Center of Mass
5 questionsKey ideaxcm = Σm·x / Σm, or ∫x dm / M for a continuous object, with dm = λ dx.
Common mistakeUsing the geometric midpoint when the density is not uniform.
Covers: Newton’s laws, resistive forces and circular motion. Quiz one topic at a time or take the whole unit.
The idea each topic turns on, and the mistake students make most often.
Key ideaxcm = Σm·x / Σm, or ∫x dm / M for a continuous object, with dm = λ dx.
Common mistakeUsing the geometric midpoint when the density is not uniform.
Key ideaDraw a free-body diagram for each object, choose axes, and write ΣF = ma along each axis.
Common mistakeAdding ma to the diagram as if it were a force, or leaving out a force that acts on the object.
Key ideaKinetic friction is μk·N. Static friction adjusts as needed, up to μs·N. Find N from the perpendicular equation first.
Common mistakeAssuming N = mg when an angled force or an incline changes it.
Key ideaWhen F depends on t, write m dv/dt = F(t) and integrate. The impulse is ∫F dt.
Common mistakeUsing F = ma with a single value of the force when the force changes during the interval.
Key ideaFor an object falling with drag F = −bv, the velocity approaches the terminal velocity mg/b exponentially, with time constant m/b.
Common mistakeThinking the acceleration stays g, or that terminal velocity is reached in a finite time. The acceleration starts at g and decreases toward zero; the net force is zero only at terminal velocity.
Key ideaThe net inward force is mv²/r. For an orbit, gravity provides it: GMm/r² = mv²/r.
Common mistakeAdding a separate “centripetal force” to the free-body diagram, or using the radius from the surface.
Key ideaF = −kx, measured from the spring’s natural length. Springs in series act softer than either spring alone, and springs in parallel act stiffer.
Common mistakeMeasuring x in F = −kx from the hanging-equilibrium position. The spring force uses the stretch from the natural length; only the net force (spring plus gravity) is −k times the displacement from equilibrium.
This is one of the 33 questions in the Unit 2 quiz.
A truck collides head-on with a small car. How does the force the truck exerts on the car compare with the force the car exerts on the truck?
Answer: A. They are equal in magnitude
These forces are a Newton’s third law pair, so they are always equal in magnitude and opposite in direction. The car changes velocity more because it has less mass.
AP-style FRQs for this unit. Students score themselves point by point against a rubric and see a full solution.