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Edexcel A-level Further Maths revision notes

Momentum and impulse

Section FM1-1
Both years
Both years: this holds AS subject content and content the exam board adds beyond it for the full A-level.
2 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section FM1-1

Checked against Edexcel 9FM0 section FM1-1. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

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In the exam: Formulae booklet provided · calculator allowed in every paper

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FM1-1.1

Momentum and impulse. The impulse-momentum principle. The principle of conservation of momentum applied to two spheres colliding directly.

Notes
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A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For a particle moving along a line, momentum is the signed quantity mvmv. A positive direction must therefore be fixed before velocities enter an equation.
  • Impulse is the change in momentum, I=m(v−u)I=m(v-u), and also I=∫F dtI=\int F\,dt, the signed area under a force-time graph.
  • During a direct collision of two spheres modelled as particles, total momentum is conserved when the external impulse over the short impact is negligible.
  • Write one momentum equation for the system and separate impulse equations for individual spheres when required.
  • Examiners expect signed velocities throughout; a negative result describes motion opposite to the chosen direction and must not be silently changed into a positive speed.
A direct collision drawn on one signed line, with the chosen positive direction shown.
Worked example

Sphere AA of mass 2 kg2\,\text{kg} moves right at 5 m s−15\,\text{m s}^{-1} and collides directly with sphere BB of mass 3 kg3\,\text{kg} moving left at 1 m s−11\,\text{m s}^{-1}. They coalesce. Find the common velocity and the impulse on AA.

  1. 1.Take rightwards as positive and conserve momentum: 2(5)+3(−1)=5v2(5)+3(-1)=5v.
  2. 2.Thus v=75 m s−1v=\dfrac75\,\text{m s}^{-1}.
  3. 3.For AA, I=2(75−5)=−365 N sI=2(\tfrac75-5)=-\dfrac{36}{5}\,\text{N s}.

Answer: The common velocity is 75 m s−1\dfrac75\,\text{m s}^{-1} rightwards; the impulse on AA is 365 N s\dfrac{36}{5}\,\text{N s} leftwards.

Common mistakes

  • Don't use positive speeds for both spheres even though one initial velocity is in the negative direction.
  • Don't calculate impulse as m(u−v)m(u-v) instead of the final momentum minus the initial momentum.
  • Don't conserve kinetic energy when the spheres coalesce.

Exam tip

State the positive direction above the first momentum equation and interpret every negative final value against it.

Tier 1 · Easy

ORIGINAL

1.

Rightwards is positive. A 3 kg3\,\text{kg} particle changes velocity from 4 m s−14\,\text{m s}^{-1} rightwards to 2 m s−12\,\text{m s}^{-1} leftwards. Find its impulse.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Two spheres collide head-on and remain together. The first has mass 2 kg2\,\text{kg} and velocity 7 m s−17\,\text{m s}^{-1}; the second has mass 3 kg3\,\text{kg} and velocity −1 m s−1-1\,\text{m s}^{-1}. Calculate their common velocity.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

Sphere AA, of mass m kgm\,\text{kg}, moves rightwards at 8 m s−18\,\text{m s}^{-1} behind a 2 kg2\,\text{kg} sphere BB moving rightwards at 2 m s−12\,\text{m s}^{-1}. After their direct collision, the velocities of AA and BB are respectively 1 m s−11\,\text{m s}^{-1} and 5 m s−15\,\text{m s}^{-1} rightwards. Find mm and the impulse on AA.

(4)

(Total for Question 1 is 4 marks)

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FM1-1.2

Momentum as a vector. The impulse-momentum principle in vector form.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Momentum is a vector: p=mv\mathbf p=m\mathbf v. The impulse-momentum principle is J=m(v−u)\mathbf J=m(\mathbf v-\mathbf u), so v=u+J/m\mathbf v=\mathbf u+\mathbf J/m.
  • Work component by component in a fixed orthogonal basis, retaining signs. Only after the new velocity or impulse vector is known should its magnitude and direction be calculated using Pythagoras and an angle with a quadrant check.
  • A stated impulse magnitude alone does not determine its components; direction information or another condition is needed.
  • Examiners expect vector units, a complete i,j\mathbf i,\mathbf j answer and any requested scalar speed distinguished from the velocity vector.
  • Perpendicularity is tested by a zero scalar product.
Vector impulse added to initial momentum to produce the final momentum.
Worked example

A 5 kg5\,\text{kg} particle initially moves with velocity (2i+3j) m s−1(2\mathbf i+3\mathbf j)\,\text{m s}^{-1} and receives impulse (15i−20j) N s(15\mathbf i-20\mathbf j)\,\text{N s}. Find its final velocity and speed.

  1. 1.v=u+J/m=(2i+3j)+(3i−4j)\mathbf v=\mathbf u+\mathbf J/m=(2\mathbf i+3\mathbf j)+(3\mathbf i-4\mathbf j).
  2. 2.Therefore v=(5i−j) m s−1\mathbf v=(5\mathbf i-\mathbf j)\,\text{m s}^{-1}.
  3. 3.The speed is ∣v∣=52+(−1)2=26 m s−1|\mathbf v|=\sqrt{5^2+(-1)^2}=\sqrt{26}\,\text{m s}^{-1}.

Answer: Final velocity (5i−j) m s−1(5\mathbf i-\mathbf j)\,\text{m s}^{-1}; speed 26 m s−1\sqrt{26}\,\text{m s}^{-1}.

Common mistakes

  • Don't add the impulse vector directly to velocity without dividing by the mass.
  • Don't use the impulse magnitude as both its i\mathbf i and j\mathbf j components.
  • Don't report 26 m s−1\sqrt{26}\,\text{m s}^{-1} as a velocity rather than a speed.

Exam tip

Keep the calculation in vector form until the final line, then calculate magnitude or direction only if requested.

Tier 1 · Easy

ORIGINAL

1.

A 2 kg2\,\text{kg} particle changes velocity from (3i−4j) m s−1(3\mathbf i-4\mathbf j)\,\text{m s}^{-1} to (5i+j) m s−1(5\mathbf i+\mathbf j)\,\text{m s}^{-1}. Find the impulse vector and its magnitude.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

A 4 kg4\,\text{kg} particle initially has velocity (2i−j) m s−1(2\mathbf i-\mathbf j)\,\text{m s}^{-1}. It receives an impulse (−6i+8j) N s(-6\mathbf i+8\mathbf j)\,\text{N s}. Determine its new speed and the acute angle its velocity makes above the positive i\mathbf i direction.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

A 3 kg3\,\text{kg} particle has initial velocity (4i−2j) m s−1(4\mathbf i-2\mathbf j)\,\text{m s}^{-1}. An impulse of magnitude 15 N s15\,\text{N s} leaves its final velocity perpendicular to its initial velocity. Find all possible final velocities and the corresponding impulse vectors.

(5)

(Total for Question 1 is 5 marks)

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