Ch 4 · Momentum and Impulse · IGCSE Physics 0625
Topic 4 of 5 in Motion, Forces and Energy — IGCSE Physics 0625 — 19 lessons.
Introduction
A slow-moving lorry is much harder to stop than a fast tennis ball, even though the ball is moving faster. The quantity that captures "how hard something is to stop" is momentum. Momentum is especially powerful because, in any collision or explosion, the total momentum stays the same. It also explains why airbags, crumple zones and bending your knees on landing all reduce injuries.
Key ideas
1. Momentum. The momentum p of an object is its mass multiplied by its velocity:
🧑🏫 Teacher's explanation
Think of momentum as 'mass in motion.' The heavier an object is and the faster it moves, the harder it is to stop. Imagine a heavy truck and a small bicycle both moving at 5 m/s; the truck has much more momentum because its mass is much larger. Because momentum depends on direction, we must treat it as a vector—if one object moves right (+), we call that positive, and if another moves left (-), we call that negative.
📖 Definitions
🔍 Worked example — Easy
A 0.50 kg football is kicked with a velocity of 20 m/s. Calculate its momentum.
🔍 Worked example — Standard
A car of mass 1200 kg moving at 15 m/s hits a wall and comes to a complete stop in 0.60 s. Calculate the average resultant force exerted by the wall on the car.
🔍 Worked example — Exam-style
A 2.0 kg trolley moving at 3.0 m/s towards the right collides with a stationary 1.0 kg trolley. After the collision, they stick together and move as one mass. Calculate their common velocity.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked example
Example 1 — impulse and force. A 0.40 kg football at rest is kicked and leaves the boot at 25 m/s. The boot is in contact with the ball for 0.020 s.
🔍 Worked example — Easy
A 0.40 kg football is at rest. It is kicked and leaves the boot at 25 m/s. The boot is in contact with the ball for 0.020 s. Calculate the average force exerted on the ball.
🔍 Worked example — Standard
A 2.0 kg trolley moving at 3.0 m/s collides with a stationary 1.0 kg trolley. They stick together after the collision. Calculate their common velocity.
🔍 Worked example — Exam-style
A 1.5 kg ball moving right at 4.0 m/s hits a 0.50 kg ball moving left at 6.0 m/s. They stick together. Calculate the velocity of the combined mass. (Take right as positive).
⚠️ Common errors
Common mistakes and examiner tips
Practice
Answers: (1) p = 1200 × 15 = 18 000\ kg m/s = 1.8 × 10^4\ kg m/s. (2) impulse = 0.060 × 50 = 3.0\ N s; F = 3.0 / 0.0040 = 750\ N. (3) Δ p = 900 × 20 = 18 000\ kg m/s; (a) F = 18 000/0.10 = 1.8 × 10^5\ N; (b) F = 18 000/0.50 = 3.6 × 10^4\ N. (4) 3.0 × 2.0 = 6.0\ kg m/s; v = 6.0/4.0 = 1.5\ m/s. (5) 40 × 1.5 = 60\ kg m/s, so 60v = -60 and v = 1.0\ m/s in the opposite direction. (6) Taking the rebound direction as positive: Δ p = 0.20 × 3.0 - 0.20 × (-5.0) = 1.6\ N s; F = 1.6/0.040 = 40\ N. (7) Her change in momentum is the same on either surface, but the mat increases the time taken to stop, so the rate of change of momentum, and therefore the force on her, is smaller.