Laws of Motion · Lesson 14 of 14
Chapter Summary and Practice
“Bring Newton’s laws, force diagrams, friction, momentum and circular-motion reasoning together in worked practice.”
• Choose a system, frame and force model before calculating. • Apply Newton’s laws and momentum balance with their conditions. • Check contact, no-slip and string assumptions after solving. • Explain the most common force and motion misconceptions.
The three laws work together
The first law identifies unaccelerated motion when the resultant external force vanishes in an inertial frame. The second connects the resultant to momentum change, giving F_net = ma for fixed mass. The third identifies equal and opposite forces on the two participants in an interaction. Together they connect individual body diagrams with system momentum balance. Keep the system boundary and frame visible throughout the reasoning.
| Quantity | Relationship | SI unit and dimension |
|---|---|---|
| Linear momentum | p = mv | kg m/s; [M L T⁻¹] |
| Net force | F_net = dp/dt = ma for fixed mass | N = kg m/s²; [M L T⁻²] |
| Impulse | J = Δp = integral of F dt | N s; [M L T⁻¹] |
| Weight near Earth | W = mg | N; [M L T⁻²] |
| Spring constant | F_x = −kx | N/m; [M T⁻²] |
| Friction coefficient | μ_s or μ_k | Dimensionless |
| Tension or normal force | Determined by interaction and constraints | N; [M L T⁻²] |
A compact map of conditions
Worked practice: compare possible friction states
Problem
A 5 kg block rests on a horizontal floor with μ_s = 0.40 and μ_k = 0.30. Take g = 10 m/s². Compare horizontal pulls of 15 N and 25 N.
- 1.N = 50 N, so the maximum static friction is 20 N.
- 2.At 15 N pull, static friction is 15 N opposite the pull and acceleration is zero.
- 3.At 25 N pull, the initial no-slip balance would require more than 20 N, so sliding begins. Then kinetic friction is 15 N and acceleration is (25 − 15)/5 = 2 m/s².
- 4.The actual static force in the first case is not μ_sN. The sliding force in the second case uses μ_k rather than μ_s.
Worked practice: combined and separate diagrams
Problem
A 2 kg block and a 3 kg block are connected on a smooth horizontal table. A 20 N force pulls the 3 kg block away from the 2 kg block. Find acceleration and tension.
- 1.For the combined 5 kg system, a = 20/5 = 4 m/s².
- 2.The only horizontal force on the trailing 2 kg block is T, so T = 2 × 4 = 8 N.
- 3.Check the pulled block: 20 − 8 = 3 × 4. Internal string forces cancel only in the combined equation; each block still has a tension force on its own diagram.
Worked practice: an impulse and a full reversal
Problem
A 0.20 kg ball initially moves at +8 m/s and leaves a collision at −6 m/s. The interaction lasts 0.010 s. Find net impulse and average resultant force.
- 1.Δp = 0.20(−6 − 8) = −2.8 kg m/s.
- 2.Net impulse is −2.8 N s; average resultant force is −2.8/0.010 = −280 N.
- 3.The negative direction is opposite the original motion. Using only the speed decrease 8 − 6 would miss the reversal and greatly underestimate the momentum change.
Worked practice: a force limit on a circular path
Problem
A 1000 kg car takes a level turn of radius 50 m at 10 m/s. Let μ_s = 0.30 and g = 10 m/s². Find required lateral friction and test no slip.
- 1.Required inward force is mv²/R = 1000 × 100/50 = 2000 N.
- 2.The available static-friction magnitude is at most μ_smg = 3000 N.
- 3.Because 2000 ≤ 3000, the ideal no-slip turn is possible. The actual lateral friction is 2000 N, not 3000 N.
- 4.The speed bound is √(0.30 × 50 × 10) = √150 = 12.25 m/s. Additional braking or acceleration would create extra friction demand beyond this purely lateral example.
Concept checks with explanations
| Claim | Assessment and reason |
|---|---|
| Zero velocity means zero net force | False: a freely thrown ball at its top can have zero vertical velocity and nonzero weight. |
| A moving object carries a force from earlier acceleration | False: current interactions determine current force. |
| Weight and normal reaction form a third-law pair | False: both act on one body; each has a partner on a different body. |
| Static friction always equals μ_sN | False: that is only the limiting magnitude. |
| Friction always points opposite ground-frame velocity | False: it opposes relative slip or its tendency at the contact. |
| Centripetal force is an extra force to add | False: it names the inward role of the actual net force. |
| Equal mutual forces imply equal accelerations | False: masses can differ. |
| Internal forces can accelerate an isolated system’s centre of mass | False under the Newtonian paired-interaction model: total external force controls centre-of-mass acceleration. |
| A body released from circular motion starts radially outward | False: its initial free velocity is tangent to the path. |
| The feeling of being pushed outward proves an outward interaction | False: the description depends on the frame and contact forces; draw the actual forces in the chosen frame. |
An end-to-end checking routine
Read the motion interval and identify all phase changes. State an inertial frame, system boundary and axes. Draw only forces on the system and name each agent. Write Newton’s equations and constraints before substituting numbers. If using momentum conservation, justify the neglected external impulse. After solving, check units, vector directions, tension signs, contact conditions and static-friction bounds. If a result violates the assumed state, change the model rather than keeping an impossible force.
To practise independently, cover the worked solutions and redraw every free-body diagram. Explain which third-law partner is absent from each selected-body diagram and why. For connected bodies, solve once using separate equations and once using the combined system. For circular motion, replace the word “centripetal” with the name of the real agent. These checks help make the force model explicit before the algebra becomes complicated.
Quiz
A block stays at rest on a table while a horizontal pull is gradually increased below the slipping threshold. What happens to static friction?
Key Takeaways
• Choose a system and inertial frame before applying the laws. • Net external force changes momentum; real interactions supply every force arrow. • Third-law partners act on different bodies and cancel internally only in a combined system. • Friction, contact and string assumptions must be checked against the solved motion. • Use explicit geometry, directions and time intervals in collision and circular-motion problems.