Explanation: Passengers tend to continue in uniform motion (Newton I — inertia). When bus brakes, bus decelerates but passengers' bodies resist change and move forward relative to bus. Safety measures: seat belts restrain passenger and provide net backward force to change momentum over controlled time; airbags and headrests reduce head/thorax injury by increasing stopping time and contact area.

Explanation: Plates tend to remain at rest due to inertia. A very quick pull applies a short horizontal impulse to cloth, insufficient to overcome static friction and inertia of plates substantially, so plates remain nearly stationary. The practical demonstration relies on minimizing horizontal force transfer and short contact time.

Explanation: Using F = m·a, for same net applied force F, acceleration a = F/m. The empty car (smaller mass) has larger a; loaded car (greater mass) has smaller a. Thus less massive car accelerates faster because inertia (mass) is smaller.

Explanation: Let passenger mass m, upward acceleration a. Forces: normal R up, weight mg down. Apply ΣF = m·a upward: R − mg = m·a → R = m(g + a). Scale reads R which is larger than mg when a > 0, so passenger feels heavier. For downward acceleration, R = m(g − a) gives lighter feeling.

Solution: a = F/m = 20/4 = 5 m/s². From rest, s = ½ a t² = 0.5×5×36 = 90 m. Velocity after 6 s: v = at = 5×6 = 30 m/s.

Solution: a = (v − u)/t = (30 − 10)/8 = 20/8 = 2.5 m/s². Net force F = m·a = 1200×2.5 = 3000 N.

Solution: Net force = 30 − 10 = 20 N (right). a = F/m = 20/4 = 5 m/s² to the right.

Explanation: For variable mass, F_net = dp/dt = m·a + v_rel·(dm/dt). Thrust results from expelled mass; as onboard mass m decreases (dm/dt negative), for similar thrust the term m·a must increase so a increases. Alternatively, for approximately constant thrust, a = F_net/m increases as m decreases (inverse proportionality).

Solution: a = (0 − 6)/3 = −2 m/s². F = m·a = 2×(−2) = −4 N. Magnitude 4 N opposite motion (negative sign indicates direction opposite chosen positive).

Explanation: The swimmer exerts backward force on water (action); water exerts equal and opposite reaction on swimmer propelling forward. Effective propulsion requires that the swimmer applies force against water and that the water provides reaction — this is analogous to friction/traction in walking but here involves fluid reaction forces. Frictional drag acts opposite motion; swimmer must overcome drag to maintain speed.

Solution: Before firing total momentum zero. After firing: m_b v_b + m_g v_g = 0 → v_g = −(m_b v_b)/m_g = −(0.02×400)/2 = −4/2 = −2 m/s. Gun recoils backward at 2 m/s. Action: expanding gas pushes bullet forward; reaction pushes gun backward.

Explanation: Action: oar pushes water backward. Reaction: water pushes oar/boat forward. Internal forces between man and canoe transmit the reaction to the canoe. Momentum transfer to water produces forward motion; no ground contact needed because action–reaction works via fluid.

Explanation: For same braking force F, deceleration a = F/m. Loaded truck (larger m) → smaller |a| → longer stopping distance. Stopping distance s = v²/(2|a|) is larger if a smaller. Friction at brakes and tires provides braking force; heavier mass requires greater braking force for same deceleration.

Explanation: Static friction adjusts up to maximum f_s(max) = μ_s N. Until applied horizontal force ≤ f_s(max), box remains at rest. If push > f_s(max), static friction is overcome and box moves. Example: mass 10 kg, μ_s = 0.4 → N = 98 N → f_s(max) = 39.2 N. Any push above ~39.2 N starts motion.

Explanation & Measures: Icy roads have low μ_k leading to small frictional braking and traction forces. Measures: reduce speed, use tires with better grip (winter tires), increase contact friction (chains), avoid hard braking (increase stopping distance), use ABS to prevent lock and allow controlled braking. Increasing normal force (not practical) would increase friction; better solution is increasing μ via surface treatment.

Solution & Explanation: Δp = m(v_f − v_i) = 0.16(0 − 30) = −4.8 kg·m/s. Average force ≈ Δp/Δt = −4.8/0.2 = −24 N (opposite motion). Glove increases contact time Δt (padding), reducing average force magnitude and risk of injury.

Explanation: In elastic collision, both momentum and kinetic energy are conserved. For equal masses and opposite velocities of equal magnitude, they exchange velocities — post-collision velocities reverse directions if magnitudes equal. Use conservation equations to find exact values if magnitudes differ.

Explanation: For given change in momentum Δp, average force F_avg = Δp/Δt. Crumple zone increases time over which car's momentum reduces to zero, decreasing average force transmitted to occupants. Airbags and seat belts also increase Δt and spread force, reducing peak accelerations and injuries.

Solution: Let m2 descend with acceleration a. For m2: m2 g − T = m2 a. For m1: T − m1 g = m1 a (upwards). Add equations: (m2 − m1) g = (m1 + m2) a → a = (m2 − m1) g / (m1 + m2) = (5 − 3)×9.8 / 8 = 2×9.8/8 = 19.6/8 = 2.45 m/s². T from T = m1(g + a) = 3(9.8 + 2.45) = 3×12.25 = 36.75 N.

Checklist (Stepwise):

1. Read problem carefully, list given & unknown quantities (with units).
2. Draw clear diagram and free-body diagram labeling all forces and directions.
3. Choose coordinate axes and sign convention; resolve forces into components if needed.
4. Decide assumptions (frictionless, massless rope, constant g, rigid bodies).
5. Write ΣF = m·a separately along axes (or use momentum form if mass variable).
6. Solve algebraically, keep units consistent (SI preferred).
7. Check limiting cases and units; verify answer plausibility (signs, magnitudes).

Tip: Writing units at each step and drawing diagrams gains method marks in exams.