These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. This is the algebra-based course, so no calculus is needed. Use g = 9.8 m/s². Ignore friction and air resistance unless told otherwise. Round final answers to 2 significant figures unless told otherwise.
Question 1 (multiple choice · foundation)
A pendulum bob swings back and forth through small angles. Consider the bob–Earth system, with h = 0 at the lowest point. At the instant the bob reaches the highest point of its swing, which statement is correct?
- (A) The kinetic energy is at its largest and the potential energy is at its smallest.
- (B) The kinetic energy is zero and the potential energy is at its largest.
- (C) The kinetic energy equals the potential energy.
- (D) The kinetic energy and the potential energy are both zero.
Answer and explanation
Answer: (B). At the highest point the bob is momentarily at rest, so K = 0, the smallest possible value. All the system’s energy is gravitational potential energy, which is at its largest because h is at its largest.
- (A) describes the lowest point of the swing.
- (C) is true only at one in-between height (half the maximum height), not at the top.
- (D) would mean the system has no energy at all; the energy has not disappeared, it is stored as potential energy.
Question 2 (multiple choice · core)
An object on a spring oscillates on a frictionless surface. Its amplitude is changed from A to 3A. By what factor does the total energy of the spring–object system change?
- (A) 3
- (B) 9
- (C) √3
- (D) 1 (no change)
Answer and explanation
Answer: (B). E = ½kA². Replacing A with 3A gives ½k(3A)² = 9 × ½kA².
- (A) treats energy as proportional to amplitude. That is true for the maximum speed, not the energy.
- (C) takes a square root where there is a square.
- (D) confuses energy with period, which really is unchanged.
Question 3 (multiple choice · core)
A spring–object system oscillates with amplitude A. When the object is at x = A/2, what fraction of the total energy is kinetic?
- (A) 1/2
- (B) 1/4
- (C) 3/4
- (D) √3/2
Answer and explanation
Answer: (C). U = ½k(A/2)² = ¼ × ½kA² = ¼E. So K = E − ¼E = ¾E.
- (A) assumes energy is shared equally halfway out. Equal sharing happens at x = A/√2, not A/2.
- (B) is the potential-energy fraction, not the kinetic one.
- (D) is the ratio v/v_max at x = A/2 (the square root of ¾), not an energy fraction.
Question 4 (calculation · core)
A 0.50 kg block on a frictionless track is attached to a spring with k = 50 N/m. It passes through the equilibrium position at 0.80 m/s.
(a) Find the total energy of the block–spring system. (b) Find the amplitude. (c) Find the block’s speed when it is 0.060 m from equilibrium.
Worked solution
(a) At equilibrium U = 0, so E = K = ½ × 0.50 × 0.80² = 0.16 J.
(b) At a turning point E = ½kA², so A = √(2E/k) = √(2 × 0.16 ÷ 50) = √0.0064 = 0.080 m.
(c) U = ½ × 50 × 0.060² = 0.090 J. K = 0.16 − 0.090 = 0.070 J. v = √(2 × 0.070 ÷ 0.50) = √0.28 = 0.53 m/s.
Suggested mark points (4): 1 for E at equilibrium; 1 for A from ½kA²; 1 for U at 0.060 m subtracted from E; 1 for the speed.
Common error: assuming speed falls in proportion to distance, (0.020 ÷ 0.080) × 0.80 = 0.20 m/s. Speed does not fall linearly with x.
Question 5 (graph · core)
An energy–position graph for a 2.0 kg object on a spring shows a U-shaped spring potential energy curve that passes through (−0.10 m, 0.50 J), (0, 0) and (+0.10 m, 0.50 J). A horizontal line shows the system’s total energy, 0.18 J.
(a) Find the spring constant. (b) Find the amplitude. (c) Find U and K at x = +0.030 m. (d) Find the maximum speed of the object. (e) Describe the graph of K against x on the same axes.
Worked solution
(a) U = ½kx²: 0.50 = ½ × k × 0.10², so k = 100 N/m.
(b) The turning points are where U meets the total-energy line: ½ × 100 × A² = 0.18, so A = 0.060 m.
(c) U = ½ × 100 × 0.030² = 0.045 J; K = 0.18 − 0.045 = 0.135 J (0.14 J to 2 significant figures).
(d) At x = 0 all 0.18 J is kinetic: v_max = √(2 × 0.18 ÷ 2.0) = 0.42 m/s.
(e) An upside-down parabola: 0.18 J at x = 0, falling to zero at x = ±0.060 m. At every x, K is the gap between the U curve and the total-energy line. It does not exist beyond ±0.060 m, because the object never goes there.
Suggested mark points (5): one for each part.
Question 6 (constructed response · stretch)
A student attaches a 0.25 kg cart to a spring on a level, low-friction track. She releases it from rest at different amplitudes A and uses a photogate at the equilibrium position to measure its maximum speed:
| A (m) | 0.020 | 0.040 | 0.060 | 0.080 | 0.100 |
|---|---|---|---|---|---|
| v_max (m/s) | 0.25 | 0.51 | 0.76 | 1.01 | 1.27 |
Her partner claims: “Doubling the amplitude doubles the energy of the system.”
(a) Using energy conservation, explain what quantities she should plot to get a straight line whose slope gives the spring constant. (b) Process the data and find the spring constant. (c) Use the data to evaluate the partner’s claim.
Worked solution and suggested Marlbridge rubric
(a) At release all the energy is ½kA²; at equilibrium it is all ½mv_max². So ½mv_max² = ½kA². Plot K_max = ½mv_max² (vertical) against A² (horizontal): a straight line through the origin with slope ½k.
(b)
| A² (m²) | 0.0004 | 0.0016 | 0.0036 | 0.0064 | 0.0100 |
|---|---|---|---|---|---|
| K_max (J) | 0.0078 | 0.0325 | 0.0722 | 0.1275 | 0.2016 |
Slope of the best-fit line ≈ 20 J/m², so k = 2 × 20 = 40 N/m.
(c) The claim is refuted. From A = 0.040 m to 0.080 m, K_max rises from 0.0325 J to 0.1275 J, a factor of about 3.9, close to 4, not 2. The same is true from 0.020 m to 0.040 m. Energy is proportional to A², as E = ½kA² predicts.
| Point | What earns it |
|---|---|
| 1 | States ½mv_max² = ½kA² from energy conservation |
| 1 | Chooses axes that give a straight line (K_max against A², or v_max against A) |
| 1 | Processes the data correctly |
| 1 | k ≈ 40 N/m with working |
| 1 | Refutes the claim using a ratio from the data (about 4, not 2) |
Alternative method. Plotting v_max against A gives a straight line of slope √(k/m) ≈ 12.7 s⁻¹, so k = m × slope² ≈ 40 N/m. This earns full credit.
Question 7 (constructed response · stretch)
Two pendulums have the same length. Bob P has mass 0.10 kg and bob Q has mass 0.30 kg. Each is released from rest 0.045 m above its lowest point. A student says: “Q has three times as much energy, so Q will be moving faster at the bottom.”
(a) Calculate the total energy of each bob–Earth system. (b) Calculate the speed of each bob at the bottom of its swing. (c) Evaluate the student’s claim. (d) For bob P, describe energy bar charts (K and U) at the release point, at a height of 0.0225 m and at the bottom. (e) Over many swings, the real amplitude slowly decreases. Explain why, in terms of energy.
Model answer and suggested Marlbridge rubric
(a) E = mgh: P: 0.10 × 9.8 × 0.045 = 0.044 J. Q: 0.30 × 9.8 × 0.045 = 0.13 J.
(b) ½mv² = mgh, so v = √(2gh) = √(2 × 9.8 × 0.045) = 0.94 m/s for both bobs.
(c) The first part is correct: Q’s system has three times the energy. The conclusion is wrong. Q also has three times the mass, so it needs three times the kinetic energy for the same speed. The mass cancels, and both bobs reach 0.94 m/s.
(d) Release: K = 0 (a line on the zero mark), U = 0.044 J. At 0.0225 m (half the height): K = U = 0.022 J. Bottom: U = 0 (a line on the zero mark), K = 0.044 J. The total height is the same, 0.044 J, in all three.
(e) Air resistance and friction at the pivot do negative work on the system, turning some mechanical energy into thermal energy. With less total energy, the bob cannot rise as high, so the amplitude decreases.
| Point | What earns it |
|---|---|
| 1 | Both energies correct |
| 1 | Same speed, 0.94 m/s, with the mass cancelling |
| 1 | Claim judged wrong because greater energy is matched by greater mass |
| 1 | Bar charts: K = 0 and U = E at release; U = 0 and K = E at the bottom |
| 1 | Bar charts: equal K and U at half height, with constant total |
| 1 | Energy transferred out of the system as thermal energy, so amplitude falls |
How did you do?
- Q1 or Q3 wrong: revisit “Trading energy back and forth” and Figure 1 in the study guide.
- Q2 wrong: see “Changing the amplitude”: energy goes with A².
- Q4 or Q5 wrong: redo Worked example 1, using ½mv² + ½kx² = ½kA².
- Q6 incomplete: link the plot to the equation ½mv_max² = ½kA² (Worked example 3).
- Q7 incomplete: see Worked example 2 and “Energy bar charts”.
Then tick off the topic checklist.