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 where gravity appears. 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 0.50 kg ball rolls across a floor at 6.0 m/s. What is its kinetic energy?
- (A) 1.5 J
- (B) 3.0 J
- (C) 9.0 J
- (D) 18 J
Answer and explanation
Answer: (C). K = ½mv² = ½ × 0.50 kg × (6.0 m/s)² = ½ × 0.50 × 36 = 9.0 J.
- (A) uses ½mv without squaring the speed: ½ × 0.50 × 6.0 = 1.5. Its unit would be kg·m/s, which is not an energy unit.
- (B) uses mv = 3.0. That is momentum (Unit 4), not kinetic energy.
- (D) squares the speed but forgets the ½: 0.50 × 36 = 18.
Question 2 (multiple choice · core)
Cart A has mass m and speed v. Cart B has mass 3m and speed v/3. What is the ratio K_A / K_B of their kinetic energies?
- (A) 1/3
- (B) 1
- (C) 3
- (D) 9
Answer and explanation
Answer: (C). K_A = ½mv². K_B = ½(3m)(v/3)² = ½ × 3m × v²/9 = ⅓ × ½mv². So K_A / K_B = 3.
- (A) inverts the ratio: it gives K_B / K_A.
- (B) assumes the factor of 3 in mass cancels the factor of 3 in speed. That is true for mv (momentum), but speed is squared in K, so the speed change wins.
- (D) squares the speed factor but ignores the mass factor: (3)² = 9.
Question 3 (multiple choice · core)
A 50 kg skateboarder rolls east at 4.0 m/s relative to the ground. A cyclist rides beside her, also at 4.0 m/s east relative to the ground. Which statement about the skateboarder’s kinetic energy is correct?
- (A) It is 400 J for every observer, because kinetic energy is a scalar.
- (B) It is 400 J for an observer standing on the ground and 0 J for the cyclist, and both values are correct.
- (C) It is 400 J. The cyclist’s value of 0 J is wrong because the cyclist is moving.
- (D) It is 400 J for the ground observer and −400 J for the cyclist, because the ground moves west in the cyclist’s frame.
Answer and explanation
Answer: (B). In the ground frame, K = ½ × 50 × (4.0)² = 400 J. In the cyclist’s frame the skateboarder is at rest, so K = 0. Kinetic energy depends on the frame, and each observer is right in their own frame.
- (A) confuses “scalar” with “the same for everyone”. A scalar has no direction, but it can still depend on the frame.
- (C) treats the ground as the only valid frame. A frame moving at constant velocity is just as valid.
- (D) gives K a sign. Kinetic energy is never negative; besides, in the cyclist’s frame the skateboarder’s speed is zero.
Question 4 (calculation · core)
A 1500 kg car has 2.7 × 10⁵ J of kinetic energy.
(a) Calculate the car’s speed. (b) The driver brakes until the car’s kinetic energy is half its original value. Calculate the new speed. (c) A passenger says, “Half the kinetic energy means half the speed.” Use your answers to comment.
Worked solution
(a) v = √(2K / m) = √(2 × 2.7 × 10⁵ J ÷ 1500 kg) = √360 m²/s² = 19 m/s (18.97 m/s unrounded).
(b) v = √(2 × 1.35 × 10⁵ ÷ 1500) = √180 = 13 m/s (13.4 m/s unrounded).
(c) The claim is wrong. The new speed is 13.4 ÷ 18.97 = 0.71 of the old speed, which is 1/√2, not ½. Because K ∝ v², halving the speed would leave only one quarter of the kinetic energy (6.75 × 10⁴ J).
Suggested mark points (3): 1 for rearranging to v = √(2K / m) and getting 19 m/s; 1 for 13 m/s; 1 for explaining the factor 1/√2 using K ∝ v².
Question 5 (graph · core)
Take +y upward. A 0.20 kg ball is thrown straight up at 9.8 m/s and caught at the same height 2.0 s later. Ignore air resistance.
(a) Calculate the ball’s kinetic energy at t = 0, 0.50 s and 1.0 s. (b) Sketch a graph of kinetic energy against time from 0 to 2.0 s. Label the values at t = 0, 1.0 s and 2.0 s. (c) A student says: “At 0.50 s the ball is halfway through its rise, so it has half its starting kinetic energy.” Evaluate this claim.
Worked solution
(a) The velocity is v_y = 9.8 − 9.8t (in m/s).
| t (s) | v_y (m/s) | K = ½ × 0.20 × v_y² (J) |
|---|---|---|
| 0 | +9.8 | 9.6 |
| 0.50 | +4.9 | 2.4 |
| 1.0 | 0 | 0 |
(b) The sketch is a U-shaped curve. It starts at 9.6 J, falls to 0 at t = 1.0 s (the top, where the ball is momentarily at rest), then rises back to 9.6 J at 2.0 s. It is symmetric about t = 1.0 s. The curve is not two straight lines: it is steep at the start and flattens as it touches zero, because K depends on v², and v changes at a steady rate.
(c) The claim is incorrect. At 0.50 s the speed is half its starting value, so K = (½)² × 9.6 J = 2.4 J, one quarter of the starting kinetic energy, not half.
Suggested mark points (5): 1 for the three values in (a); 1 for K = 0 at t = 1.0 s in the sketch; 1 for symmetry about 1.0 s with 9.6 J at both ends; 1 for a curved shape that is flat (touching zero) at the minimum; 1 for (c) with the reasoning “speed halves, so K is multiplied by ¼”.
Question 6 (constructed response · stretch)
A spring launcher pushes carts of different mass along a level, low-friction track. A student claims the launcher always gives a cart the same kinetic energy, whatever its mass. A photogate measures each cart’s speed just after launch:
| m (kg) | 0.25 | 0.50 | 0.75 | 1.00 |
|---|---|---|---|---|
| v (m/s) | 2.01 | 1.41 | 1.16 | 0.99 |
(a) If the claim is true, what quantities should be plotted to give a straight line through the origin? Explain why, starting from K = ½mv². (b) Process the data and find the kinetic energy from the slope of your line. (c) State whether the data support the claim, with a reason. (d) Describe one source of uncertainty in this experiment and how to reduce it.
Worked solution and suggested Marlbridge rubric
(a) Rearranging K = ½mv² gives v² = 2K × (1/m). If K is the same for every cart, v² is proportional to 1/m. Plot v² (vertical, m²/s²) against 1/m (horizontal, kg⁻¹). The line should pass through the origin with slope 2K.
(b)
| 1/m (kg⁻¹) | 4.00 | 2.00 | 1.33 | 1.00 |
|---|---|---|---|---|
| v² (m²/s²) | 4.04 | 1.99 | 1.35 | 0.98 |
Slope from the two end points: (4.04 − 0.98) ÷ (4.00 − 1.00) = 1.02 J. A best-fit line through the origin gives a slope of about 1.0 J. So 2K ≈ 1.0 J and K ≈ 0.50 J. (Any answer from 0.49 J to 0.52 J with a correct method is fine.)
(c) The data support the claim. The points lie close to one straight line through the origin, so v² is proportional to 1/m. Checking each cart directly gives K = 0.51, 0.50, 0.50 and 0.49 J, all within about 2% of 0.50 J.
(d) Any one, with its fix: the photogate measures speed over a short distance, so a cart already slowed by friction gives a low value (place the gate right after the launcher); the launcher may not be compressed by exactly the same amount each time (lock it at a marked position and repeat each run three times, then average); friction varies along the track (clean the track or tilt it very slightly to balance friction).
| Point | What earns it |
|---|---|
| 1 | Rearranges to v² = 2K/m and names the axes v² against 1/m |
| 1 | Processes all four points correctly |
| 1 | Finds a slope near 1.0 J and K ≈ 0.50 J, with unit |
| 1 | Supports the claim because the points fit a straight line through the origin (or each K is about the same) |
| 1 | Names a specific uncertainty and a matching way to reduce it |
Alternative method. Calculating K for each cart and showing all four values agree earns the points for (b) and (c) if the spread is discussed. Plotting v against m gives a curve and earns no credit for (a).
Question 7 (constructed response · stretch)
Take +x to the right. Two identical 0.30 kg pucks slide towards each other on smooth ice. Puck A moves at +2.0 m/s and puck B moves at −2.0 m/s.
A student says: “The two velocities add to zero, so the total kinetic energy of the two-puck system is zero.”
(a) Calculate the total kinetic energy of the two-puck system in the ice frame. (b) Evaluate the student’s claim. (c) Calculate the total kinetic energy of the system in a frame moving with puck A. Explain why it differs from (a).
Model answer and suggested Marlbridge rubric
(a) Each puck has K = ½ × 0.30 × (2.0)² = 0.60 J. Total = 0.60 + 0.60 = 1.2 J.
(b) The claim is incorrect. Kinetic energy is a scalar that is never negative. It depends on v², so the direction of motion does not change its sign. The kinetic energies of the two pucks add; they cannot cancel. Only the velocities (vectors) add to zero.
(c) In puck A’s frame, A is at rest (K = 0) and B moves at −2.0 − (+2.0) = −4.0 m/s, so K_B = ½ × 0.30 × (4.0)² = 2.4 J. The system total is 2.4 J. It differs from 1.2 J because kinetic energy depends on the reference frame: the speeds are different in this frame, and K depends on the square of the speed.
| Point | What earns it |
|---|---|
| 1 | 1.2 J, from 0.60 J per puck |
| 1 | States that K is a scalar and never negative (v² removes the sign) |
| 1 | Concludes the claim is wrong because kinetic energies add while velocities can cancel |
| 1 | Finds B’s speed of 4.0 m/s in A’s frame and a total of 2.4 J |
| 1 | Explains the difference by frame dependence of speed and so of K |
How did you do?
- Q1 or Q4 wrong: re-read “Energy of motion” and “How K depends on mass and speed” in the study guide.
- Q2 wrong: work through the factor-of-change table and Worked example 1.
- Q3 or Q7(c) wrong: revisit “Kinetic energy depends on the observer” and Worked example 2.
- Q5 incomplete: study Figure 1 and the limiting case “Turning point”.
- Q6 incomplete: re-read the paragraph on linearising after Figure 1.
- Q7(b) incomplete: see “Kinetic energy is a scalar”.
Then tick off the topic checklist.