OxfordAQA A Level Physics: Motion Along a Straight Line and Newton's Laws — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Motion Along a Straight Line and Newton’s Laws study guide, Motion Along a Straight Line and Newton’s Laws revision notes

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Section A

1. State the four SUVAT equations for uniform acceleration. [4]

2. State Newton’s three laws of motion in full. [3]

Section B

3. A car accelerates uniformly from 5 m/s to 20 m/s in 6 s.

(a) Calculate its acceleration. [2] (b) Calculate the distance travelled during this time. [2]

4. A velocity-time graph shows an object decelerating uniformly from 15 m/s to 0 m/s over 5 s, then remaining at rest for a further 3 s.

(a) Calculate the deceleration during the first phase. [2] (b) Calculate the total displacement over the full 8 s. [3]

5. Explain how a graph of s against t² can be used to determine g in the required practical for this sub-topic, including why this graph is chosen over a graph of s against t. [4]

6. A block of mass 4 kg is pushed along a frictionless surface with a constant resultant force of 12 N.

(a) Calculate its acceleration. [2] (b) Explain why F = ma could not be applied directly if the block were losing mass during the motion (for example, leaking sand from a hole). [2]

7. A student writes: “A Newton’s third law pair is two forces that balance each other on the same object.” Explain why this statement is incorrect, and give a correct example of a genuine third law pair. [4]

8. State Newton’s first law, and use it to explain why passengers in a car feel pushed back into their seats when the car accelerates forward. [3]

9. Explain the difference between what a gradient represents and what an area represents on a velocity-time graph, using a specific example of each. [4]

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Answers

1. v = u + at [1]. s = ½(u + v)t [1]. s = ut + ½at² [1]. v² = u² + 2as [1].

2. First law: an object remains at rest or moves at constant velocity unless acted on by a resultant force [1]. Second law: the resultant force acting on an object equals its mass multiplied by its acceleration, F = ma, applying only where mass is constant [1]. Third law: for every action force there is an equal and opposite reaction force, acting on a different object [1].

3. (a) a = (v − u) / t = (20 − 5) / 6 = 2.5 m/s² [2]. (b) Using s = ut + ½at²: s = (5)(6) + 0.5(2.5)(6²) = 30 + 45 = 75 m [2].

4. (a) Deceleration = (0 − 15) / 5 = −3 m/s², i.e. a deceleration of 3 m/s² [2]. (b) Displacement in first phase (area of triangle) = ½ × 5 × 15 = 37.5 m [1]; displacement in second phase (at rest, zero velocity) = 0 m [1]; total displacement = 37.5 m [1].

5. Since s = ut + ½at² with u = 0 for an object released from rest, this simplifies to s = ½gt² [1]. Plotting s against t² therefore gives a straight line through the origin, whereas s against t would give a curve, which is much harder to extract a precise value of g from [1] [1]. The gradient of the straight line equals g/2, so g is found as 2 × gradient [1].

6. (a) a = F/m = 12/4 = 3 m/s² [2]. (b) F = ma assumes mass is constant throughout the motion [1]; if mass is changing, the correct relationship involves the rate of change of momentum (F = Δp/Δt), since simply substituting a changing mass into F = ma does not correctly account for the momentum lost with the escaping material [1].

7. The statement is incorrect because a genuine third law pair acts on two different objects, not the same object [1] — two forces balancing on one object (such as an object’s weight balanced by a normal contact force) is an application of Newton’s first law, not the third [1]. A correct example: when a person pushes on a wall, the wall pushes back on the person with an equal and opposite force — one force acts on the wall, the other acts on the person, and both are of the same type (contact force) [1] [1].

8. Newton’s first law states that an object continues at rest or at constant velocity unless a resultant force acts on it [1]. When the car accelerates forward, a resultant force acts on the car (and the seat) but the passenger’s body tends to continue at its original, slower velocity due to this law, so relative to the accelerating car the passenger appears to be pushed backward into the seat, when in fact the seat is being pushed forward into the passenger [1] [1].

9. The gradient of a velocity-time graph represents acceleration — for example, a constant positive gradient shows uniform acceleration, such as a gradient of 2 m/s² for an object speeding up steadily [2]. The area under a velocity-time graph represents displacement — for example, the area under a graph section showing constant velocity of 10 m/s for 5 s is a rectangle giving a displacement of 50 m [2].

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Where marks are usually lost

• Selecting a SUVAT equation that requires a variable not given in the question, forcing an unnecessary extra calculation step.
• Confusing the physical meaning of a gradient versus an area on a motion graph.
• Quoting F = ma without noting that it applies only when mass is constant.
• Stating only Newton’s second law when a question asks about “Newton’s laws” in general.
• Describing a Newton’s third law pair as two forces acting on the same object, rather than on two different objects.
• Plotting s against t instead of s against t² when analysing the free-fall required practical, missing the point of the linearisation.