Ch 11 · 1.6 Momentum · O Level Physics 5054

Topic 11 of 17 in Motion, Forces and Energy — O Level Physics 5054 — 49 lessons.

Introduction

Why do cars have crumple zones and airbags? Why does a gun recoil when it fires? Both come down to momentum, the "quantity of motion" an object carries, and how forces change it. (Syllabus section 1.6.)

Key ideas — Momentum

p = mv

🧑‍🏫 Teacher's explanation

Imagine you are walking through a crowded hallway. A heavy, slow-moving person and a light, fast-moving person both have 'oomph' or 'impact' when they bump into you. Momentum is the mathematical way of measuring that 'oomph'. It depends on two things: how much stuff is moving (mass) and how fast it is moving (velocity).

📖 Definitions

🔍 Worked example — Easy

A 0.5 ext kg ball is moving at 4 ext m/s to the right. Calculate its momentum.

🔍 Worked example — Standard

A 2.0 ext kg trolley moving at 3 ext m/s hits a stationary 1.0 ext kg trolley. They stick together and move as one. Calculate the final velocity.

🔍 Worked example — Exam-style

A 1500 kg car moving at 20 m/s hits a wall and comes to rest in 0.5 s. Calculate the average resultant force exerted by the wall.

⚠️ Common errors

🧠 Quick check 1

🧠 Quick check 2

🎯 Exam question

Key ideas — Impulse

Impulse is force × the time for which the force acts. It equals the change in momentum:

🧑‍🏫 Teacher's explanation

Imagine you are playing football. If you kick a stationary ball with a massive force for a split second, it flies away. If you push it with that same force for several seconds, it travels much further. This 'effect' of a force acting over time is what we call impulse. It is essentially the measure of how much the motion (momentum) of an object is changed by a force.

📖 Definitions

🔍 Worked example — Easy

A ball of mass 0.5 kg is moving at 4 m/s. A player kicks it with a force of 20 N for 0.1 s. Calculate the impulse delivered to the ball.

🔍 Worked example — Standard

A 1200 kg car traveling at 20 m/s comes to a stop in 4.0 s when the driver applies the brakes. Calculate the average resultant force acting on the car.

🔍 Worked example — Exam-style

A 0.2 kg tennis ball hits a wall at 30 m/s and bounces back at 25 m/s in the opposite direction. The contact time with the wall is 0.02 s. Calculate the average force exerted by the wall on the ball.

⚠️ Common errors

🧠 Quick check 1

🧠 Quick check 2

🎯 Exam question

Key ideas — Force and the rate of change of momentum

Rearranging gives the resultant force as the change in momentum per unit time:

🧑‍🏫 Teacher's explanation

Imagine you are running and suddenly try to stop. If you hit a soft, thick foam mat, you stop slowly over a longer period. If you hit a concrete wall, you stop almost instantly. Even though your change in momentum is the same in both cases, the concrete wall exerts a much higher force because you stop much faster.

📖 Definitions

🔍 Worked example — Easy

A 0.5 kg football is kicked with an initial velocity of 0 m/s. After being struck by a foot, it moves at 20 m/s. Calculate the impulse applied to the ball.

🔍 Worked example — Standard

A car of mass 1200 kg moving at 30 m/s comes to a stop in 6.0 s when the driver applies the brakes. Calculate the average resultant force exerted by the brakes.

🔍 Worked example — Exam-style

A 0.15 kg cricket ball hits a wall with a velocity of 40 m/s and rebounds at 35 m/s in the opposite direction. The collision lasts 0.050 s. Calculate the average force exerted by the wall on the ball.

⚠️ Common errors

🧠 Quick check 1

🧠 Quick check 2

🎯 Exam question

(a) Calculate the change in momentum of the ball. [2] (b) Calculate the average resultant force exerted by the wall on the ball. [2] [4 marks]

Key ideas — Conservation of momentum

When objects collide or push apart, and no external resultant force acts, the total momentum before = total momentum after.

🧑‍🏫 Teacher's explanation

Think of momentum as 'mass in motion.' The more mass an object has, or the faster it is moving, the harder it is to stop. Imagine a heavy bowling ball and a light tennis ball rolling at the same speed; the bowling ball has much more momentum because its mass is greater.

📖 Definitions

🔍 Worked example — Easy

A 0.5 kg ball moving at 4 m/s hits a stationary 0.5 kg ball. After the collision, the first ball moves at 1 m/s. Calculate the velocity of the second ball.

🔍 Worked example — Standard

A 2.0 kg trolley moving at 3.0 m/s collides with a 1.0 kg trolley moving at -1.0 m/s (opposite direction). They stick together. Calculate their common velocity.

🔍 Worked example — Exam-style

A 0.2 kg ball hits a wall with a momentum of 1.5 kg m/s and bounces back with a momentum of -1.2 kg m/s. Calculate the impulse exerted by the wall on the ball.

⚠️ Common errors

🧠 Quick check 1

🧠 Quick check 2

🎯 Exam question

(a) Calculate the impulse applied to the ball. [2] (b) Calculate the change in velocity of the ball. [2] [4 marks]

Worked example

Example 1: momentum. A 1200 kg car moves at 15 m/s. p = mv = 1200 × 15 = 18 000\ kg m/s.

🔍 Worked example — Easy

A 1200 kg car is traveling at a constant velocity of 15 m/s. Calculate its momentum.

🔍 Worked example — Standard

A 0.40 kg football is initially at rest. It is kicked, reaching a velocity of 25 m/s. If the foot is in contact with the ball for 0.050 s, calculate the average resultant force exerted on the ball.

🔍 Worked example — Exam-style

A 2.0 kg trolley moving at 3.0 m/s collides with a stationary 1.0 kg trolley. After the collision, the two trolleys stick together and move as a single unit. Calculate the final velocity of the combined trolleys.

⚠️ Common errors

Common mistakes

Practice

Answers: (1) 70 × 8.0 = 560\ kg m/s. (2) Taking the final direction as positive: 0.15 × 30 - 0.15 × (-20) = 4.5 + 3.0 = 7.5\ kg m/s. (3) F = 7.5 / 0.010 = 750\ N. (4) 3.0 × 2.0 = 6.0 = 4.0 v, so v = 1.5\ m/s. (5) It crumples, increasing the time over which the head's momentum changes, so F = Δ p/Δ t is smaller.

Ch 11 · 1.6 Momentum · O Level Physics 5054

All chapters in Motion, Forces and Energy