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 set is for the calculus-based Physics C: Mechanics course. All data are invented for practice. Use g = 9.8 m/s² and ignore air resistance unless told otherwise; answers with g = 10 m/s² are equally acceptable. Round final answers to 2 significant figures unless told otherwise.
Question 1 (multiple choice · foundation)
Take +x to the right. A motion diagram for a cart shows its position every 0.10 s: 0, 9.0, 16.0, 21.0 and 24.0 cm. Which statement describes the cart’s motion?
- (A) v_x is positive and decreasing; a_x is constant at about −2.0 m/s².
- (B) v_x is positive and decreasing; a_x is negative and growing in size, because the gaps keep shrinking.
- (C) v_x is negative, because the cart is slowing down; a_x is constant at about −2.0 m/s².
- (D) v_x is positive and decreasing; a_x is constant at about −0.20 m/s².
Answer and explanation
Answer: (A). The gaps are 9.0, 7.0, 5.0 and 3.0 cm. The cart moves right (v_x > 0) and slows down. Each gap is 2.0 cm shorter than the one before, so the acceleration is constant: a_x = (change in gap) ÷ (Δt)² = −0.020 m ÷ (0.10 s)² = −2.0 m/s².
- (B) The gaps shrink by the same amount each step, which means constant acceleration.
- (C) confuses slowing down with moving left. The positions increase, so v_x is positive.
- (D) divides the change in gap by Δt only once. Acceleration needs (Δt)², so the unit would be wrong too.
Question 2 (multiple choice · core)
Take +x to the right, with the object at x₀ = 0 at t = 0. Its v_x–t graph is a straight line from +4.0 m/s at t = 0 to −4.0 m/s at t = 4.0 s. Which description of its x–t graph is correct?
- (A) A straight line with a negative slope, starting at x = 0.
- (B) A curve bending downward that rises to a maximum of 4.0 m at t = 2.0 s and returns to x = 0 at t = 4.0 s.
- (C) A curve bending upward that falls to a minimum at t = 2.0 s and returns to x = 0 at t = 4.0 s.
- (D) A curve bending downward that keeps rising, reaching its maximum at t = 4.0 s.
Answer and explanation
Answer: (B). The slope of the v_x–t graph is a_x = −2.0 m/s², so the x–t graph bends downward. Integrating, x = 4.0t − 1.0t². Velocity is zero at t = 2.0 s, where x = 8.0 − 4.0 = 4.0 m (the maximum). At t = 4.0 s, x = 16 − 16 = 0. The areas under v_x–t cancel.
- (A) copies the shape of the velocity graph. The slope of x–t, not its shape, gives v_x.
- (C) has the wrong curvature. A negative a_x makes x–t bend downward.
- (D) forgets that after 2.0 s the velocity is negative, so the object moves back towards the origin.
Question 3 (multiple choice · core)
Take +y upward. A drone hovering at rest releases a sensor package, which falls a height h in time T. Ignore air resistance. From what height must the package be released so that its fall lasts 3T?
- (A) √3 h
- (B) 3h
- (C) 9h
- (D) 27h
Answer and explanation
Answer: (C). Starting from rest, the drop is h = ½gT², so h ∝ T². Multiplying the time by 3 multiplies the height by 3² = 9.
- (A) inverts the relationship. It would be right for “how does the time change if the height triples?” (t ∝ √h).
- (B) assumes height ∝ time, which needs constant velocity; the package speeds up.
- (D) uses a cube. A t³ dependence needs an acceleration that grows with time.
Question 4 (calculation · core)
Take +x to the right. A particle is at x₀ = 0 with v_x0 = −3.0 m/s at t = 0. Its acceleration is:
| Time interval (s) | 0 to 3.0 | 3.0 to 5.0 | 5.0 to 7.0 |
|---|---|---|---|
| a_x (m/s²) | +2.0 | 0 | −1.0 |
Find (a) v_x at t = 3.0 s, 5.0 s and 7.0 s, (b) the time when the particle is momentarily at rest, (c) its displacement from 0 to 7.0 s, (d) the distance it travels in that time.
Worked solution
- (a) Use areas under the a_x–t graph: +2.0 × 3.0 = +6.0 m/s, then 0, then −1.0 × 2.0 = −2.0 m/s. So v_x(3.0) = −3.0 + 6.0 = +3.0 m/s, v_x(5.0) = +3.0 m/s, v_x(7.0) = 3.0 − 2.0 = +1.0 m/s.
- (b) In the first stage v_x = −3.0 + 2.0t = 0 at t = 1.5 s. After that v_x stays positive.
- (c) Areas under the v_x–t graph: 0 to 1.5 s is a triangle below the axis, −½ × 1.5 × 3.0 = −2.25 m; 1.5 to 3.0 s is a triangle above, +2.25 m; 3.0 to 5.0 s is a rectangle, +6.0 m; 5.0 to 7.0 s is a trapezium, ½ × (3.0 + 1.0) × 2.0 = +4.0 m. Displacement = −2.25 + 2.25 + 6.0 + 4.0 = +10 m.
- (d) Distance = 2.25 + 2.25 + 6.0 + 4.0 = 14.5 m (15 m to 2 s.f.).
Suggested mark points (4): 1 for the three velocities, including v_x0; 1 for t = 1.5 s; 1 for the displacement from signed areas; 1 for splitting at t = 1.5 s and adding sizes for the distance.
Common error: +4.0 m/s at 7.0 s, from leaving out v_x0.
Question 5 (constructed response · core)
Take +y upward, origin at a juggler’s hand. A ball leaves the hand moving straight up at 3.0 m/s and is caught at the same height.
(a) Sketch the y–t, v_y–t and a_y–t graphs from release to catch, on the same time axis. Label the time and the values at the top of the path and at the catch. (b) A student says: “At the top, the acceleration is zero because the ball stops.” Use your v_y–t graph to explain why this is wrong. (c) Use an area on your v_y–t graph to find the greatest height of the ball above the hand. (d) Derive an expression for the time in the air in terms of v₀ and g. Use it to state the factor by which the time in the air and the greatest height change if the launch speed is halved.
Worked solution and suggested Marlbridge rubric
(a) Top at t = v₀/g = 3.0/9.8 = 0.31 s; catch at 0.61 s.
- y–t: a curve bending downward from 0, peaking at 0.46 m at 0.31 s, back to 0 at 0.61 s, symmetric about the top.
- v_y–t: a straight line from +3.0 m/s, crossing zero at 0.31 s, reaching −3.0 m/s at 0.61 s.
- a_y–t: a flat line at −9.8 m/s² for the whole flight.
(b) The v_y–t graph is one straight line with slope −9.8 m/s² everywhere, including where it crosses zero. The velocity is still changing at the top, so the acceleration is not zero.
(c) Area of the triangle from 0 to 0.31 s: ½ × 0.306 × 3.0 = 0.46 m. (Check: v₀²/(2g) = 9.0/19.6 = 0.46 m.)
(d) At the catch, y = 0: 0 = v₀T − ½gT², so T(v₀ − ½gT) = 0. Reject T = 0 (the release), leaving T = 2v₀/g. T ∝ v₀, so halving v₀ halves the time in the air (0.31 s). The height is v₀²/(2g) ∝ v₀², so it falls to one quarter (0.11 m).
| Point | What earns it |
|---|---|
| 1 | (a) y–t curving downward, symmetric, peak at 0.31 s, back to 0 at 0.61 s |
| 1 | (a) v_y–t straight line from +3.0 to −3.0 m/s, crossing zero at the time of the top; a_y–t flat at −9.8 m/s² |
| 1 | (b) Uses the constant slope of v_y–t (or the changing velocity) to show a_y = −g at the top |
| 1 | (c) 0.46 m from the triangle area under v_y–t |
| 1 | (d) Derives T = 2v₀/g, rejecting T = 0 |
| 1 | (d) Time halves and height becomes one quarter, from T ∝ v₀ and h ∝ v₀² |
With g = 10 m/s², 0.30 s, 0.60 s and 0.45 m earn full credit.
Question 6 (constructed response · stretch)
Take +x up a ramp, origin at the cart’s position at t = 0. A student gives a cart a push up a straight ramp and lets it roll. A motion sensor records:
| t (s) | 0 | 0.40 | 0.80 | 1.20 | 1.60 | 2.00 |
|---|---|---|---|---|---|---|
| v_x (m/s) | 1.20 | 0.90 | 0.60 | 0.30 | 0.00 | −0.30 |
(a) The student claims the acceleration is constant throughout, both up and back down the ramp. Use the data to support or reject this claim. (b) Find a_x, with its unit. (c) Find how far up the ramp the cart travels before it stops. Use one method and check it with a second. (d) Find the cart’s position at t = 2.00 s. (e) Another student says the acceleration is zero at t = 1.60 s, because the cart is at rest. Use the data to respond.
Worked solution and suggested Marlbridge rubric
(a) In every 0.40 s interval, v_x falls by 0.30 m/s, including where the cart turns round. The v_x–t graph is a straight line with constant slope, so the data support the claim.
(b) Slope = (−0.30 − 1.20) ÷ (2.00 − 0) = −0.75 m/s² (down the ramp).
(c) Area under v_x–t from 0 to 1.60 s: ½ × 1.60 × 1.20 = 0.96 m. Check with the equation with no t: 0 = 1.20² + 2(−0.75)Δx gives Δx = 1.44 ÷ 1.5 = 0.96 m.
(d) x = 0 + 1.20 × 2.00 + ½(−0.75)(2.00)² = 2.40 − 1.50 = 0.90 m. (Areas: 0.96 m up, then ½ × 0.40 × 0.30 = 0.06 m back.)
(e) The velocity goes from +0.30 m/s at 1.20 s to −0.30 m/s at 2.00 s, changing at the same rate through 1.60 s. So a_x = −0.75 m/s² at 1.60 s too.
| Point | What earns it |
|---|---|
| 1 | (a) Equal velocity changes in equal times (or a straight v_x–t graph), so the claim is supported |
| 1 | (b) −0.75 m/s² with sign and unit |
| 1 | (c) 0.96 m by one method and confirmed by a second |
| 1 | (d) 0.90 m (carry forward an incorrect a_x) |
| 1 | (e) Uses the data around 1.60 s to show v_x is still changing, so a_x ≠ 0 |
Question 7 (explanation · stretch)
Take +x along a straight test track. A test vehicle starts from rest and speeds up at constant acceleration a for time T. It then brakes with constant acceleration of size 2a until it stops.
A student says: “Braking takes half as long. Distance goes as time squared, so the braking distance is a quarter of the speed-up distance.”
(a) Derive expressions, in terms of a and T, for the braking time and for each distance. (b) Evaluate the student’s claim and explain the result using a v_x–t graph or average velocities. (c) If the same vehicle speeds up for 2T with the same a, then brakes as before, by what factor does the total distance change?
Model answer and suggested Marlbridge rubric
(a) Top speed: v = aT. Braking time: 0 = aT − 2a t_b, so t_b = T/2. Speed-up distance: d₁ = ½aT². Braking distance: d₂ = (aT)² ÷ (2 × 2a) = aT²/4.
(b) d₂/d₁ = (aT²/4) ÷ (aT²/2) = ½, not ¼, so the claim is false. On a v_x–t graph both stages are triangles with the same height aT. Their areas are in the ratio of their bases, T/2 to T. Equivalently, both stages have the same average velocity, aT/2, so distance ∝ time. “Distance ∝ t²” holds only for one acceleration from rest.
(c) Total distance = ½aT² + aT²/4 = ¾aT², so it is ∝ T². Doubling T multiplies the total distance by 4.
| Point | What earns it |
|---|---|
| 1 | (a) t_b = T/2 from the velocity equation |
| 1 | (a) d₁ = ½aT² and d₂ = aT²/4, each with working |
| 1 | (b) Ratio ½, so the claim fails |
| 1 | (b) Explains with equal heights of the v_x–t triangles or equal average velocities |
| 1 | (c) Factor of 4, from total distance ∝ T² |
A numerical test with chosen a and T can earn only the (b) ratio point.
How did you do?
- Q1 or Q2 wrong: re-read “Motion diagrams” and “The graph links” in the study guide.
- Q3 or Q7(c) wrong: revisit “Functional dependence and factors of change”.
- Q4 wrong: work through Worked example 1 and Figure 2.
- Q5 or Q6 incomplete: review “Free fall near Earth” and Worked example 2. Point to a slope, an area or a data pattern.
Then tick off the topic checklist.