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How do Newton's laws and the conservation of momentum govern collisions and explosions?

Newton's laws of motion and momentum: the three laws, linear momentum and its conservation, impulse as the change in momentum, and elastic and inelastic collisions.

A focused answer to the OCR H556 forces and motion content on Newton's laws and momentum, covering the three laws of motion, linear momentum and its conservation, impulse as force times time and the change in momentum, and the distinction between elastic and inelastic collisions.

Generated by Claude Opus 4.812 min answer

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

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What this dot point is asking

OCR wants you to state and apply Newton's three laws of motion, define linear momentum and apply its conservation to collisions and explosions, use impulse as force times time equal to the change in momentum, and distinguish elastic from inelastic collisions.

The answer

Newton's three laws

The momentum form of the second law is the more general statement and is essential where mass changes or where a force acts over a time interval.

Momentum and its conservation

Set up a conservation equation by choosing a positive direction, then writing total momentum before equals total momentum after, being careful with the signs of velocities in opposite directions.

Impulse

Elastic and inelastic collisions

Examples in context

Vehicle safety design uses impulse: crumple zones and airbags extend the collision time, reducing the peak force on occupants for the same change in momentum. Rocket propulsion is conservation of momentum, with exhaust gas ejected backwards so the rocket gains forward momentum. Particle physics relies on momentum conservation to deduce the properties of unseen products in collisions. Newton's cradle demonstrates near-elastic collisions where both momentum and kinetic energy are almost conserved.

Try this

Q1. State Newton's second law in terms of momentum. [1 mark]

  • Cue. The resultant force equals the rate of change of momentum, F=ΔpΔtF = \frac{\Delta p}{\Delta t}.

Q2. A 0.40 kg0.40\ \text{kg} ball moving at 5.0 m s−15.0\ \text{m s}^{-1} is stopped in 0.10 s0.10\ \text{s}. Find the average force. [2 marks]

  • Cue. F=Δpt=0.40×5.00.10=20 NF = \frac{\Delta p}{t} = \frac{0.40 \times 5.0}{0.10} = 20\ \text{N}.

Q3. State the difference between an elastic and an inelastic collision. [2 marks]

  • Cue. Kinetic energy is conserved in an elastic collision but not in an inelastic one; momentum is conserved in both.

Exam-style practice questions

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

OCR 20194 marksA 1500 kg1500\ \text{kg} car travelling at 12 m s−112\ \text{m s}^{-1} collides with a stationary 1000 kg1000\ \text{kg} car. They lock together. Calculate their common velocity after the collision and state whether the collision is elastic.
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Conserve momentum: total before equals total after. (1500×12)+(1000×0)=(1500+1000)v(1500 \times 12) + (1000 \times 0) = (1500 + 1000)v, so 18 000=2500v18\,000 = 2500v and v=7.2 m s−1v = 7.2\ \text{m s}^{-1}.

Check kinetic energy: before =12(1500)(12)2=1.08×105 J= \frac{1}{2}(1500)(12)^2 = 1.08 \times 10^{5}\ \text{J}; after =12(2500)(7.2)2=6.48×104 J= \frac{1}{2}(2500)(7.2)^2 = 6.48 \times 10^{4}\ \text{J}. Kinetic energy has decreased, so the collision is inelastic.

Markers reward conserving momentum, the value 7.2 m s−17.2\ \text{m s}^{-1}, and a kinetic-energy comparison showing it is inelastic.

OCR 20224 marksA 0.058 kg0.058\ \text{kg} tennis ball travelling at 9.0 m s−19.0\ \text{m s}^{-1} is struck and rebounds straight back at 22 m s−122\ \text{m s}^{-1}. The contact lasts 5.0 ms5.0\ \text{ms}. Calculate 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(v−u)=0.058(22−(−9.0))=0.058×31=1.80 N s\Delta p = m(v - u) = 0.058(22 - (-9.0)) = 0.058 \times 31 = 1.80\ \text{N s}.

Impulse equals the change in momentum, 1.80 N s1.80\ \text{N s}.

Average force: F=Δpt=1.805.0×10−3=360 NF = \frac{\Delta p}{t} = \frac{1.80}{5.0 \times 10^{-3}} = 360\ \text{N}.

Markers reward including the sign change of the velocity, the impulse 1.8 N s1.8\ \text{N s}, and the force about 360 N360\ \text{N}.

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