Edexcel A-level Further Maths coverage

Momentum and impulse

Section FM1-1
2 spec leafs

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

Open the printable pack
FM1-1.1

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

  • For a particle of mass mm moving with velocity vv along a line, its momentum is mvmv; choose and state a positive direction before assigning signs.
  • Impulse equals change in momentum: I=m(v−u)I=m(v-u). It is also the area under a force-time graph, I=∫F dtI=\int F\,dt.
  • During a direct collision, total momentum is conserved when the external impulse on the colliding system is negligible.
  • A negative velocity means motion opposite to the chosen positive direction; do not replace signed velocities by speeds inside a momentum equation.

Tier 1 · Easy

2 marks
ORIGINAL

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.

Tier 2 · Standard

3 marks
ORIGINAL

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.

Tier 3 · Hard

4 marks
ORIGINAL

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.

FM1-1.2

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

  • Vector momentum is mvm\mathbf v, so its components are found by multiplying every velocity component by the mass.
  • In vector form, J=m(v−u)\mathbf J=m(\mathbf v-\mathbf u), and hence v=u+J/m\mathbf v=\mathbf u+\mathbf J/m.
  • Resolve a vector impulse into perpendicular components before finding its magnitude or the direction of the resulting velocity.
  • Do not apply the impulse magnitude separately to both components; the signed vector components, not just ∣J∣|\mathbf J|, determine the change in velocity.

Tier 1 · Easy

3 marks
ORIGINAL

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.

Tier 2 · Standard

4 marks
ORIGINAL

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.

Tier 3 · Hard

5 marks
ORIGINAL

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.