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. Positions are in m, velocities in m/s, accelerations in m/s² and t in s, so each coefficient carries the unit that makes its term correct. Use g = 9.8 m/s² and ignore air resistance. Round final answers to 2 significant figures unless told otherwise.
Question 1 (multiple choice · foundation)
Take +x east and +y north on a sheet of frictionless ice. A puck slides east at a steady 3.0 m/s. From t = 0, a constant push gives it an acceleration of 0.50 m/s² north. Which statement describes the motion after t = 0?
- (A) The east component of velocity stays 3.0 m/s, and the path curves northward as a parabola.
- (B) The puck turns until it moves due north, then carries on north.
- (C) The east component falls as the north component grows, so the speed stays 3.0 m/s.
- (D) The puck moves in a straight line pointing between east and north.
Answer and explanation
Answer: (A). a_x = 0, so v_x = 3.0 m/s for ever. Along y, v_y = 0.50t and y = 0.25t². With x = 3.0t, the path is y = 0.25(x/3.0)², a parabola. A change in the north motion does not change the east motion.
- (B) assumes the push replaces the original motion. Nothing acts along x, so v_x never falls to zero.
- (C) assumes the two components share a fixed speed. They are independent: v_x stays 3.0 m/s while v_y grows, so the speed rises.
- (D) would need the velocity and the acceleration to point the same way. Here the puck starts moving east and the acceleration is north, so v_y grows while v_x stays fixed and the direction keeps turning toward north.
Question 2 (multiple choice · core)
Take +x east and +y north. A particle’s position is r(t) = (1.0t³) i + (8.0t − 2.0t²) j. What is its speed at t = 1.0 s?
- (A) 5.0 m/s
- (B) 6.1 m/s
- (C) 7.0 m/s
- (D) 7.2 m/s
Answer and explanation
Answer: (A). v = dr/dt = (3.0t²) i + (8.0 − 4.0t) j. At t = 1.0 s, v = 3.0 i + 4.0 j m/s, so the speed is √(3.0² + 4.0²) = 5.0 m/s.
- (B) is the size of the position vector, |r(1.0)| = √(1.0² + 6.0²) ≈ 6.1. Position is not velocity.
- (C) adds the components as numbers, 3.0 + 4.0. Perpendicular components combine by Pythagoras.
- (D) differentiates twice: a = 6.0 i − 4.0 j has size ≈ 7.2 m/s², an acceleration, not a speed.
Question 3 (multiple choice · core)
Take +x horizontal and +y up. A ball rolls off a horizontal table top at speed v₀ and lands a horizontal distance D from the table’s edge. A second ball rolls off a table four times as high at speed 2v₀. How far from its table’s edge does the second ball land?
- (A) 2D
- (B) 4D
- (C) 8D
- (D) 16D
Answer and explanation
Answer: (B). The fall starts with v_y0 = 0, so h = ½gt² and t = √(2h/g). Four times the height gives √4 = 2 times the fall time. The horizontal distance is v₀t, so it changes by 2 × 2 = 4: the ball lands at 4D.
- (A) doubles the speed but ignores the longer fall. The fall time depends on the height.
- (C) assumes the fall time is proportional to the height (4 times). Because h ∝ t², the time grows only as √h.
- (D) squares both factors (2² × 4). Distance is proportional to v₀, not v₀², and to √h, not h.
Question 4 (calculation · core)
Take +x horizontal and +y up, origin at the launch point. A small drone moves in a vertical plane with velocity v(t) = (0.75t²) i + (4.0 − 2.0t) j. It is at the origin at t = 0.
(a) Find a(t), and the size and direction of the acceleration at t = 2.0 s. (b) Find the time at which the drone is moving horizontally, and its speed and position then. (c) Find its position at t = 4.0 s.
Worked solution
- (a) a = dv/dt = (1.5t) i − 2.0 j. At t = 2.0 s, a = 3.0 i − 2.0 j m/s². Size √(3.0² + 2.0²) = 3.6 m/s², at tan⁻¹(2.0 ÷ 3.0) = 34° below the +x direction.
- (b) Moving horizontally means v_y = 0: 4.0 − 2.0t = 0, so t = 2.0 s. Then v = 0.75 × 4.0 = 3.0 m/s (all horizontal).
- Integrate with r(0) = 0: x = 0.25t³ and y = 4.0t − 1.0t². At 2.0 s: r = 2.0 i + 4.0 j m.
- (c) At 4.0 s: x = 0.25 × 64 = 16 m; y = 16 − 16 = 0. So r = 16 i m: back at launch height, 16 m away.
Suggested mark points (4): 1 for a(t) and its size and direction at 2.0 s; 1 for t = 2.0 s from v_y = 0 with the speed; 1 for integrating each component with the initial condition; 1 for r at 4.0 s.
Common error: x = ½a_x t² with a_x = 3.0 m/s² (its value at 2.0 s) gives 6.0 m, not 2.0 m. a_x changes, so integrate.
Question 5 (constructed response · core)
Take +x horizontal toward a wall and +y up, origin at the launch point on level ground. A ball is kicked at 18.0 m/s, 50.0° above the horizontal, toward a vertical wall 20.0 m away.
(a) Find how high above the ground the ball strikes the wall. (b) Is the ball rising or falling when it strikes the wall? Justify your answer with a calculation. (c) Find the ball’s speed as it strikes the wall.
Worked solution and suggested Marlbridge rubric
Components. v_x0 = 18.0 cos 50.0° = 11.57 m/s; v_y0 = 18.0 sin 50.0° = 13.79 m/s. a_x = 0, a_y = −9.8 m/s².
(a) From the x-motion: t = 20.0 ÷ 11.57 = 1.73 s. Use this same t in the y-motion: y = 13.79 × 1.73 − 4.9 × 1.73² = 9.2 m.
(b) v_y = 13.79 − 9.8 × 1.73 = −3.2 m/s. The negative sign (with +y up) means the ball is falling. Equivalently, it peaked at t = 13.79 ÷ 9.8 = 1.41 s, before reaching the wall.
(c) v_x is still 11.57 m/s. Speed = √(11.57² + 3.15²) = 12 m/s.
| Point | What earns it |
|---|---|
| 1 | Resolves the launch velocity into 11.6 m/s and 13.8 m/s components |
| 1 | Finds t from the x-motion and uses the same t in the y-motion |
| 1 | Height 9.2 m |
| 1 | (b) Falling, supported by a negative v_y or by t_top < t_wall |
| 1 | (c) Speed 12 m/s from both components (carry forward an incorrect v_y) |
Alternative for (a). Putting x = 20.0 m into the trajectory equation (Question 6) gives the same height and earns the second and third points.
Question 6 (constructed response · core)
Take +x horizontal and +y up, origin at the launch point on level ground. A projectile is launched with speed v₀ at angle θ above the horizontal. Its acceleration is a = −g j.
(a) Starting from the acceleration, derive x(t) and y(t), stating the initial conditions you use. (b) Show that the path is a parabola, y = x tan θ − gx² / (2v₀² cos²θ). (c) Derive the range R on level ground in terms of v₀, θ and g. (d) A student launches a second projectile at the same angle but with a speed 30% greater. By what factor does the range change? By what factor does the flight time change?
Worked solution and suggested Marlbridge rubric
(a) x: v_x = v₀ cos θ + ∫₀ᵗ 0 dt = v₀ cos θ, so x = (v₀ cos θ)t, using x(0) = 0 and v_x(0) = v₀ cos θ. y: v_y = v₀ sin θ + ∫₀ᵗ (−g) dt = v₀ sin θ − gt, so y = (v₀ sin θ)t − ½gt², using y(0) = 0 and v_y(0) = v₀ sin θ.
(b) From the x equation, t = x / (v₀ cos θ). Substitute into y: y = x tan θ − gx² / (2v₀² cos²θ). This is y = bx − cx² with constants b and c, which is a parabola.
(c) Land when y = 0 (and t ≠ 0): t = 2v₀ sin θ / g. Then R = v₀ cos θ × 2v₀ sin θ / g = v₀² sin 2θ / g.
(d) R ∝ v₀², so the range changes by 1.30² = 1.69 (about 1.7). The flight time is proportional to v₀, so it changes by 1.3.
| Point | What earns it |
|---|---|
| 1 | (a) Integrates both components with the stated initial conditions |
| 1 | (b) Eliminates t correctly to reach the trajectory equation |
| 1 | (b) Identifies the y = bx − cx² form as a parabola |
| 1 | (c) Flight time from y = 0 and range v₀² sin 2θ / g (2v₀² sin θ cos θ / g is equally acceptable) |
| 1 | (d) Range factor 1.69 and time factor 1.3, each linked to the power of v₀ |
Alternative for (c). Solving the trajectory equation for the non-zero x with y = 0 also earns the point.
Question 7 (data analysis and design · stretch)
Take +x horizontal and +y up, origin at the launch point. A student films a ball launched upward at an angle and measures its position from the video every 0.10 s, starting 0.10 s after launch:
| t (s) | 0.10 | 0.20 | 0.30 | 0.40 | 0.50 | 0.60 |
|---|---|---|---|---|---|---|
| x (m) | 0.322 | 0.639 | 0.963 | 1.278 | 1.601 | 1.918 |
| y (m) | 0.404 | 0.702 | 0.913 | 1.013 | 1.027 | 0.932 |
(a) Explain how the x data show that a_x = 0, and find v_x. (b) Assuming a_y = −g, show that y/t = v_y0 − (g/2)t. (c) Calculate y/t for each reading. Use a graph of y/t against t to find g and v_y0. (d) Describe an experiment, using the same video method, to test the claim that a ball’s vertical motion does not depend on its horizontal speed.
Worked solution and suggested Marlbridge rubric
(a) Each 0.10 s step increases x by about 0.32 m, so x rises linearly with t. A straight x–t graph has constant slope, so v_x is constant and a_x = 0. The slope (best-fit line, or x/t for each reading) gives v_x = 3.2 m/s.
(b) Integrating a_y = −g twice with y(0) = 0 gives y = v_y0 t − ½gt². Divide by t: y/t = v_y0 − (g/2)t, a straight line with slope −g/2 and intercept v_y0.
(c) y/t values: 4.04, 3.51, 3.04, 2.53, 2.05, 1.55 m/s. These fall on a straight line. The best-fit slope is about −4.95 m/s², so g ≈ 9.9 m/s² (accept 9.6 to 10.2 m/s²). The intercept is v_y0 ≈ 4.5 m/s. The value of g is within about 1% of 9.8 m/s², which supports a_y = −g.
(d) Example procedure. Roll the ball off a ramp whose end is horizontal and at a fixed height. Release it from different points on the ramp to change the launch speed. Film each run against a metre scale, camera square to the plane of motion. Find v_x from the x–t slope and the fall time from the y data. If the fall time is the same for every launch speed while the landing distance grows in proportion to v_x, the evidence supports independence. Repeat each speed and average.
| Point | What earns it |
|---|---|
| 1 | (a) Links equal x-steps (linear x–t) to constant v_x, and gives v_x ≈ 3.2 m/s |
| 1 | (b) Derives y/t = v_y0 − (g/2)t from integration or the constant-acceleration equation |
| 1 | (c) Correct y/t values and g from −2 × slope, in the accepted range |
| 1 | (c) v_y0 ≈ 4.5 m/s from the intercept |
| 1 | (d) Varies horizontal speed at fixed height, keeps other factors the same, and states what result would support the claim |
Alternative for (c). The first and last y/t values give a slope of (1.55 − 4.04) ÷ 0.50 = −4.98 m/s², so g ≈ 10.0 m/s². This earns the point, but a best-fit line uses all the data.
How did you do?
- Q1 or Q3 wrong: re-read “One motion, two one-dimensional problems” and the patterns under “Projectile motion: a special case” in the study guide.
- Q2 or Q4 wrong: revisit “Position, velocity and acceleration as vectors” and Worked example 2. Differentiate or integrate one component at a time.
- Q5 or Q6 incomplete: work through Worked example 1 and the projectile derivation. Use the same t in both components.
- Q7 incomplete: see “Investigating motion in a plane”. Say what you vary, what you keep the same and what result would support the claim.
Then tick off the topic checklist.