Tracked MC checkpoint across the full Dynamics sequence, with written-response prompts below for separate practice. Use it to test Forces, Energy, and Momentum before deciding your next repair move.
Covers all of L01–L15 | No notes | Show all working | Define positive direction for all vector problems
Phase 1: Forces (L01–L05) · Phase 2: Energy (L06–L10) · Phase 3: Momentum (L11–L15)
Section 1 — Multiple Choice
A 12 kg box sits on a horizontal surface. The normal force acting on the box is:
A 5 kg object accelerates at 3 m/s² on a flat surface. The net force on the object is:
According to Newton's Third Law, the reaction force to "Earth pulls the Moon gravitationally" is:
A 20 kg box is pulled up a frictionless 30° incline by a force parallel to the slope. The component of gravity acting along the slope (pulling it back down) is:
A velocity-time graph shows a straight line with a negative gradient. This means the object is:
A 300 N force acts at 60° to the horizontal, moving a box 8 m along a flat floor. The work done by this force is:
A 3 kg ball is dropped from 5 m height (frictionless). Using conservation of mechanical energy, its speed just before hitting the ground is closest to:
A car travelling at constant 25 m/s has engine power 50 kW. The resistance force acting on the car is:
A 500 kg roller-coaster car starts from rest at the top of a 30 m frictionless drop. Its speed at the bottom is:
A student uses KE₁ + U₁ = KE₂ + U₂ on a rough incline and calculates v = 12 m/s. The actual speed will be:
A 4 kg ball moves east at 5 m/s. Defining east as positive, its momentum is:
An airbag reduces injury in a car collision. Which statement correctly explains this using the impulse-momentum theorem?
A 6 kg ball at +4 m/s collides with a stationary 2 kg ball. They stick together. The final velocity is:
After a collision, KE_before = 100 J and KE_after = 60 J. The collision is best classified as:
A 3 kg ball at +6 m/s collides with a stationary 3 kg block and they stick together. The combined mass slides on a surface with μk = 0.2. Which correctly gives the slide distance?
Section 2 — Short Answer
A 15 kg box sits on a slope inclined at 25° to the horizontal. The coefficient of static friction is μs = 0.5. (a) Draw a free-body diagram labelling all forces. (b) Calculate the normal force. (c) Calculate the maximum static friction force and determine whether the box slides or stays put.
Show all calculations. State whether the box slides.
Draw FBD in your book. Show all calculations.
A 1000 kg car accelerates from rest to 20 m/s in 8 s on a flat road. The resistance force is 400 N throughout. (a) Calculate the net force on the car. (b) Calculate the driving force produced by the engine. (c) Calculate the work done by the engine over the 8 s (use average speed × time for distance).
A 25 kg box slides from rest down a slope (height 6 m, road distance 14 m, μk = 0.22). (a) Calculate the normal force. (b) Calculate the friction force and the work done by friction. (c) Using W_net = ΔKE, find the speed at the bottom.
A 700 kg car rolls from rest down a frictionless 15 m hill. At the bottom it travels along a flat road with engine power 28 kW and resistance force 700 N. (a) Find the speed at the bottom of the hill using conservation. (b) Find the constant cruise speed on the flat road. (c) Explain why the cruise speed differs from the speed at the bottom of the hill.
A 0.2 kg ball travelling at +15 m/s strikes a wall and bounces back at −12 m/s. Contact time = 0.006 s. (a) Apply the Vector Protocol. Calculate the change in momentum. (b) Calculate the impulse. (c) Calculate the average force on the ball. (d) A second identical ball travelling at +15 m/s sticks to the wall. Compare the Δp of the two balls and explain which delivers a larger force to the wall and why.
A 2 kg ball at +8 m/s collides with a stationary 6 kg block. They stick together and slide on a surface with μk = 0.18. (a) Find the velocity of the combined mass just after the collision. (b) Calculate the KE lost in the collision. (c) Find the slide distance using W_net = ΔKE. (d) Calculate the total energy dissipated as heat from the moment of collision to the moment the block stops. Verify it equals the original KE of the ball.
Section 3 — Extended Response
A 40 kg student on a skateboard (total mass 45 kg) is pushed from rest by a force of 120 N at 20° below the horizontal for 3 s on a flat surface with μk = 0.15.
Show all five parts with full working.
Answer in your book — draw FBD, show all five parts.
A 60 kg cyclist freewheels (no pedalling) from rest down a 20 m hill (road distance 60 m, μk = 0.08). At the bottom they cycle along a flat road against 80 N of air resistance with power output 200 W.
Show full working for all five parts.
Answer in your book — all five parts.
A 0.01 kg bullet is fired from a 4 kg rifle at rest. The bullet strikes and embeds in a 3 kg block resting on a rough surface (μk = 0.25). The block slides 0.4 m before stopping.
Label each stage. Show the full chain from slide distance back to rifle recoil.
Answer in your book — label all stages, work backward from Stage 3 to Stage 1.
Q1 — C (117.6 N upward). Newton 1: box stationary → N = mg = 12 × 9.8 = 117.6 N upward. N balances weight.
Q2 — B (15 N). F_net = ma = 5 × 3 = 15 N. Weight (49 N) is vertical — irrelevant to horizontal acceleration.
Q3 — A. Newton 3 pairs act on different objects. Earth pulls Moon → Moon pulls Earth. Same type of force (gravitational), equal magnitude, opposite direction.
Q4 — D (98 N). W‖ = mg sinθ = 20 × 9.8 × sin30° = 196 × 0.5 = 98 N. Option A (196 N) is the full weight. Option B is mg cos30°.
Q5 — C. Negative gradient on v-t graph = negative acceleration. If velocity is positive and decreasing → deceleration. If velocity is already negative → accelerating in negative direction. The gradient IS the acceleration — negative means the velocity is becoming more negative over time.
Q6 — B (1200 J). W = Fs cosθ = 300 × 8 × cos60° = 300 × 8 × 0.5 = 1200 J. Option A (2400) uses cos0°. Option C uses sin60°.
Q7 — A (9.9 m/s). v = √(2gh) = √(2 × 9.8 × 5) = √98 = 9.9 m/s. Mass cancels in conservation.
Q8 — D (2000 N). Constant v → F_drive = F_resist. P = F × v → F = P/v = 50 000/25 = 2000 N.
Q9 — C (24.2 m/s). v = √(2gh) = √(2 × 9.8 × 30) = √588 = 24.2 m/s. Mass cancels.
Q10 — B. Friction converts mechanical energy to heat — KE at bottom is less than the idealised (frictionless) value. Actual speed < 12 m/s. Option D is wrong — we can determine it will be less; we just need μk to find the exact value.
Q11 — C (+20 N s). p = mv = 4 × 5 = 20 N s. East = positive → p = +20 N s.
Q12 — A. Airbag extends contact time Δt. J = FΔt = Δp is constant. Larger Δt → smaller F. Δp is not reduced — same mass, same velocity change.
Q13 — D (+3 m/s). (6+2)v_f = 6×4 = 24. v_f = 24/8 = +3 m/s.
Q14 — B (Inelastic). KE_after < KE_before → inelastic. Objects separated (not perfectly inelastic). Momentum still conserved.
Q15 — C. Stage 1: v_f = 3×6/(3+3) = 18/6 = +3 m/s. KE_f = ½×6×9 = 27 J. f_k = 0.2×6×9.8 = 11.76 N. s = 27/11.76 = 2.30 m. Option A uses initial speed (6 m/s), not v_f. Option B same error.
Q16: (a) FBD: weight mg = 147 N vertically down; Normal force F_N perpendicular to slope (up-left); Static friction f_s up the slope. (b) F_N = mg cos25° = 15 × 9.8 × 0.906 = 133.2 N. (c) f_max = 0.5 × 133.2 = 66.6 N. Gravity down slope = mg sin25° = 15 × 9.8 × 0.423 = 62.2 N. Since f_max (66.6 N) > mg sin25° (62.2 N), the maximum static friction is sufficient to hold the box — it stays put.
Q17: (a) a = Δv/Δt = 20/8 = 2.5 m/s². F_net = 1000 × 2.5 = 2500 N. (b) F_drive = F_net + F_resist = 2500 + 400 = 2900 N. (c) Average speed = (0+20)/2 = 10 m/s. d = 10 × 8 = 80 m. W_engine = F_drive × d = 2900 × 80 = 232 000 J = 232 kJ.
Q18: sinθ = 6/14 = 0.429, cosθ = 0.903. (a) F_N = 25 × 9.8 × 0.903 = 221.2 N. (b) f_k = 0.22 × 221.2 = 48.7 N. W_friction = −48.7 × 14 = −681.5 J. (c) W_gravity = 25 × 9.8 × 6 = 1470 J. W_net = 1470 − 681.5 = 788.5 J. ½ × 25 × v² = 788.5. v = √(63.1) = 7.94 m/s.
Q19: (a) v = √(2 × 9.8 × 15) = √294 = 17.1 m/s. (b) Cruise: P = F_resist × v → v = 28 000/700 = 40 m/s. (c) At the bottom of the hill, speed = 17.1 m/s. To maintain 17.1 m/s on flat, engine would need P = 700 × 17.1 = 11 970 W. Actual engine power is 28 000 W — so at 17.1 m/s, F_drive = 28 000/17.1 = 1637 N >> F_resist = 700 N. Net force is forward → cyclist accelerates. Speed increases until P/v = 700 N → v = 40 m/s cruise.
Q20: Positive = toward wall. (a) Δp = 0.2 × (−12 − 15) = 0.2 × (−27) = −5.4 N s. (b) J = −5.4 N s. (c) F = −5.4/0.006 = −900 N (away from wall). (d) Ball that sticks: Δp = 0.2 × (0 − 15) = −3 N s. Bouncing ball has |Δp| = 5.4 N s vs 3 N s — 80% larger. By Newton 3, force on wall equals force on ball in magnitude. Bouncing ball delivers larger force (900 N vs 500 N) because the direction reversal doubles the velocity change compared to merely stopping.
Q21:
Q22:
Q23:
(e) At the bottom of the hill the cyclist moves at 17.4 m/s — far above the cruise speed of 2.5 m/s. On the flat road at 17.4 m/s: F_drive = P/v = 200/17.4 = 11.5 N. F_resist = 80 N. F_net = 11.5 − 80 = −68.5 N (negative = opposing motion). By Newton 2: a = −68.5/60 = −1.14 m/s² — the cyclist decelerates. As speed decreases, F_drive = P/v increases. At v = 2.5 m/s: F_drive = 80 N = F_resist → F_net = 0 → Newton 1 → constant velocity at cruise speed.
Q24:
(5) Newton 3 evaluation: The statement "recoil force equals force on bullet" is exactly correct by Newton 3 — the force the rifle exerts on the bullet is equal in magnitude and opposite in direction to the force the bullet exerts on the rifle, at every instant during firing. Both forces act for the same contact time (the firing duration). By the impulse-momentum theorem, the impulse on each object is equal and opposite → |Δp_rifle| = |Δp_bullet| exactly. This is verified: p_bullet = 0.01 × 421.4 = 4.21 N s; p_rifle = 4 × 1.054 = 4.22 N s (small rounding difference). The forces are equal — but the accelerations are very different: a_bullet = F/0.01 (enormous); a_rifle = F/4 (small). Newton 3 constrains forces, not accelerations.
Tick when you have finished the multiple-choice checkpoint and checked your answers.
You have finished all of Dynamics — Forces, Energy, and Momentum.