Ch 3 · 4.3 Momentum · A Level Mathematics 9709
Topic 3 of 5 in Mechanics — A Level Mathematics 9709 — 58 lessons.
Introduction
Snooker balls, railway wagons coupling together, a skater pushing off a wall: when objects collide or push apart, the total momentum stays the same. That single law lets you find speeds after a collision without knowing the forces during it. (Syllabus section 4.3, Paper 4.)
Key ideas — Linear momentum
The momentum of a body of mass m moving with velocity v is
🧑🏫 Teacher's explanation
Imagine you are playing a game of pool. A heavy, fast-moving cue ball hits a stationary ball. The cue ball's motion changes, and the target ball starts moving. Momentum is essentially the 'quantity of motion' an object carries. It depends on two things: how much stuff is in the object (mass) and how fast it is moving (velocity).
📖 Definitions
🔍 Worked example — Easy
A ball of mass 0.5 kg moves with a velocity of 4 m s^-1 to the right. Calculate its momentum, taking right as the positive direction.
🔍 Worked example — Standard
A 2 kg trolley moving at 3 m s^-1 to the right hits a stationary 1 kg trolley. After the collision, they move together as a single unit. Calculate their common velocity.
🔍 Worked example — Exam-style
A 5 kg object moving at 4 m s^-1 to the left hits a 3 kg object moving at 2 m s^-1 to the right. After the impact, the 5 kg object moves at 1 m s^-1 to the left. Find the velocity of the 3 kg object after the impact.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Conservation of momentum
When two bodies collide directly (along the line joining them) and no external force acts along that line, the total momentum before equals the total momentum after:
🧑🏫 Teacher's explanation
Imagine you are playing pool. When the white cue ball hits a colored ball, the cue ball slows down or stops, and the colored ball zooms away. Even though they look like they are doing different things, the 'amount of motion' (momentum) is simply being transferred from one to the other. In physics, we say the total amount of motion stays the same if no outside force, like friction, interferes.
📖 Definitions
🔍 Worked example — Easy
A 2 kg trolley moving at 3 m s^-1 collides with a stationary 1 kg trolley. After the collision, the 2 kg trolley moves at 1 m s^-1. Find the velocity of the 1 kg trolley.
🔍 Worked example — Standard
A 0.5 kg ball moving at 4 m s^-1 to the right hits a 1.5 kg ball moving at 2 m s^-1 to the left. They stick together after impact. Find their common velocity.
🔍 Worked example — Exam-style
A 5 kg object moving at 2 m s^-1 hits a 10 kg object moving at 1 m s^-1 in the opposite direction. After collision, the 5 kg object rebounds at 3 m s^-1 in its original direction. Find the final velocity of the 10 kg object.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Bodies that coalesce
If the bodies stick together ("coalesce"), they move off with one common velocity v:
🧑🏫 Teacher's explanation
Imagine two different sized cars crashing into each other. If they are just glancing blows, they bounce off separately. But if they are designed to crumple and lock together, they become one single, larger object moving at a new speed. This is what we call 'coalescing'.
📖 Definitions
🔍 Worked example — Easy
A 2 kg mass moving at 3 m/s collides with a stationary 4 kg mass. If they stick together, find their common velocity.
🔍 Worked example — Standard
A 5 kg trolley moving at 4 m/s to the right hits a 3 kg trolley moving at 2 m/s to the left. They coalesce. Find the velocity of the combined mass.
🔍 Worked example — Exam-style
A 1200 kg car moving at 20 m/s hits a stationary 3000 kg truck. They stick together. Calculate the velocity of the combined vehicles.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Collisions in a sequence
For three or more particles in a line, deal with one collision at a time, using the velocities after the first collision as the "before" velocities of the next. Questions may ask whether a further collision happens: compare the speeds and directions of the particles afterwards.
🧑🏫 Teacher's explanation
Imagine a line of billiard balls. If you hit the first one, it hits the second, which then hits the third. You cannot calculate the movement of the third ball by looking at the first one and the third one simultaneously. You must treat the interaction as a sequence of individual events. Think of it like a relay race: the first runner passes the baton to the second, and only after that handoff can the second runner start their sprint to pass it to the third.
📖 Definitions
🔍 Worked example — Easy
Three particles, A, B, and C, with masses 2 kg, 3 kg, and 1 kg respectively, are moving along a straight line. Initially, A moves at 6 m/s towards B, and B and C are at rest. A hits B, and they coalesce. Calculate the velocity of the combined mass AB immediately after the first collision.
🔍 Worked example — Standard
Particle A (4 kg) moving at 5 m/s hits particle B (6 kg) which is moving at 2 m/s in the opposite direction. After the collision, A moves at 1 m/s in its original direction. Find the velocity of B after the collision.
🔍 Worked example — Exam-style
Three particles P, Q, and R with masses m, 2m, and m are in a line. P moves at 10 m/s towards Q. Q is at rest. After P hits Q, they coalesce and move together. This combined mass then hits R, which is at rest. If the final velocity of the combined mass PQ after hitting R is 2 m/s, find the mass of R in terms of m.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked examples — Example 1: coalescing
A 3 kg trolley moving at 4 m s⁻¹ hits a stationary 2 kg trolley and they stick together. 3(4) + 2(0) = 5v ⇒ v = 2.4 m s⁻¹.
🔍 Worked example — Easy
A 5 kg object moving at 2 m s^-1 hits a stationary 3 kg object. They stick together. Calculate the final velocity.
🔍 Worked example — Standard
A 3 kg trolley moving at 4 m s^-1 hits a stationary 2 kg trolley and they stick together. Calculate the loss of kinetic energy during the collision.
🔍 Worked example — Exam-style
A 10 kg block moving at 3 m s^-1 collides with a 5 kg block moving in the opposite direction at 2 m s^-1. They stick together. Find the velocity of the combined mass.
⚠️ Common errors
Worked examples — Example 2: moving towards each other
Sphere A (2 kg, 6 m s⁻¹) and sphere B (3 kg, 2 m s⁻¹) move towards each other and collide. After the collision A moves back at 1.5 m s⁻¹. Find B's velocity.
🔍 Worked example — Easy
A 5 kg object moving at 4 m s⁻¹ hits a stationary 3 kg object. Find the total momentum before the collision.
🔍 Worked example — Standard
Sphere A (2 kg, 6 m s⁻¹) and sphere B (3 kg, 2 m s⁻¹) move towards each other. After collision, A moves back at 1.5 m s⁻¹. Find B's velocity, taking A's original direction as positive.
🔍 Worked example — Exam-style
A 4 kg trolley moving at 5 m s⁻¹ collides with a 6 kg trolley moving at 2 m s⁻¹ in the opposite direction. They stick together after the collision. Find their common velocity.
⚠️ Common errors
Worked examples — Example 3: a chain of collisions
Particles A (1 kg), B (2 kg) and C (3 kg) lie in a straight line. A is projected at 5 m s⁻¹ towards B. After A hits B, A is at rest. B then hits C and they coalesce.
🔍 Worked example — Easy
A 2 kg mass moving at 4 m s^-1 hits a stationary 2 kg mass. After the collision, the first mass is at rest. Calculate the velocity of the second mass.
🔍 Worked example — Standard
Particles A (1 kg), B (2 kg) and C (3 kg) lie in a straight line. A is projected at 5 m s^-1 towards B. After A hits B, A is at rest. B then hits C and they coalesce. Calculate the final velocity of the B-C unit.
🔍 Worked example — Exam-style
A 5 kg object moving at 3 m s^-1 collides with a 10 kg object moving at 1 m s^-1 in the opposite direction. They coalesce on impact. Calculate the velocity of the combined mass.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) 12000 = 2500v ⇒ v = 4.8 m s⁻¹. (2) 2 = 0.4 + 0.6v ⇒ v = 2.67 m s⁻¹. (3) 12 - 10 = 6v ⇒ v = tfrac13 m s⁻¹ in A's original direction. (4) 18 + 25 - tfrac12(6) (tfrac13 )^2 = 43 - tfrac13 = 42.7 J. (5) 0 = 0.02(300) - 2v ⇒ v = 3 m s⁻¹.