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What are momentum and impulse, and how do you use conservation of momentum to analyse collisions and impulsive forces?

Define linear momentum and impulse; relate impulse to change of momentum; apply conservation of linear momentum to direct collisions; and handle impulsive tensions in connected bodies.

A focused answer to the SQA Advanced Higher Mathematics of Mechanics content on momentum and impulse, covering linear momentum, the impulse-momentum principle, the impulse of a variable force as an integral, conservation of momentum in direct collisions, and impulsive tensions in connected particles.

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  1. What this dot point is asking
  2. Momentum and impulse
  3. Conservation of momentum in collisions
  4. Impulsive tensions
  5. Try this

What this dot point is asking

Momentum is the quantity that is conserved when bodies interact, and impulse is the change it undergoes. The SQA wants you to define both, relate impulse to the change of momentum (including for a force that varies with time, via an integral), and use conservation of linear momentum to analyse direct collisions and impulsive tensions in connected bodies.

Momentum and impulse

Momentum measures "quantity of motion": it depends on both mass and velocity, so a slow heavy body can carry as much momentum as a fast light one. Impulse is what changes it.

The impulse-momentum principle is just Newton's second law in integrated form: since F=dpdt\mathbf{F} = \dfrac{d\mathbf{p}}{dt}, integrating over time gives the change in momentum. This is the right tool for short, sharp forces (a kick, a hammer blow) where the force is hard to track instant by instant but its overall effect is what matters.

Conservation of momentum in collisions

When two bodies collide and no external horizontal force acts during the (very short) impact, the internal forces are equal and opposite (Newton's third law), so the total momentum is unchanged.

The method for a direct (head-on) collision is always the same: fix a positive direction, write the conservation-of-momentum equation, and solve for the unknown. If asked about energy, compare 12mv2\tfrac{1}{2}mv^2 totals before and after.

Impulsive tensions

When a string connecting two bodies suddenly becomes taut, it transmits an impulsive tension - a large force acting for a very short time, modelled by its impulse. The same impulse acts on each connected body (in opposite senses), changing their momenta so that they end with a common speed if the string is inextensible.

Try this

Q1. A 55 kg trolley moving at 33 m sβˆ’1^{-1} collides and sticks to a stationary 11 kg trolley. Find their common speed. [3 marks]

  • Cue. 5(3)+1(0)=(5+1)v5(3) + 1(0) = (5 + 1)v, so v=156=2.5v = \dfrac{15}{6} = 2.5 m sβˆ’1^{-1}.

Q2. A constant force of 1212 N acts on a body for 0.50.5 s. Find the impulse and the change of momentum. [2 marks]

  • Cue. J=Ft=12Γ—0.5=6J = Ft = 12\times 0.5 = 6 N s, which equals the change of momentum.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AH style: collision5 marksA 33 kg ball moving at 44 m sβˆ’1^{-1} collides directly with a 22 kg ball moving in the same direction at 11 m sβˆ’1^{-1}. After the collision the 33 kg ball moves at 22 m sβˆ’1^{-1}. Find the speed of the 22 kg ball and state whether kinetic energy is conserved.
Show worked answer β†’

Conservation of momentum: 3(4)+2(1)=3(2)+2v3(4) + 2(1) = 3(2) + 2v, so 14=6+2v14 = 6 + 2v (2 marks).

Solving, 2v=82v = 8, giving v=4v = 4 m sβˆ’1^{-1} for the 22 kg ball (1 mark).

KE before =12(3)(16)+12(2)(1)=24+1=25= \tfrac{1}{2}(3)(16) + \tfrac{1}{2}(2)(1) = 24 + 1 = 25 J. KE after =12(3)(4)+12(2)(16)=6+16=22= \tfrac{1}{2}(3)(4) + \tfrac{1}{2}(2)(16) = 6 + 16 = 22 J (1 mark).

Since 22<2522 < 25, kinetic energy is not conserved, so the collision is inelastic; 33 J is lost (1 mark). Markers reward applying conservation of momentum, solving for the unknown speed, and comparing the kinetic energies.

AH style: impulse of a force4 marksA force F=6tF = 6t N acts on a particle of mass 22 kg, initially at rest, for 33 seconds. Find the impulse of the force and the final speed.
Show worked answer β†’

The impulse of a variable force is J=∫03F dt=∫036t dtJ = \displaystyle\int_0^3 F\,dt = \int_0^3 6t\,dt (1 mark).

J=[3t2]03=27J = \left[3t^2\right]_0^3 = 27 N s (1 mark).

Impulse equals change of momentum: J=mvβˆ’mu=2vβˆ’0J = m v - m u = 2v - 0, so 27=2v27 = 2v (1 mark).

Hence v=13.5v = 13.5 m sβˆ’1^{-1} (1 mark). Markers reward integrating the force for the impulse and equating impulse to the change in momentum.

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