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. Any scientific calculator is fine.
Question 1 (multiple choice · foundation)
Which list contains only vector quantities?
- (A) distance, displacement, velocity
- (B) position, velocity, acceleration
- (C) speed, velocity, acceleration
- (D) displacement, speed, time
Answer and explanation
Answer: (B). Position, velocity and acceleration all need a direction as well as a size, so all three are vectors.
- (A) includes distance, which is the total path length. It has no direction, so it is a scalar.
- (C) includes speed. Speed is the magnitude of velocity, with no direction, so it is a scalar.
- (D) includes two scalars, speed and time. Only displacement is a vector.
Question 2 (multiple choice · foundation)
Take +y upward. A lift (elevator) moves 15 m up, then 24 m down, then 6.0 m up. What is the lift’s total displacement?
- (A) +45 m
- (B) −3.0 m
- (C) +3.0 m
- (D) −9.0 m
Answer and explanation
Answer: (B). Give each move a sign from the axis and add: (+15 m) + (−24 m) + (+6.0 m) = −3.0 m. The lift ends 3.0 m below where it started.
- (A) adds the sizes of the moves. That is the total distance, 45 m, a scalar. It also gives it a direction it does not have.
- (C) has the right size but the wrong sign. It treats the downward move as positive and the upward moves as negative, which reverses the stated axis.
- (D) stops after the first two moves (15 − 24 = −9.0 m) and forgets the last 6.0 m up.
Question 3 (multiple choice · core)
Take +x to the right. Four carts on a straight track have these velocity components at one instant:
| Cart | P | Q | R | S |
|---|---|---|---|---|
| v_x (m/s) | +3.0 | −5.0 | +4.0 | −2.0 |
Which statement is correct?
- (A) Cart R is the fastest, because +4.0 m/s is the largest value.
- (B) Cart Q is the fastest, and it is moving to the left.
- (C) Cart Q is the slowest, because −5.0 m/s is the smallest value.
- (D) Cart S is slowing down, because its velocity is negative.
Answer and explanation
Answer: (B). Speed is the magnitude of velocity: |−5.0| = 5.0 m/s is the largest of 3.0, 5.0, 4.0 and 2.0 m/s. The negative sign tells you Q moves opposite to +x, which is to the left.
- (A) ranks the signed numbers as if they were sizes. The sign is a direction, not part of the size.
- (C) makes the same error the other way: “most negative” does not mean “slowest”.
- (D) confuses direction with speeding up or slowing down. A negative velocity only means “moving left”. To know whether S is slowing down you would need its acceleration as well (Topic 1.2).
Question 4 (representation · core)
Take +x to the right. A student draws velocity arrows to a scale of 1.0 cm = 2.0 m/s.
(a) Describe the arrows (length and direction) for cart A with v_x = +6.0 m/s and cart B with v_x = −3.0 m/s. (b) Cart C’s arrow is 4.5 cm long and points to the left. Write cart C’s velocity component. (c) How many times greater is cart C’s speed than cart A’s speed? Explain why the direction of the arrows does not affect your answer.
Worked solution
(a) Cart A: 6.0 ÷ 2.0 = 3.0 cm, pointing right. Cart B: 3.0 ÷ 2.0 = 1.5 cm, pointing left.
(b) Size = 4.5 cm × 2.0 m/s per cm = 9.0 m/s. Left is opposite to +x, so v_x = −9.0 m/s.
(c) Speed of C ÷ speed of A = 9.0 ÷ 6.0 = 1.5 times. Speed is a magnitude, so it is shown by arrow length alone (4.5 cm ÷ 3.0 cm = 1.5). Direction only decides which way the arrow points.
Suggested mark points (4): 1 for both arrow lengths in (a); 1 for both directions in (a); 1 for −9.0 m/s with the sign in (b); 1 for the factor 1.5 with the reason that speed depends on length (magnitude) only.
Common error: writing +9.0 m/s in (b). The size is right, but the left-pointing arrow needs a negative component.
Question 5 (calculation · core)
Take +x east. A trolley is moving west at 3.0 m/s. A constant acceleration of 1.5 m/s² east then acts on it for 4.0 s. Use v_x = v_x0 + a_x t.
(a) Write v_x0 and a_x with their signs. (b) Find v_x at t = 2.0 s and at t = 4.0 s, and describe the motion at each time in words. (c) A student says: “The trolley’s velocity at 4.0 s is the same as at the start, because the speed is 3.0 m/s both times.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) West is opposite to +x, so v_x0 = −3.0 m/s. The acceleration is east, so a_x = +1.5 m/s².
(b) At 2.0 s: v_x = (−3.0) + (1.5)(2.0) = 0 m/s. The trolley is momentarily at rest. At 4.0 s: v_x = (−3.0) + (1.5)(4.0) = +3.0 m/s. The trolley is moving east at 3.0 m/s.
(c) The claim is incorrect. The speeds are equal (|−3.0| = |+3.0| = 3.0 m/s), but velocity is a vector, so its direction must match too. At the start the trolley moves west (−3.0 m/s); at 4.0 s it moves east (+3.0 m/s). The velocity has changed by (+3.0) − (−3.0) = +6.0 m/s.
| Point | What earns it |
|---|---|
| 1 | Opposite signs for v_x0 and a_x, consistent with +x east |
| 1 | v_x = 0 at 2.0 s, described as momentarily at rest |
| 1 | v_x = +3.0 m/s at 4.0 s, described as 3.0 m/s east |
| 1 | Rejects the claim because velocity needs the same size and direction; the directions differ |
Common error: taking a_x = −1.5 m/s² “because the trolley is moving west”. That gives v_x = −3.0 − 6.0 = −9.0 m/s at 4.0 s. The sign of a_x comes from the direction of the acceleration, not the direction of motion.
Question 6 (constructed response · stretch)
A cart rolls down a straight ramp. At t = 0 it is already moving down the ramp at 0.50 m/s, and its acceleration is 2.0 m/s² down the ramp. Two students find its velocity at t = 1.5 s using v_x = v_x0 + a_x t.
- Student A takes +x down the ramp and gets v_x = +3.5 m/s.
- Student B takes +x up the ramp and gets v_x = −3.5 m/s.
A third student says: “They got different answers, so one of them must be wrong.”
(a) Show the substitution each student made. (b) Evaluate the third student’s claim. (c) Student B then says: “My final velocity, −3.5 m/s, is less than my starting velocity, −0.50 m/s, so the cart has slowed down.” Evaluate this claim, comparing the speeds at t = 0 and t = 1.5 s.
Model answer and suggested Marlbridge rubric
(a) Student A: v_x = (+0.50) + (+2.0)(1.5) = 0.50 + 3.0 = +3.5 m/s. Student B: v_x = (−0.50) + (−2.0)(1.5) = −0.50 − 3.0 = −3.5 m/s.
(b) The claim is incorrect. Both answers describe the same vector: 3.5 m/s down the ramp. Student A’s + means “along +x”, which is down the ramp. Student B’s − means “opposite to +x”, and +x is up the ramp, so it also means down the ramp. Choosing the opposite axis flips every sign but not the physics.
(c) The claim is incorrect. On a number line −3.5 is less than −0.50, but the signs only show direction. The speed rises from |−0.50| = 0.50 m/s to |−3.5| = 3.5 m/s, so the cart is moving 7 times as fast. The cart has sped up, as you would expect for a cart rolling down a ramp with its acceleration down the ramp.
| Point | What earns it |
|---|---|
| 1 | Both substitutions shown with signs consistent with each student’s axis |
| 1 | States that both answers mean 3.5 m/s down the ramp |
| 1 | Explains that reversing the axis reverses every sign but not the physical direction |
| 1 | Compares speeds as magnitudes: 0.50 m/s rising to 3.5 m/s |
| 1 | Concludes the cart speeds up, so Student B’s claim is wrong |
An answer to (c) that says only “the minus sign is a direction”, without comparing the two speeds, earns at most the fifth point.
Question 7 (data analysis · stretch)
A robot vacuum moves back and forth along a straight hallway. Its app records each move as a distance and a direction relative to the front door. The charger is at the origin, and the front door is at the east end of the hallway, beyond every point the robot reaches.
| Move | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Distance (m) | 2.4 | 3.6 | 1.5 | 2.0 |
| Direction | towards door | away from door | towards door | away from door |
A student claims: “At the end of move 4, the robot is farther from its charger than it was after any earlier move.”
(a) Choose an axis and write each move as a signed displacement. (b) Find the robot’s total displacement and the total distance it travelled. (c) Use the robot’s position after each move to evaluate the student’s claim.
Model answer and suggested Marlbridge rubric
(a) Take +x east (towards the door), origin at the charger. The moves are +2.4 m, −3.6 m, +1.5 m and −2.0 m.
(b) Total displacement = 2.4 − 3.6 + 1.5 − 2.0 = −1.7 m (1.7 m west of the charger). Total distance = 2.4 + 3.6 + 1.5 + 2.0 = 9.5 m.
(c) Keep a running sum of the displacements:
| After move | Position x (m) | Distance from charger (m) |
|---|---|---|
| 1 | +2.4 | 2.4 |
| 2 | −1.2 | 1.2 |
| 3 | +0.3 | 0.3 |
| 4 | −1.7 | 1.7 |
The claim is not supported. After move 4 the robot is 1.7 m from the charger, but after move 1 it was 2.4 m away. How far the robot is from the charger is the magnitude of its position, and the largest magnitude came after move 1. The total distance of 9.5 m does not tell you how far the robot ends up from the charger, because moves in opposite directions cancel.
| Point | What earns it |
|---|---|
| 1 | Axis stated, with opposite signs for “towards” and “away from” the door |
| 1 | Total displacement −1.7 m with direction (west, or away from the door) |
| 1 | Total distance 9.5 m as a scalar sum |
| 1 | Correct positions after each move, found as running vector sums |
| 1 | Rejects the claim because the magnitude after move 1 (2.4 m) is greater than after move 4 (1.7 m) |
Taking +x west instead flips every sign (final position +1.7 m) and earns full credit if the axis is stated.
How did you do?
- Q1 wrong: re-read “Two kinds of quantity” in the study guide.
- Q2 or Q7 wrong: revisit “Adding vectors in one dimension” and Worked example 1 (Figure 2). Keep vector sums and distances apart.
- Q3 wrong: go back to “Comparing vectors”. Compare sizes with absolute values.
- Q4 wrong: look again at “Drawing vectors as arrows” and Figure 1.
- Q5 or Q6 incomplete: work through Worked example 2 and “Choosing an axis”. Every sign comes from the axis.
Then tick off the topic checklist.