CARA'S TIPS
Further Mathematics · Lesson 20

Momentum, impulse, work–energy & friction (mechanics)

WAECNECO

What the syllabus asks

WAEC: linear momentum and impulse; conservation of linear momentum; direct impact of two bodies, coalescence, recoil; Newton’s law of restitution (introduction); work, energy (kinetic and potential), power; work–energy principle; friction: limiting friction, coefficient of friction, angle of friction, motion on rough horizontal and inclined planes

NECO: linear momentum and impulse; conservation of linear momentum; direct impact of two bodies, coalescence, recoil; Newton’s law of restitution (introduction); work, energy (kinetic and potential), power; work–energy principle; friction: limiting friction, coefficient of friction, angle of friction, motion on rough horizontal and inclined planes

What you will learn

Key points

Start here: Linear momentum

A slow lorry and a fast football can both knock you over. A heavy lorry moving slowly is hard to stop; so is a light bullet moving very fast. The quantity that measures this "quantity of motion" is momentum.

Linear momentum p = mass × velocity = mv. Unit: kg m/s (equivalently N s). Momentum is a vector: it has the direction of the velocity.

Because momentum is a vector, choose a positive direction and give motion the other way a negative sign.

Worked example 20.1

Find the momentum of (a) a 2000 kg car at 15 m/s, (b) a 0.02 kg bullet at 400 m/s, (c) a 0.5 kg ball moving at 6 m/s to the left (take right as positive).

  1. (a) p = 2000 × 15 = 30,000 kg m/s.
  2. (b) p = 0.02 × 400 = 8 kg m/s.
  3. (c) p = 0.5 × (−6) = −3 kg m/s (3 kg m/s to the left).
30,000; 8; −3 kg m/s
Worked example 20.2 (change in momentum)

A 0.4 kg ball moving at 20 m/s hits a wall and rebounds at 15 m/s. Find the change in momentum.

  1. Take the original direction as positive: u = +20, v = −15.
  2. Δp = m(v − u) = 0.4(−15 − 20) = 0.4 × (−35) = −14 kg m/s.
  3. The magnitude is 14 kg m/s, directed away from the wall.
14 N s away from the wall
⚠️ Common mistake: subtracting speeds (20 − 15 = 5) when a body rebounds. The velocity changes sign, so the speeds add: 20 + 15 = 35.
Try it yourself. A 60 kg athlete runs at 9 m/s. Find his momentum.
Show solution
60 × 9 = 540 kg m/s.

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Practice questions

1. Find the momentum of a 1500 kg bus moving at 12 m/s.
Show worked answer
18,000 kg m/s.
2. A 0.3 kg ball at 10 m/s is stopped in 0.02 s. Find the average force.
Show worked answer
3/0.02 = 150 N.
3. A 6 kg body at 5 m/s collides with a 4 kg body at rest; they move together. Find the common velocity.
Show worked answer
30 = 10v → 3 m/s.
4. A 3 kg gun fires a 0.015 kg bullet at 500 m/s. Find the recoil speed.
Show worked answer
7.5/3 = 2.5 m/s.
5. Find the work done when a 40 N force at 60° to the horizontal moves a box 10 m horizontally.
Show worked answer
40 × 10 × 0.5 = 200 J.

Key words

Momentum
Mass × velocity; a vector.
Impulse
Force × time; equals change in momentum.
Conservation of momentum
Total momentum of a system is constant if no external force acts.
Coalesce
Two bodies join and move together after impact.
Recoil
The backward motion of a gun or body when it fires or throws something forward.
Coefficient of restitution e
Speed of separation ÷ speed of approach in a direct impact; 0 ≤ e ≤ 1.
Work
Force × distance moved in the direction of the force; unit joule.
Kinetic energy
Energy of motion, ½mv².
Potential energy
Energy due to position, mgh.
Power
Rate of doing work; unit watt (J/s).
Friction
A force opposing relative motion between surfaces in contact.
Limiting friction
The maximum frictional force, μR, when sliding is about to start.
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