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 and ignore air resistance unless a question says otherwise. Round final answers to 2 significant figures unless told otherwise, and keep unrounded values in your calculator until the end. Any scientific calculator is fine.
Question 1 (multiple choice · foundation)
A coconut falls from a tree. A student chooses the coconut alone as the system. Which statement is correct while it falls?
- (A) The system’s gravitational potential energy is converted into kinetic energy, so its total energy stays constant.
- (B) Earth’s gravitational force does positive work on the system, so the system’s kinetic energy and total energy increase.
- (C) The system’s mechanical energy stays constant because gravity is a conservative force.
- (D) The system has no energy until the coconut hits the ground.
Answer and explanation
Answer: (B). A single-object system can only have kinetic energy. Earth is outside the system, so its pull is an outside force. It points the same way as the motion, so it does positive work, and the system’s energy (all kinetic) increases by that amount.
- (A) describes the coconut + Earth system. The coconut alone has no potential energy.
- (C) Gravity being conservative lets the coconut + Earth system store potential energy. For the coconut alone, gravity is still an outside force doing work.
- (D) The coconut has kinetic energy the moment it starts moving.
Question 2 (multiple choice · core)
A 1.0 kg block and a 3.0 kg block are released from rest at the same height, 0.80 m above the floor, on two different smooth ramps. Ramp X is steep and short; ramp Y is gentle and long. How do their speeds at the bottom compare?
- (A) Both reach about 4.0 m/s.
- (B) The 3.0 kg block is faster, because it has more potential energy at the start.
- (C) The 1.0 kg block is faster, because it has less inertia.
- (D) The block on the steeper ramp is faster, whatever its mass.
Answer and explanation
Answer: (A). For the block + Earth system with no friction, mgh = ½mv², so v = √(2gh) = √(2 × 9.8 × 0.80) = 4.0 m/s for both. Mass cancels, and the shape of a smooth ramp does not matter: only the drop in height does.
- (B) The heavier block does start with three times the potential energy, but it also needs three times the kinetic energy for the same speed.
- (C) The smaller inertia is matched by a smaller gravitational force; the energy equation shows mass cancels.
- (D) A steeper ramp gets the block to the bottom sooner, but not faster. Time and speed are different questions.
Question 3 (multiple choice · core)
A spring launcher fires a ball straight up to a maximum height h above its starting position. The spring is now compressed twice as far and the same ball is launched. Ignore air resistance. The new maximum height above the starting position is
- (A) 2h
- (B) 4h
- (C) √2 h
- (D) 8h
Answer and explanation
Answer: (B). For the ball + spring + Earth system, ½kΔx² = mgh, so h ∝ Δx². Doubling Δx multiplies the stored energy, and so the height, by 2² = 4.
- (A) assumes the stored energy is proportional to Δx, as the spring force is.
- (C) applies a square root that belongs to speed, not height: the launch speed doubles (v ∝ Δx), it does not rise by √2.
- (D) cubes the factor; no quantity here goes as Δx³.
Question 4 (multiple choice · core)
Three identical balls are thrown from the top of a tower with the same speed: P straight up, Q horizontally, R at 30° below the horizontal. Ignore air resistance. Which correctly compares their speeds just before they hit the ground?
- (A) P is fastest, because it rises higher before falling.
- (B) R is fastest, because it is aimed downward.
- (C) Q is slowest, because none of its launch speed is vertical.
- (D) All three have the same speed.
Answer and explanation
Answer: (D). For each ball + Earth system, the starting K and U_g are the same (same mass, speed and height), and each ends at ground level. Kinetic energy is a scalar, so the direction of the throw does not matter: each lands with the same K and therefore the same speed.
- (A) P does rise higher, but it falls back past its starting point with its launch speed again, then falls the same tower height as the others.
- (B) R reaches the ground soonest, but speed depends on energy, not on aim.
- (C) treats speed as only its vertical part. Q’s horizontal velocity still counts in ½mv².
Question 5 (calculation · core)
A 350 kg roller-coaster car passes point A, 22 m above the ground, at 3.0 m/s. Treat the track as frictionless and use the system car + Earth.
(a) Calculate the car’s speed at point B, 8.0 m above the ground. (b) Calculate its kinetic energy at B. (c) How far above A could the car rise, at most, further along the track? (d) A second car of mass 700 kg passes A at the same speed. State its speed at B and explain.
Worked solution
(a) ½mv_A² + mgy_A = ½mv_B² + mgy_B, so v_B = √(v_A² + 2g(y_A − y_B)) = √(3.0² + 2 × 9.8 × 14) = √283.4 = 17 m/s (16.8 m/s).
(b) K_B = ½ × 350 × 283.4 = 5.0 × 10⁴ J.
(c) At the highest possible point K = 0, so all of A’s kinetic energy becomes extra U_g: ½mv_A² = mgΔy, Δy = v_A² / (2g) = 9.0 ÷ 19.6 = 0.46 m above A.
(d) 17 m/s, the same. Mass cancels from every term of the energy equation, so the speed at B does not depend on the car’s mass.
Suggested mark points (5): 1 for an energy equation with both K and U_g at A and B; 1 for 17 m/s; 1 for (b); 1 for (c); 1 for (d) with the mass-cancels reason.
Common error: leaving out the 3.0 m/s at A gives √(2 × 9.8 × 14) = 16.6 m/s. That treats the car as starting from rest. At 2 significant figures it still rounds to 17 m/s, so the error hides in (a), but it shows in (b): 4.8 × 10⁴ J instead of 5.0 × 10⁴ J.
Question 6 (constructed response · stretch)
A 0.80 kg block on a horizontal table is pushed against a spring (k = 500 N/m), compressing it 0.10 m. When released, the block leaves the spring at the spring’s relaxed length with a measured speed of 2.2 m/s. The table is rough.
(a) Choose a system for which the total energy stays constant during the launch. Justify your choice. (b) Draw energy bar charts for the moment of release and the moment the block leaves the spring, labelled with K, U_s and ΔE_thermal. (c) Calculate how much mechanical energy is dissipated. (d) Predict the speed at which the block would leave the spring on a frictionless table.
Worked solution and suggested Marlbridge rubric
(a) Block + spring + table. The friction between block and table is then an interaction inside the system, so the thermal energy it produces stays inside. The wall holding the spring does not move, so it does no work, and the table is horizontal, so U_g does not change. With no outside work, the total energy is constant. (Earth may be included too; it makes no difference here.)
(b) Release: U_s = 2.5 J; K = 0; ΔE_thermal = 0. Leaving the spring: U_s = 0; K = 1.9 J; ΔE_thermal = 0.56 J. The two bars at the second moment add to the same 2.5 J as the first.
(c) U_s = ½ × 500 × (0.10)² = 2.5 J. K = ½ × 0.80 × 2.2² = 1.9 J (1.936 J). Dissipated = 2.5 − 1.936 = 0.56 J, about 23% of the stored energy.
(d) All 2.5 J would become K: v = √(2 × 2.5 ÷ 0.80) = 2.5 m/s.
| Point | What earns it |
|---|---|
| 1 | A system that includes the rough surface, with the reason that friction is then internal and no outside force does work |
| 1 | Bar charts with equal totals at both moments and a thermal-energy bar at the second |
| 1 | U_s = 2.5 J and K = 1.9 J |
| 1 | 0.56 J dissipated, described as thermal energy (and sound) |
| 1 | 2.5 m/s for the frictionless case |
An answer choosing block only can still earn the calculation points, but (a) needs the reason that the system’s energy is constant; for the block alone it is not, because the spring and friction do work on it from outside.
Question 7 (constructed response · stretch)
A student releases a cart from rest at different heights h on a ramp and measures its speed v at the bottom with a light gate. The cart has light, low-friction wheels.
| h (m) | 0.10 | 0.20 | 0.30 | 0.40 | 0.50 |
|---|---|---|---|---|---|
| v (m/s) | 1.33 | 1.88 | 2.30 | 2.65 | 2.97 |
The student claims: “The mechanical energy of the cart–Earth system is conserved.”
(a) Use conservation of energy to show which quantity should be plotted against h to give a straight line, and predict its slope if the claim is true. (b) Process the data and find the slope. (c) Evaluate the claim using your results. (d) Suggest where any missing energy has gone.
Worked solution and suggested Marlbridge rubric
(a) If mechanical energy is conserved, mgh = ½mv², so v² = 2gh. A graph of v² against h should be a straight line through the origin with slope 2g = 19.6 m/s².
(b)
| h (m) | 0.10 | 0.20 | 0.30 | 0.40 | 0.50 |
|---|---|---|---|---|---|
| v² (m²/s²) | 1.77 | 3.53 | 5.29 | 7.02 | 8.82 |
The points lie on a straight line through (almost exactly) the origin. Slope = (8.82 − 1.77) ÷ (0.50 − 0.10) = 18 m/s² (17.6 m/s²; a best-fit line gives the same).
(c) The data do not fully support the claim. The line is straight and passes through the origin, so v² is proportional to h, as energy conservation predicts. But the slope is about 17.6 m/s², only about 90% of 19.6 m/s². So the cart keeps only about 90% of the mechanical energy at every height; about 10% is dissipated. The claim is a reasonable approximation, not exactly true.
(d) Friction in the wheel bearings and between the wheels and the ramp, and air resistance, turn some mechanical energy into thermal energy and sound. (The light wheels make the energy in their spinning small; energy of rotation comes later in the course.)
| Point | What earns it |
|---|---|
| 1 | Derives v² = 2gh from mgh = ½mv² |
| 1 | Predicts slope 2g = 19.6 m/s² |
| 1 | Correct v² values and a slope of about 18 m/s², with unit |
| 1 | Compares slope with 2g and concludes only about 90% is kept |
| 1 | Names a non-conservative interaction and thermal energy or sound |
Alternative method. Calculating ½v² / (gh) for each row gives about 0.90 every time; this earns the comparison point if the constant ratio is linked to a constant fraction of energy kept.
How did you do?
- Q1 or Q6(a) wrong: re-read “Choosing the system” in the study guide.
- Q2, Q4 or Q5 wrong: go back to Worked example 1 and “Comparing scenarios with energy”. Mass and direction cancel.
- Q3 wrong: revisit Worked example 2 and Figure 1. Stored spring energy goes as Δx².
- Q6 incomplete: work through Worked example 3 on friction and thermal energy.
- Q7 incomplete: your reasoning needs the why: compare the slope with 2g, then name where the energy went.
Then tick off the topic checklist.