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How is momentum conserved in collisions and explosions, and how is impulse linked to force and time?

Conservation of momentum in one dimension, elastic and inelastic collisions, explosions, and the link between impulse, force and the change in momentum.

An SQA Higher Physics answer on collisions, momentum and impulse, covering the conservation of momentum in one dimension, the difference between elastic and inelastic collisions and explosions, and how impulse links force, time and the change in momentum.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this key area is asking
  2. Conservation of momentum
  3. Elastic and inelastic collisions
  4. Impulse
  5. Examples in context
  6. Try this

What this key area is asking

The SQA wants you to apply the conservation of momentum to one-dimensional collisions and explosions, distinguish elastic from inelastic collisions using kinetic energy, and use the impulse relationship to link force, contact time and the change in momentum.

Conservation of momentum

For two objects moving along a line, taking one direction as positive:

An explosion is the reverse of a collision: an object initially at rest (zero total momentum) breaks into parts that fly apart with equal and opposite momenta, so the total stays zero. This is how recoil works, and how a rocket gains forward momentum by ejecting exhaust backwards.

Elastic and inelastic collisions

To decide which type a collision is, calculate the total kinetic energy before and after using Ek=12mv2E_k = \tfrac{1}{2}mv^2. If it is unchanged the collision is elastic; if it has decreased it is inelastic.

Impulse

When a force acts for a time it changes an object's momentum. The product of force and time is the impulse, equal to the change in momentum.

The impulse equals the area under a force-time graph, which matters when the force varies during a short impact. For a fixed change in momentum, increasing the contact time tt reduces the average force FF. This is the physics of safety design: crumple zones, airbags, padded gloves and bending your knees on landing all lengthen the stopping time to lower the peak force.

Examples in context

A car crumple zone is engineered to deform on impact, extending the time over which the occupants are brought to rest and so reducing the peak force on them for the same change in momentum. Airbags do the same for the head and chest. In snooker, a near-elastic collision between the cue ball and an object ball transfers momentum with very little kinetic energy lost. Rocket propulsion and a rifle recoil are explosions in the momentum sense: the forward momentum of the projectile or exhaust equals the backward momentum of the rocket or gun, conserving the zero total. Newton's cradle demonstrates near-elastic transfer along a line of balls.

Try this

Q1. State the principle of conservation of momentum. [1 mark]

  • Cue. With no external force, the total momentum before an interaction equals the total momentum after.

Q2. A 2.0 kg2.0\ \text{kg} trolley at 3.0 m s−13.0\ \text{m s}^{-1} collides and sticks to a 4.0 kg4.0\ \text{kg} stationary trolley. Calculate the common velocity afterwards. [3 marks]

  • Cue. 2.0×3.0=(2.0+4.0)v2.0 \times 3.0 = (2.0 + 4.0)v, so v=6.06.0=1.0 m s−1v = \frac{6.0}{6.0} = 1.0\ \text{m s}^{-1}.

Q3. Explain how an airbag reduces the force on a passenger during a crash. [2 marks]

  • Cue. It lengthens the time to stop the passenger; for the same change in momentum, a longer time means a smaller average force (F=ΔptF = \frac{\Delta p}{t}).

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.

SQA Higher 20185 marksA trolley of mass 0.80 kg moving at 3.0 m per second collides with and sticks to a stationary trolley of mass 1.2 kg. Calculate the velocity of the combined trolleys after the collision, and by calculating the total kinetic energy before and after, determine whether the collision is elastic or inelastic.
Show worked answer →

Conservation of momentum, taking the initial direction as positive.

Relationship: m1u1+m2u2=(m1+m2)vm_1 u_1 + m_2 u_2 = (m_1 + m_2)v.

Substitution: 0.80×3.0+1.2×0=(0.80+1.2)v0.80 \times 3.0 + 1.2 \times 0 = (0.80 + 1.2)v.

Answer: 2.4=2.0v2.4 = 2.0 v, so v=1.2v = 1.2 m per second.

Kinetic energy before: Ek=12×0.80×3.02=3.6E_k = \tfrac{1}{2} \times 0.80 \times 3.0^2 = 3.6 J. Kinetic energy after: Ek=12×2.0×1.22=1.44E_k = \tfrac{1}{2} \times 2.0 \times 1.2^2 = 1.44 J. Since kinetic energy has decreased, the collision is inelastic.

Markers reward momentum conservation, the correct combined velocity, both kinetic energy values, and the inelastic conclusion.

SQA Higher 20214 marksA ball of mass 0.15 kg travelling at 12 m per second strikes a wall at right angles and rebounds at 8.0 m per second in the opposite direction. The contact lasts 0.020 s. Calculate the impulse on the ball and the average force exerted by the wall on the ball.
Show worked answer →

Take the initial direction as positive, so the rebound velocity is negative.

Impulse equals change in momentum. Relationship: impulse=mv−mu\text{impulse} = mv - mu.

Substitution: impulse=0.15×(−8.0)−0.15×12=−1.2−1.8=−3.0\text{impulse} = 0.15 \times (-8.0) - 0.15 \times 12 = -1.2 - 1.8 = -3.0 kg m per second.

Magnitude of impulse is 3.0 N s.

Average force: relationship F=ΔpΔtF = \frac{\Delta p}{\Delta t}, substitution F=3.00.020F = \frac{3.0}{0.020}, answer F=150F = 150 N (directed away from the wall).

Markers reward treating the rebound velocity as negative, the impulse magnitude, and the force with unit.

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