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1.3.2States of Matter

Forces and Momentum — Cambridge IGCSE Chemistry 0620 Revision Notes & Topical Questions

5 min read 5 parts Lesson 25 of 30 Unit quiz

📚 Cambridge IGCSE Chemistry 0620 Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 5Key Relationships

A force changes momentum, and the size of the force depends on how quickly the change happens.

Change in momentum
  • Change in momentum The final momentum minus the initial momentum, written as Δp = mv − mu.
Rate of change
  • Rate of change The change in momentum divided by the time taken, Δp ÷ t, which equals the resultant force.
Force equation
  • Force equation F = Δp ÷ t, where F is in newtons, Δp in kg m/s and t in seconds.
Link to F = ma
  • Link to F = ma Since Δp ÷ t equals m × Δv ÷ t, this equation is consistent with F = ma.

Part 2 of 5Calculating a Force

This method works for any collision or push where momentum changes.

1
Find the initial and final velocities, and assign a positive direction.
2
Calculate the momentum before and after, then the change in momentum Δp.
3
Write down the time taken for the change, converting any milliseconds into seconds.
4
Divide the change in momentum by the time to find the force, and state its direction.

Part 3 of 5Real-life Applications

Real-life Applications
Safety
  • Safety Designs that lengthen the time of impact reduce the force on a passenger during a crash.
Sport
  • Sport A cricketer moves the hands back when catching a ball to increase the time and reduce the force.
Vehicle design
  • Vehicle design Crumple zones and airbags spread out the change in momentum over a longer time.
Rockets
  • Rockets A rocket produces a force by expelling gas backwards, which changes its momentum.

Part 4 of 5Worked Example

Question

  • A 0.15 kg ball hits a bat at 20 m/s and rebounds at 10 m/s in the opposite direction.
  • The contact lasts 0.01 s.
  • Calculate the average force on the ball, taking the initial direction as positive.

  1. Initial momentum = 0.15 × 20 = 3.0 kg m/s.
  2. Final momentum = 0.15 × (−10) = −1.5 kg m/s.
  3. Change in momentum Δp = −1.5 − 3.0 = −4.5 kg m/s.
  4. Force F = Δp ÷ t = −4.5 ÷ 0.01 = −450 N.

Answer The average force is 450 N, acting in the opposite direction to the ball's initial motion.

Part 5 of 5Exam Checklist & Tips

1.3.2 – Key Points for the Exam

  • Rate of change of momentum The change in momentum per second, which equals the resultant force.
  • Impact A brief collision in which a large force acts over a short time.
  • Force A push or pull that changes the momentum of an object, measured in newtons.
  • Newton The unit of force, equal to a rate of change of momentum of 1 kg m/s per second.

Common Mistakes

  • Forgetting the sign of the change in momentum when the object reverses direction.
  • Using change in speed instead of change in velocity, which loses the direction information.
  • Entering a time in milliseconds directly into F = Δp ÷ t, which gives a force 1000 times too large.
  • Stating that a larger time of contact increases the force; a longer time reduces the force.

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