This diagnostic is for the algebra-based Physics 1 course, Unit 2 (Force and Translational Dynamics). It shows you which topics to revisit. Allow about 30 minutes, then use the table at the end.
These are original Marlbridge practice questions, not past exam questions. They have not been calibrated against real exam results, so your total does not predict an exam score. Use a scientific calculator in degree mode and g = 9.8 m/s² (9.8 N/kg). Ignore air resistance, and treat strings and springs as ideal.
Question 1 (multiple choice · 2.1)
Three small weights are glued to a light right-angled bracket. Take the origin at the corner, +x to the right and +y up. The weights are 2.0 kg at (0, 0), 1.0 kg at (0.90 m, 0) and 3.0 kg at (0, 0.60 m). Where is the center of mass?
- (A) (0.30 m, 0.60 m)
- (B) (0.30 m, 0.20 m)
- (C) (0.15 m, 0.30 m)
- (D) (0.45 m, 0.10 m)
Answer and explanation
Answer: (C). Total mass 6.0 kg. x_cm = (1.0 × 0.90) ÷ 6.0 = 0.15 m. y_cm = (3.0 × 0.60) ÷ 6.0 = 0.30 m.
- (A) divides each weighted sum by 3, the number of weights, instead of 6.0 kg, the total mass.
- (B) is the geometric centre of the triangle. It ignores the masses.
- (D) swaps the positions of the 1.0 kg and 3.0 kg weights.
If you missed this: read “The equation” and Worked example 2 in the Topic 2.1 study guide.
Question 2 (multiple choice · 2.2)
A ball hangs at rest from a string tied to a hook on a smooth vertical wall. The ball rests against the wall, so the string is tilted. Which set of arrows belongs on the ball’s free-body diagram?
- (A) Gravity; separate vertical and horizontal arrows for the tension; normal force
- (B) Gravity; tension; the push of the ball on the wall, towards the wall
- (C) Gravity; tension; normal force; friction (wall on ball), up
- (D) Gravity (Earth on ball), down; tension (string on ball), towards the hook; normal force (wall on ball), away from the wall
Answer and explanation
Answer: (D). Earth, the string and the wall each exert one force, drawn as one straight arrow from the dot.
- (A) draws components. Only whole forces go on the diagram.
- (B) includes a force the ball exerts. It acts on the wall.
- (C) adds friction, but the wall is smooth.
If you missed this: read “What a free-body diagram shows” in the Topic 2.2 study guide.
Question 3 (multiple choice · 2.3)
A camera drone hovers at a fixed point. Its rotors push air downward. Which is a Newton’s third-law pair?
- (A) Air on drone (up) and Earth on drone (gravity, down)
- (B) Drone on air (down) and air on drone (up)
- (C) Earth on drone (gravity, down) and drone on air (down)
- (D) Air on drone (up) and drone on Earth (gravity, up)
Answer and explanation
Answer: (B). Swap the objects and keep the type: “drone on air” becomes “air on drone”. Same type, opposite directions, different objects.
- (A) is the trap. These forces are equal only because the drone is in equilibrium. Both act on the drone, and they are different types.
- (C) and (D) each mix two different interactions (drone–Earth and drone–air). The partner of Earth’s pull on the drone is the drone’s pull on Earth.
If you missed this: read “How to find a third-law pair” in the Topic 2.3 study guide.
Question 4 (multiple choice · 2.4)
A 50 N shop sign hangs at rest from two light cords. The left cord rises at 30° above the horizontal; the right cord at 60°. What are the tensions?
- (A) Left (30°) cord 25 N, right (60°) cord 43 N
- (B) Left (30°) cord 43 N, right (60°) cord 25 N
- (C) 25 N in each cord
- (D) 50 N in each cord
Answer and explanation
Answer: (A). At rest, so ΣF_x = 0 and ΣF_y = 0. Horizontal: T_L cos 30° = T_R cos 60°. Vertical: T_L sin 30° + T_R sin 60° = 50 N. Solving gives T_L = 25 N and T_R = 43 N. Check: 12.5 + 37.5 = 50 N vertically.
- (B) swaps the values. The steeper cord, closer to vertical, does more of the lifting.
- (C) shares the weight equally, which ignores the angles.
- (D) makes each cord hold the whole weight.
If you missed this: read “Translational equilibrium” and Worked example 2 in the Topic 2.4 study guide.
Question 5 (multiple choice · 2.5)
Take +x east and +y north, looking down on frictionless ice. Two horizontal forces act on a 2.0 kg puck: 6.0 N east and 8.0 N north. What is the puck’s acceleration?
- (A) 7.0 m/s² at 53° north of east
- (B) 5.0 m/s² at 53° north of east
- (C) 3.0 m/s² east
- (D) 4.0 m/s² north
Answer and explanation
Answer: (B). ΣF = √(6.0² + 8.0²) = 10 N at tan⁻¹(8.0 ÷ 6.0) = 53° north of east. a = 10 ÷ 2.0 = 5.0 m/s², along the net force.
- (A) adds the sizes. Perpendicular forces combine by Pythagoras.
- (C) and (D) each use one force only.
If you missed this: read “Newton’s second law” in the Topic 2.5 study guide.
Question 6 (multiple choice · 2.6)
At what height above Earth’s surface is the gravitational field strength one ninth of its value at the surface? R is Earth’s radius.
- (A) 8R
- (B) 3R
- (C) 2R
- (D) 9R
Answer and explanation
Answer: (C). g = GM ÷ r², with r from Earth’s centre. One ninth needs r = 3R, which is 2R above the surface.
- (A) forgets to square r (so r = 9R), then subtracts R.
- (B) is the distance from the centre, not the height.
- (D) forgets to square r and measures from the centre.
If you missed this: read “The gravitational field” and Figure 2 in the Topic 2.6 study guide.
Question 7 (multiple choice · 2.7)
Take +x in the direction of the pull. A 4.0 kg block rests on a level table. For the block and table, μ_s = 0.50 and μ_k = 0.30. A horizontal pull of 15 N is applied. What is the block’s acceleration?
- (A) −1.2 m/s²
- (B) +0.81 m/s²
- (C) +3.8 m/s²
- (D) 0
Answer and explanation
Answer: (D). Static friction can supply up to μ_s mg = 0.50 × 39.2 = 19.6 N. The 15 N pull is less, so the block stays at rest, with 15 N of static friction in the −x direction.
- (A) treats μ_s F_n as the actual friction. It is only a maximum.
- (B) uses kinetic friction (11.8 N), which acts only once the surfaces slide.
- (C) ignores friction.
If you missed this: read “Static friction” and Worked example 1 in the Topic 2.7 study guide.
Question 8 (multiple choice · 2.8)
A 1.2 kg mass hangs at rest from two identical springs side by side. Each spring is stretched 3.0 cm from its relaxed length. What is the spring constant of one spring?
- (A) 2.0 × 10² N/m
- (B) 3.9 × 10² N/m
- (C) 2.0 N/m
- (D) 20 N/m
Answer and explanation
Answer: (A). At rest, 2kΔx = mg, so k = (1.2 × 9.8) ÷ (2 × 0.030) = 196 N/m.
- (B) lets one spring hold the whole 11.8 N weight.
- (C) leaves the stretch in centimetres.
- (D) uses the mass (1.2) instead of the weight (11.8 N).
If you missed this: read “Hooke’s law” and “Comparing springs and situations” in the Topic 2.8 study guide.
Question 9 (multiple choice · 2.9)
A car turns on a flat, level road along a curve of radius 40 m. For the tyres and road, μ_s = 0.80. What is the greatest speed at which the car can follow the curve without skidding?
- (A) 20 m/s
- (B) 18 m/s
- (C) 22 m/s
- (D) 5.7 m/s
Answer and explanation
Answer: (B). On a flat road, static friction supplies the whole inward force. At the limit, μ_s mg = mv² ÷ r, so v = √(μ_s g r) = √(0.80 × 9.8 × 40) = 18 m/s. The mass cancels.
- (A) is √(gr), which leaves out μ_s.
- (C) divides by μ_s instead of multiplying.
- (D) leaves out g.
If you missed this: read “Where does the inward force come from?” in the Topic 2.9 study guide.
Question 10 (short answer · 2.5)
Take +y up. A student pulls up with 60 N on a 2.0 kg block. A light string hangs from it to a 3.0 kg block below.
(a) Find the acceleration of the blocks. (b) Find the tension in the string. (c) Another student says the tension must be 3.0 × 9.8 = 29 N, the lower block’s weight. Explain the mistake.
Worked answer
(a) System: both blocks (string forces internal). ΣF_y = 60 − 5.0 × 9.8 = 11 N, so a_y = +2.2 m/s².
(b) System: the lower block. T − m₂g = m₂a_y, so T = 3.0(9.8 + 2.2) = 36 N. Check on the top block: 60 − 19.6 − 36 = 4.4 N = 2.0 × 2.2. ✓
(c) T = m₂g only if the lower block does not accelerate. It speeds up upward, so the string pulls with more than its weight; the extra 6.6 N gives it 2.2 m/s².
Check yourself: 1 point each for 2.2 m/s² from external forces only, choosing the lower block alone, 36 N, and linking (c) to the acceleration.
If you missed this: read “Choosing the system: external and internal forces” and Worked example 2 in the Topic 2.5 study guide.
Question 11 (short answer · 2.9)
A 30 kg child on a swing moves on a circle of radius 2.5 m (treat the two ropes as one). At the lowest point the child moves at 3.0 m/s.
(a) Find the child’s acceleration at the lowest point, with its direction. (b) Find the tension in the rope there. (c) A parent says the tension there equals the child’s weight, because the child is neither speeding up nor slowing down. Evaluate this claim.
Worked answer
(a) a_c = v² ÷ r = 3.0² ÷ 2.5 = 3.6 m/s², upward, towards the centre of the circle.
(b) Take “towards the centre” (up) as positive. F_T − mg = ma_c, so F_T = 30(9.8 + 3.6) = 4.0 × 10² N (402 N). The weight is 294 N.
(c) Incorrect. The speed is not changing, but the direction is. The acceleration points up, so the net force points up and the tension exceeds the weight by ma_c = 108 N.
Check yourself: 1 point each for 3.6 m/s² towards the centre, tension and weight opposite in the equation, 402 N, and rejecting the claim because of a_c.
If you missed this: read “Vertical loops and the minimum speed at the top” (the paragraph on the bottom of the loop) in the Topic 2.9 study guide.
Your next step
Mark each question right or wrong. Count Questions 10 and 11 as right only if you earned every point.
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 2.1 Systems and center of mass | 1 | Topic 2.1 study guide |
| 2.2 Forces and free-body diagrams | 2 | Topic 2.2 study guide |
| 2.3 Newton’s third law | 3 | Topic 2.3 study guide |
| 2.4 Newton’s first law | 4 | Topic 2.4 study guide |
| 2.5 Newton’s second law | 5, 10 | Topic 2.5 study guide |
| 2.6 Gravitational force | 6 | Topic 2.6 study guide |
| 2.7 Kinetic and static friction | 7 | Topic 2.7 study guide |
| 2.8 Spring forces | 8 | Topic 2.8 study guide |
| 2.9 Circular motion | 9, 11 | Topic 2.9 study guide |
How to use your result
- Missed one question in a topic? Re-read the section named in that answer, then try the topic’s practice set.
- Missed two or more questions in a topic? Start again with that study guide, including its worked examples, before doing any mixed practice.
- Guessed? Treat the question as missed.
- Look for patterns. Errors in Questions 2 and 3 usually mean you skipped “on what, by what?”. Errors in Questions 4, 7, 10 and 11 often mean a missing free-body diagram.
- All correct? Move straight to the mixed unit review, which combines the topics in longer exam-style questions.