Short on time? This page is the recap for the calculus-based course (separate from Physics 1, which uses no calculus). For explanations, diagrams and worked examples, use the full study guide.
Recap
- Choose the system first. Forces from inside it are internal; forces from outside are external.
- Unbalanced forces means ΣF ≠ 0. Only then does the velocity of the center of mass change (in size, direction or both).
- ΣF = 0 means constant velocity, which may or may not be zero.
- Internal forces cancel in third-law pairs, so they never change the motion of the system’s center of mass.
- Acceleration points along the net force, whatever the direction of the velocity.
Key relationships
| Relationship | Units | Meaning |
|---|---|---|
| a_cm = ΣF / m_sys | m/s² | size ∝ ΣF, ∝ 1/m; direction of ΣF |
| ΣF_x = m a_x, ΣF_y = m a_y | N | one equation per axis |
| 1 N = 1 kg·m/s² | — | definition of the newton |
| a_x(t) = ΣF_x(t)/m | m/s² | holds at every instant |
| v_x(t) = v_x0 + (1/m)∫₀ᵗ ΣF_x dt | m/s | Δv_x = (area under ΣF_x–t graph) ÷ m |
| a against F is a line, slope 1/m; a against 1/m is a line, slope ΣF | — | lab tests of the law |
Assumptions behind the numbers
- Inertial reference frame, and the system’s mass does not change.
- “Smooth” or “low-friction” means friction is negligible; “light” bar or string means its mass is negligible.
- g = 9.8 m/s² for weight, mg.
Mistakes to avoid
- Adding ma to the free-body diagram. It is the result of the forces.
- Leaving out the weight (or another external force) in one component.
- Counting internal forces when the system contains both objects.
- Dividing by the wrong mass. Use the mass of the system you chose.
- Thinking force sets velocity. Force sets acceleration; the velocity remembers all earlier forces.
- Constant-a equations with F(t). Integrate a_x(t) instead.
- Forgetting v_x0 when integrating.
Quick self-check
- A net force of 6.0 N acts on a 1.5 kg puck. What is its acceleration? (4.0 m/s², along the net force)
- The net force doubles and the mass is multiplied by 4. By what factor does a change? (× 0.5)
- A 2.0 kg cart starts from rest. The net force is F_x = (6.0 N/s)t. What is v_x at 2.0 s? (a_x = 3.0t, so v_x = 1.5t² = 6.0 m/s)
Next: practice questions.