centripetal acceleration in the perpendicular direction before solving for N.
In a free-body diagram for a ball on a string at the top of a vertical circle, the student draws tension, weight, and a third arrow labeled centripetal force, then either sets the forces to zero or double-counts the inward force.
The student treats centripetal as a force type that must be added, instead of recognizing it as the name for the net inward component of the real forces.
What to do instead
Never draw a centripetal-force arrow. Identify the real forces, resolve them along the inward radial direction, and set the sum of inward components equal to mv²/r. Add tangential components separately if the speed is changing.
While a rough patch acts on a block, the student writes Ki + Ui = Kf + Uf and obtains a final speed that is too high because the thermal energy is omitted.
The slogan energy is conserved is applied without distinguishing mechanical energy from total energy, so nonconservative work by friction is silently dropped.
What to do instead
Use ΔK + ΔU = W_nc. For kinetic friction, W_nc = −fk d, or add a thermal-energy term fk d to the final side. Check that the final mechanical energy is lower than the initial value when friction acts.
The student computes work as W = Fd even when the force is at an angle to the displacement, or uses the displacement of the block for a force applied at a different point on the system.
The student collapses the scalar product into a product of magnitudes and does not check which displacement belongs to the point where the force is applied.
What to do instead
Write W = F d cos θ and identify the force vector and the displacement of its point of application. For a variable force, integrate F · dx. Use the sign of cos θ to decide whether the work transfers energy into or out of the system.
After a clay ball sticks to a block, the student sets 1/2 mv² before equal to 1/2 (m + M)V² after and solves for V, treating a perfectly inelastic collision as if kinetic energy were conserved.
The student conflates momentum conservation with energy conservation and assumes that any collision problem requires an energy equation.
What to do instead
Use momentum conservation for the collision: mv = (m + M)V. Kinetic energy is conserved only in an elastic collision. After finding V from momentum, compute the lost kinetic energy separately if the problem asks for it.
For two pucks colliding at right angles, the student adds m1v1 + m2v2 and sets the sum equal to (m1 + m2)V without separating x and y components.
One-dimensional momentum habits carry over, and the student treats momentum as a scalar magnitude rather than a vector.
What to do instead
Resolve momenta into x and y components before and after the collision. Apply conservation separately to each direction, then combine the final components to find the magnitude and direction of the total momentum.
The student calculates torque from a force directed along a rod as τ = rF, or computes a disk’s rotational inertia as MR² about its center instead of 1/2 MR².
The formulas are recalled in shortened form, and the moment-of-inertia table is not tied to the chosen axis or to the perpendicular distance from the axis to the force line.
What to do instead
Use τ = rF sin θ = F d_perp, where d_perp is the perpendicular distance from the axis to the line of action. Choose the axis first, then select or derive I about that axis, using the parallel-axis theorem when the axis is shifted.
When a disk is dropped onto a rotating turntable and they rotate together, the student conserves kinetic energy as well as angular momentum, or uses L = Iω about a point that is not fixed.
The student mirrors linear collision rules and does not track which axis or point remains fixed, so the axis ambiguity allows an invalid energy equation.
What to do instead
Angular momentum is conserved about a fixed point or fixed axis when the external torque about that axis is zero. A perfectly inelastic rotational collision loses kinetic energy, so compute Li = I1ω1 and Lf = (I1 + I2)ωf with the same axis, then compare energies.
For a simple pendulum, the student writes T = 2π√(m/k), or claims that the period of any oscillating system depends on amplitude.
The spring-mass period formula is overgeneralized, and the small-angle condition that makes simple-harmonic motion amplitude-independent is not stated or checked.
What to do instead
Derive ω from the restoring force or torque. For a spring, ω = √(k/m); for a simple pendulum, ω = √(g/L), so T = 2π√(L/g). State the small-angle approximation and note that amplitude independence holds only in the simple-harmonic limit.