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What is conserved when objects collide?

Linear momentum as the product of mass and velocity, conservation of momentum in collisions and explosions, impulse as the change in momentum, and elastic versus inelastic collisions.

A focused answer to the Edexcel 9PH0 momentum content, covering linear momentum, the conservation of momentum in collisions and explosions, impulse as the change in momentum, and the distinction between elastic and inelastic collisions.

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  1. What this dot point is asking
  2. The answer
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What this dot point is asking

Edexcel wants you to define linear momentum, apply conservation of momentum to collisions and explosions in one dimension, use impulse as the change in momentum (and the area under a force-time graph), and distinguish elastic collisions (kinetic energy conserved) from inelastic collisions (kinetic energy not conserved).

The answer

Linear momentum

Newton's second law in its fundamental form is F=ΔpΔtF = \frac{\Delta p}{\Delta t}, the rate of change of momentum. For constant mass this reduces to F=maF = ma, but the momentum form is the more general statement and is needed for variable-mass problems such as rockets.

Conservation of momentum

Conservation of momentum applies equally to collisions (objects coming together) and explosions (a body flying apart). In an explosion the total momentum is zero before and after, so the fragments carry equal and opposite momenta. Always set a positive direction first and treat velocities with consistent signs.

Impulse

Impulse explains crash safety: extending the contact time Δt\Delta t for a given change in momentum reduces the average force. Crumple zones, airbags, and bending the knees on landing all lengthen the impact time and so cut the peak force.

Elastic and inelastic collisions

In an elastic collision, both momentum and total kinetic energy are conserved (as in idealised gas molecule collisions or snooker balls to a good approximation). In an inelastic collision, momentum is still conserved but kinetic energy is not, the lost kinetic energy becoming internal energy, sound or permanent deformation. A perfectly inelastic collision is one where the objects stick together and move off with a common velocity, the maximum loss of kinetic energy consistent with momentum conservation.

Examples in context

Rocket and jet propulsion conserve momentum by expelling exhaust gas backwards. Car safety design uses the impulse relation: airbags and crumple zones extend the collision time to lower the force on occupants. Newton's cradle approximates a series of elastic collisions, transferring momentum and kinetic energy along the line of balls. Recoil of a gun, the kick of a fired cannon, and the splitting of a radioactive nucleus into a recoiling daughter and an alpha particle are all explosion-type momentum problems.

Try this

Q1. Define linear momentum and give its unit. [1 mark]

  • Cue. Mass times velocity, p=mvp = mv, in kilogram metres per second.

Q2. A 2.02.0 kg trolley at 3.03.0 m per second collides and sticks to a stationary 1.01.0 kg trolley. Find their common speed. [2 marks]

  • Cue. 2.0×3.0=3.0×v2.0 \times 3.0 = 3.0 \times v, so v=2.0v = 2.0 m per second.

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

  • Cue. It increases the time over which the passenger's momentum changes, and since impulse =FΔt= F\Delta t is fixed, a longer Δt\Delta t means a smaller average force.

Exam-style practice questions

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

Edexcel 20194 marksA 12001200 kg car travelling at 1515 m per second collides with a stationary 800800 kg car and they move off together. Calculate their common velocity immediately after the collision.
Show worked answer →

Conservation of momentum: total momentum before equals total momentum after.

Before: p=1200×15+800×0=18000p = 1200 \times 15 + 800 \times 0 = 18000 kg m per second.

After (combined mass 20002000 kg moving at vv): 2000v=180002000v = 18000, so v=9.0v = 9.0 m per second.

Markers reward applying conservation of momentum, the combined mass after a sticking collision, and the value 9.09.0 m per second.

Edexcel 20225 marksA 0.0580.058 kg tennis ball strikes a racket at 3030 m per second and rebounds at 2424 m per second in the opposite direction. The contact lasts 8.08.0 ms. Determine the impulse on the ball and the average force exerted by the racket.
Show worked answer →

Take the rebound direction as positive. Change in momentum: Δp=m(vu)=0.058×(24(30))=0.058×54=3.13\Delta p = m(v - u) = 0.058 \times (24 - (-30)) = 0.058 \times 54 = 3.13 kg m per second.

Impulse equals change in momentum, so impulse =3.13= 3.13 N s.

Average force: F=ΔpΔt=3.138.0×103=3.9×102F = \frac{\Delta p}{\Delta t} = \frac{3.13}{8.0 \times 10^{-3}} = 3.9 \times 10^{2} N.

Markers reward treating velocity as a vector (sign change on rebound), impulse equal to change in momentum, and the force about 390390 N.

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