Quantities, units and the language of kinematics - Edexcel A-Level Maths (Paper 3) Exam Questions

Practise 16 exam-style questions on Quantities, units and the language of kinematics for Edexcel A-Level Maths (Paper 3). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

16 questions
2–9 marks each
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16 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

A cyclist and her bicycle are modelled as a particle when her motion along a straight road is analysed.

Sam says, "Modelling the cyclist and her bicycle as a particle means that their mass is ignored."

Explain why Sam is wrong, and state what the particle model does ignore.

Model answer

Sam is wrong: a particle has mass. In the model the whole mass of the cyclist and bicycle is concentrated at a single point, and this mass is still used, for example in F = ma.
What the particle model ignores is the size (the dimensions and shape) of the cyclist and bicycle, so effects such as rotation of the wheels, or how the rider's shape affects air resistance, are not included.

Explanation

A particle is an object whose mass is concentrated at a single point, so its mass is kept but its dimensions are ignored. This lets us treat all the forces as acting at one point and ignore rotation. Examiners' reports note that students lose marks by saying that a particle model ignores mass or weight.

Mark scheme

1 mark
B1 Sam is wrong because a particle still has mass: the whole mass of the cyclist and bicycle is taken to be concentrated at a single point (and it is used, e.g. in F = ma)
Needs the idea that the mass is kept (concentrated at a point). B0 for "Sam is wrong" with no reason.
1 mark
B1 The particle model ignores the size, shape and dimensions of the cyclist and bicycle (so, for example, rotation of the wheels and the effect of the rider's shape on air resistance are ignored)
Accept "dimensions", "size" or "shape". B0 if mass or weight is said to be ignored. Examiners' reports note that "the particle has no mass" is a common wrong answer.

Question 2

4 marks

The newton (N) is the SI unit of force.

Express each of the following units in terms of the SI base units kg, m and s:

the newton, N

the newton metre, N m, the unit of the moment of a force

the newton second, N s, the unit of impulse.

Hence explain why momentum, mass × velocity, can be measured in N s.

Model answer

From \(F = ma\): \(1\ \text{N} = 1\ \text{kg} \times 1\ \text{m s}^{-2} = 1\ \text{kg m s}^{-2}\)
\(1\ \text{N m} = 1\ \text{kg m s}^{-2} \times \text{m} = 1\ \text{kg m}^2\,\text{s}^{-2}\)
\(1\ \text{N s} = 1\ \text{kg m s}^{-2} \times \text{s} = 1\ \text{kg m s}^{-1}\)
Momentum = mass × velocity has units \(\text{kg} \times \text{m s}^{-1} = \text{kg m s}^{-1}\), which is the same as N s, so momentum can be measured in N s.

Explanation

Newton's second law, F = ma, defines the newton as the force that gives a mass of 1 kg an acceleration of 1 \(\text{m s}^{-2}\), so N = kg m s⁻². Multiplying by metres or seconds gives the units of a moment and of impulse, and the result for N s matches the units of mass × velocity. These equivalent units (N s and kg m s⁻¹) are named in the specification guidance.

Mark scheme

1 mark
B1 \(1\ \text{N} = 1\ \text{kg m s}^{-2}\)
From F = ma: mass (kg) × acceleration (\(\text{m s}^{-2}\)). Accept kg m/s² or any clear typed form.
1 mark
B1 \(1\ \text{N m} = 1\ \text{kg m}^2\,\text{s}^{-2}\)
Follow through their unit for N multiplied by m. Accept kg m²/s² or any clear typed form.
1 mark
B1 \(1\ \text{N s} = 1\ \text{kg m s}^{-1}\)
Follow through their unit for N multiplied by s. Accept kg m/s.
1 mark
B1 Mass × velocity has units \(\text{kg} \times \text{m s}^{-1} = \text{kg m s}^{-1}\), which is the same as N s, so the two units are equivalent
Must show the units of mass × velocity as kg m s⁻¹ and link them to their N s result. B0 for a bare statement that the units are equal.

Question 3

4 marks

The speed of a train moving along a straight track increases uniformly from 36 \(\text{km h}^{-1}\) to 99 \(\text{km h}^{-1}\) in 50 seconds.

Find the acceleration of the train in \(\text{m s}^{-2}\).

Express this acceleration in \(\text{km h}^{-2}\).

Model answer

\(36\ \text{km h}^{-1} = \dfrac{36}{3.6} = 10\ \text{m s}^{-1}\), \(99\ \text{km h}^{-1} = \dfrac{99}{3.6} = 27.5\ \text{m s}^{-1}\)
Acceleration \(= \dfrac{27.5 - 10}{50} = 0.35\ \text{m s}^{-2}\)
In \(\text{km h}^{-2}\): \(0.35\ \text{m s}^{-2} = 0.35 \times \dfrac{1}{1000}\ \text{km} \times \dfrac{1}{(1/3600)^2\ \text{h}^2} = 0.35 \times \dfrac{3600^2}{1000} = 4536\ \text{km h}^{-2}\)

Explanation

Acceleration is the rate of change of velocity, so with both speeds in \(\text{m s}^{-1}\) the change of 17.5 \(\text{m s}^{-1}\) is divided by 50 s. To change \(\text{m s}^{-2}\) into \(\text{km h}^{-2}\), divide by 1000 for the length and multiply by \(3600^2\) because seconds appear squared. The usual error is to multiply by 3600 only once.

Mark scheme

1 mark
M1 Converts both speeds to \(\text{m s}^{-1}\): \(36\ \text{km h}^{-1} = 10\ \text{m s}^{-1}\) and \(99\ \text{km h}^{-1} = 27.5\ \text{m s}^{-1}\), and uses change in velocity ÷ time; or correct final answers of 0.35 \(\text{m s}^{-2}\) and 4536 \(\text{km h}^{-2}\) with no working
An answer of 0.35 \(\text{m s}^{-2}\) and 4536 \(\text{km h}^{-2}\) with no working scores 4 marks. Allow converting the change in speed (63 km h⁻¹ = 17.5 m s⁻¹) instead of each speed.
1 mark
A1 \(a = \dfrac{27.5 - 10}{50} = 0.35\ \text{m s}^{-2}\)
cao. Accept 7/20. Units are given in the question, so ignore missing units here.
1 mark
M1 Converts their acceleration: \(0.35 \times \dfrac{3600^2}{1000}\) (multiplies by \(3600^2\) for the time unit and divides by 1000 for the length unit)
Alternative method: \(\dfrac{99 - 36}{50/3600}\) in \(\text{km h}^{-2}\). Multiplying by 3600 only once is M0 (a common error: the time unit is squared).
1 mark
A1 4536 \(\text{km h}^{-2}\)
cao. Accept 4540 or 4.54 × 10³ (3 s.f.).

Question 4

7 marks

A delivery robot moves on horizontal ground along two straight paths. It travels 120 m due east at a constant speed of 2.4 \(\text{m s}^{-1}\), and then 50 m due north at a constant speed of 2 \(\text{m s}^{-1}\). The robot is modelled as a particle, and the time taken to turn is ignored.

Find

the total distance that the robot travels

the magnitude of the displacement of the robot from its starting point, and its direction as a bearing

the average speed of the robot

the magnitude of the average velocity of the robot.

Give your answers to 3 significant figures where appropriate.

Model answer

Total distance \(= 120 + 50 = 170\) m
Displacement: \(\sqrt{120^2 + 50^2} = \sqrt{16\,900} = 130\) m
Direction: \(\tan\theta = \dfrac{120}{50}\), so \(\theta = 67.4^\circ\) east of north: bearing 067°
Time: \(\dfrac{120}{2.4} + \dfrac{50}{2} = 50 + 25 = 75\) s
Average speed \(= \dfrac{170}{75} = 2.27\ \text{m s}^{-1}\) (3 s.f.)
Average velocity: magnitude \(= \dfrac{130}{75} = 1.73\ \text{m s}^{-1}\) (3 s.f.), on a bearing of 067°

Explanation

Distance is a scalar, so the two path lengths are simply added, but displacement is a vector, so the two perpendicular parts combine by Pythagoras to give 130 m on a bearing of 067°. Both averages use the total time of 75 s: average speed uses the distance and average velocity uses the displacement. Averaging the two speeds is wrong because the robot spends different times on the two paths.

Mark scheme

1 mark
B1 Total distance \(= 120 + 50 = 170\) m. Correct answers with no working score full marks.
An answer of 170 m, 130 m on a bearing of 067°, 2.27 \(\text{m s}^{-1}\) and 1.73 \(\text{m s}^{-1}\) with no working scores 7 marks.
1 mark
M1 Uses Pythagoras for the displacement: \(\sqrt{120^2 + 50^2}\)
Adding 120 and 50 for the displacement is M0.
1 mark
A1 Magnitude of displacement 130 m
cao.
1 mark
B1 Bearing 067° (from \(\tan\theta = \dfrac{120}{50}\), θ measured clockwise from north; θ = 67.4° to 3 s.f., awrt 67.4, i.e. in [67.35, 67.45), giving 067°)
Accept 67°, 67.4° or 067.4°. A bearing is asked for, so a direction given in another form (e.g. "N 67.4° E" or "22.6° north of east") scores B0, as does a bearing of 022.6° or 023° (the angle with east given as a bearing).
1 mark
M1 Finds the total time \(\dfrac{120}{2.4} + \dfrac{50}{2} = 50 + 25 = 75\) s and divides a distance or displacement by it
Averaging the two speeds ((2.4 + 2) ÷ 2 = 2.2) is M0 for the average speed. The two times (50 s and 25 s) are exact.
1 mark
A1 Average speed \(= \dfrac{170}{75} = 2.27\ \text{m s}^{-1}\) (awrt 2.27, i.e. in [2.265, 2.275))
Accept \(\dfrac{34}{15}\). Ignore units if clearly m s⁻¹.
1 mark
A1 Magnitude of average velocity \(= \dfrac{130}{75} = 1.73\ \text{m s}^{-1}\) (awrt 1.73, i.e. in [1.725, 1.735))
Depends on the M1 for the total time. Accept \(\dfrac{26}{15}\). Using the distance 170 m here is A0.

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