Newton's second law and weight - Edexcel A-Level Maths (Paper 3) Exam Questions

Practise 18 exam-style questions on Newton's second law and weight 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.

18 questions
2–10 marks each
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18 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

4 marks

A van of mass 2100 kg moves along a straight horizontal road. The engine produces a constant driving force of 3.2 kN and the resistances to motion total 950 N. The van passes a point \(A\) with speed \(9\ \text{m s}^{-1}\).

Find the speed of the van when it has travelled 150 m beyond \(A\).

[Constant acceleration: \(v = u + at\), \(s = ut + \dfrac{1}{2}at^2\), \(s = vt - \dfrac{1}{2}at^2\), \(v^2 = u^2 + 2as\), \(s = \dfrac{1}{2}(u + v)t\)]

Model answer

Resultant force \(= 3200 - 950 = 2250\) N
\(F = ma\): \(2250 = 2100a\), \(a = \dfrac{15}{14} = 1.071\ \text{m s}^{-2}\)
\(v^2 = u^2 + 2as = 9^2 + 2 \times \dfrac{15}{14} \times 150 = 402.4\)
\(v = 20.1\ \text{m s}^{-1}\) (3 s.f.)

Explanation

Convert the driving force to newtons, use the resultant force in \(F = ma\), and then \(v^2 = u^2 + 2as\), as no time is given. Keep the acceleration exact or unrounded until the end. Common errors are mixing kN and N, and rounding the acceleration to 1.1 too early.

Mark scheme

1 mark
M1 Uses \(F = ma\) with the resultant force: \(3200 - 950 = 2100a\); or a correct final answer of \(20.1\ \text{m s}^{-1}\) with no working
An answer of \(20.1\ \text{m s}^{-1}\) (or 20) with no working scores 4 marks. M0 if 3.2 kN is used as 3.2 N in a way that is dimensionally inconsistent with 950 N, or if the resistance is ignored.
1 mark
A1 \(a = \dfrac{2250}{2100} = \dfrac{15}{14} = 1.07\ \text{m s}^{-2}\)
Accept awrt 1.07, i.e. [1.065, 1.075), or the exact fraction.
1 mark
M1 Uses \(v^2 = u^2 + 2as\) with \(u = 9\), \(s = 150\) and their \(a\)
Or any complete suvat method.
1 mark
A1 \(v = 20.1\ \text{m s}^{-1}\) (awrt 20.1, i.e. [20.05, 20.15))
No \(g\) is involved; accept 20 or 20.06. Using a rounded \(a = 1.1\) gives 20.3: A0 (premature approximation).

Question 2

5 marks

A particle \(P\) of mass 0.5 kg moves under the action of two forces only, \(\mathbf{F}_1\) and \(\mathbf{F}_2\). The force \(\mathbf{F}_1 = (1.3\mathbf{i} + 0.4\mathbf{j})\) N. The acceleration of \(P\) has magnitude \(5\ \text{m s}^{-2}\) and acts in the direction of the vector \((-7\mathbf{i} + 24\mathbf{j})\).

Find \(\mathbf{F}_2\), giving your answer in the form \((a\mathbf{i} + b\mathbf{j})\) N, and find the exact magnitude of \(\mathbf{F}_2\).

Model answer

\(|-7\mathbf{i} + 24\mathbf{j}| = \sqrt{49 + 576} = 25\), so \(\mathbf{a} = \dfrac{5}{25}(-7\mathbf{i} + 24\mathbf{j}) = -1.4\mathbf{i} + 4.8\mathbf{j}\)
Resultant force \(= m\mathbf{a} = 0.5(-1.4\mathbf{i} + 4.8\mathbf{j}) = -0.7\mathbf{i} + 2.4\mathbf{j}\)
\(\mathbf{F}_2 = (-0.7\mathbf{i} + 2.4\mathbf{j}) - (1.3\mathbf{i} + 0.4\mathbf{j}) = (-2\mathbf{i} + 2\mathbf{j})\) N
\(|\mathbf{F}_2| = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2}\) N

Explanation

The acceleration is its magnitude times the unit vector in its direction, and Newton's second law then gives the resultant force, \(0.5\mathbf{a}\); subtracting \(\mathbf{F}_1\) leaves \(\mathbf{F}_2\). The exact magnitude comes from Pythagoras, left as a surd. Common errors are multiplying the direction vector by 5 without dividing by its length 25, and setting \(\mathbf{F}_2\) alone equal to \(m\mathbf{a}\).

Mark scheme

1 mark
M1 Writes the acceleration as 5 times the unit vector in the direction of \(-7\mathbf{i} + 24\mathbf{j}\): \(\mathbf{a} = \dfrac{5}{25}(-7\mathbf{i} + 24\mathbf{j})\); or correct answers of \(\mathbf{F}_2 = (-2\mathbf{i} + 2\mathbf{j})\) N and \(2\sqrt{2}\) N with no working
Correct answers of \(\mathbf{F}_2 = (-2\mathbf{i} + 2\mathbf{j})\) N and \(|\mathbf{F}_2| = 2\sqrt{2}\) N with no working score 2 marks (the last two A marks). M0 for \(\mathbf{a} = 5(-7\mathbf{i} + 24\mathbf{j})\) or \(\mathbf{a} = -7\mathbf{i} + 24\mathbf{j}\) (the direction vector not made a unit vector). Accept column vectors in the working.
1 mark
A1 \(\mathbf{a} = (-1.4\mathbf{i} + 4.8\mathbf{j})\ \text{m s}^{-2}\)
1 mark
M1 Uses \(\mathbf{F}_1 + \mathbf{F}_2 = m\mathbf{a}\) with \(m = 0.5\) and their \(\mathbf{a}\)
M0 if the mass is omitted (\(\mathbf{F}_1 + \mathbf{F}_2 = \mathbf{a}\)) or if \(\mathbf{F}_2\) alone is set equal to \(m\mathbf{a}\). Condone sign slips for the M mark.
1 mark
A1 \(\mathbf{F}_2 = (-2\mathbf{i} + 2\mathbf{j})\) N
A column vector as the final answer is A0 (the form is asked for).
1 mark
A1 \(|\mathbf{F}_2| = \sqrt{(-2)^2 + 2^2} = 2\sqrt{2}\) N
Depends on the second M mark. ft the exact magnitude of their \(\mathbf{F}_2\). Exact answer required: accept \(\sqrt{8}\); 2.83 alone is A0 (isw if \(2\sqrt{2}\) or \(\sqrt{8}\) is seen first). No \(g\) is involved.

Question 3

6 marks

A wheeled recycling bin of mass 24 kg is pulled along a straight horizontal path by its handle. The pull on the handle has magnitude 60 N and is directed upwards at \(\theta\) to the horizontal, in the vertical plane containing the direction of motion. The bin's motion is opposed by a constant horizontal resistance of magnitude 30 N. The bin is modelled as a particle.

The bin moves with constant acceleration \(0.6\ \text{m s}^{-2}\).

Find the value of \(\theta\), giving your answer in degrees to one decimal place, and the magnitude of the normal reaction between the bin and the path.

Use \(g = 9.8\ \text{m s}^{-2}\) and give the normal reaction to 2 or 3 significant figures.

Model answer

Newton's second law horizontally: \(60\cos\theta - 30 = 24 \times 0.6 = 14.4\)
\(\cos\theta = \dfrac{44.4}{60} = 0.74\), so \(\theta = 42.3^\circ\) (1 d.p.)
Resolving vertically: \(R + 60\sin\theta = 24g\)
\(R = 235.2 - 60\sin 42.27^\circ = 235.2 - 40.36 = 194.8\)
Normal reaction \(= 195\) N (3 s.f.)

Explanation

Only the horizontal component of the pull, \(60\cos\theta\), accelerates the bin against the resistance, so Newton's second law horizontally gives \(\cos\theta\). The vertical component of the pull reduces the normal reaction, so \(R = 24g - 60\sin\theta\), not \(24g\). Common errors are resolving with sin instead of cos and leaving out the resistance.

Mark scheme

1 mark
M1 Newton's second law horizontally: \(60\cos\theta - 30 = 24 \times 0.6\) (three terms, pull resolved); or correct final answers of \(\theta = 42.3^\circ\) and \(R = 195\) N with no working
Correct answers of \(\theta = 42.3^\circ\) and \(R = 195\) N (or 190 N) with no working score 2 marks (the two final A marks). Condone sin/cos confusion for the M mark only: \(60\sin\theta - 30 = 14.4\) gives \(\theta = 47.7^\circ\) (M1 A0 A0). An omitted mass in the ma term is a method error. The final answers are marked at their own A marks, where the accepted values are given.
1 mark
A1 \(\cos\theta = 0.74\) oe
Accept \(\cos\theta = \dfrac{44.4}{60}\) or \(\dfrac{37}{50}\).
1 mark
A1 \(\theta = 42.3^\circ\)
Depends on the first M mark. Accept awrt 42.3, i.e. [42.25, 42.35).
1 mark
M1 Resolves vertically: \(R + 60\sin\theta = 24g\) (three terms, pull resolved)
A common error is \(R = 24g = 235.2\), ignoring the vertical component of the pull: M0. An omitted g is an accuracy error.
1 mark
A1 \(R = 24g - 60\sin\theta\) with their \(\theta\)
Correct expression or value (\(R = 235.2 - 40.36\)).
1 mark
A1 \(R = 195\) N (or 190 N)
Depends on the second M mark. Accept awrt 195, i.e. [194.5, 195.5), or 190. The answer comes from g = 9.8, so give it to 2 or 3 s.f.: 194.8 is over-accurate (penalised once). Use of g = 9.81 is penalised once in the question.

Question 4

8 marks

A skydiver and her equipment have a total mass of 75 kg. She falls vertically, and her parachute opens when her speed is \(50\ \text{m s}^{-1}\). From the moment the parachute opens, the air resistance on the skydiver is modelled as a constant force of 1100 N, and the skydiver is modelled as a particle.

Find the time taken, after the parachute opens, for her speed to fall to \(8\ \text{m s}^{-1}\), and the distance she falls in this time.

Comment on the assumption that the air resistance is constant.

Use \(g = 9.8\ \text{m s}^{-2}\) and give your answers to 2 or 3 significant figures.

[Constant acceleration: \(v = u + at\), \(s = ut + \dfrac{1}{2}at^2\), \(s = vt - \dfrac{1}{2}at^2\), \(v^2 = u^2 + 2as\), \(s = \dfrac{1}{2}(u + v)t\)]

Model answer

Taking downwards as positive, N2L: \(75g - 1100 = 75a\)
\(735 - 1100 = 75a\), \(a = -4.867\ \text{m s}^{-2}\) (a deceleration of \(4.87\ \text{m s}^{-2}\))
\(v = u + at\): \(8 = 50 - 4.867t\), \(t = 8.630\)
Time \(= 8.63\) s (3 s.f.)
\(s = \dfrac{1}{2}(u + v)t = \dfrac{1}{2}(50 + 8) \times 8.630 = 250.3\)
Distance \(= 250\) m (3 s.f.)
The assumption is not realistic: air resistance increases with speed, so it would be larger than 1100 N just after the parachute opens and smaller as she slows down.

Explanation

After the parachute opens the air resistance is greater than her weight, \(75g = 735\) N, so the resultant force is upwards and she decelerates; Newton's second law gives the deceleration, and suvat gives the time and distance. Keep the signs consistent, with downwards positive and a negative acceleration. In reality air resistance depends on speed, so a constant value is only a rough model.

Mark scheme

1 mark
M1 Uses \(F = ma\) with the weight and the air resistance: \(75g - 1100 = 75a\) (or \(1100 - 75g = 75 \times\) deceleration); or correct answers of 8.63 s and 250 m with no working
Correct answers of 8.6 s (or 8.63 s) and 250 m with no working score 2 marks (the A marks for the two answers); the comment is marked separately. An omitted g is an accuracy error; omitting the weight altogether is M0.
1 mark
A1 Correct equation
1 mark
A1 Deceleration \(= \dfrac{365}{75} = 4.87\ \text{m s}^{-2}\)
Depends on the first M mark. Accept awrt 4.87, i.e. [4.865, 4.875), or \(\dfrac{73}{15}\).
1 mark
M1 Uses \(v = u + at\) with \(u = 50\), \(v = 8\) and their deceleration
Signs must be consistent: an acceleration of \(+4.87\) with \(u = 50\), \(v = 8\) gives a negative time (M1 A0).
1 mark
A1 \(t = 8.63\) s or 8.6 s
Answers from g = 9.8 to 2 or 3 s.f. Use of g = 9.81 is penalised once in this question.
1 mark
M1 Complete method for the distance, e.g. \(s = \dfrac{1}{2}(u + v)t\) or \(v^2 = u^2 + 2as\)
With their values.
1 mark
A1 Distance \(= 250\) m
250.3 is over-accurate (penalise over-accuracy once in the question).
1 mark
B1 Not realistic: air resistance depends on speed, so it would be larger just after the parachute opens (when she is fast) and smaller as she slows down
Must refer to the resistance varying with speed (or with time). 'It is not constant' with no reason is B0. Comments about her mass, or about wind, are B0.

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