Lesson 11: Work, Energy & Momentum | A Level Mathematics

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1. Work Done by a Force M1

Work Done: Energy transferred to or from an object by a force. Work is a scalar quantity measured in Joules (J). Only the component of force in the direction of motion does work.

Work Done Formulae

Work = Force × distance moved in direction of force

W = F · d = F d cosθ (θ = angle between force and displacement)

Work done against gravity (height h): W_gravity = mgh

Work done by friction (distance d): W_friction = −μRd (always negative — energy lost)

Work done by variable force: W = ∫F dx (use calculus)

📐 Worked Example 1 — Work Done by Multiple Forces

A block of mass 5 kg is dragged 8 m up a rough incline at 20° to the horizontal. A force of 60 N acts parallel to the slope. μ = 0.25. Find: (a) work done by the applied force (b) work done against gravity (c) work done against friction (d) total work done.

(a) W_applied = 60 × 8 = 480 J

(b) h = 8 sin20° = 2.736 m
W_gravity = mgh = 5 × 9.8 × 2.736 = 134 J (against gravity)

(c) R = 5 × 9.8 × cos20° = 46.0 N. F_friction = μR = 0.25 × 46.0 = 11.5 N
W_friction = 11.5 × 8 = 92.0 J (against friction)

(d) Net work = 480 − 134 − 92 = 254 J (this equals the gain in KE)

2. Kinetic and Potential Energy M1

Energy Formulae

Kinetic Energy: KE = ½mv²

Gravitational PE: GPE = mgh (h = height above reference level)

Elastic PE (spring): EPE = ½ke² (k = spring constant, e = extension)

Work-Energy Theorem: Net work done = Change in KE = ½mv² − ½mu²

Conservation of Mechanical Energy (no friction): KE + GPE = constant

Energy equation (with friction): Initial ME = Final ME + Energy lost to friction

½mu² + mgh₁ = ½mv² + mgh₂ + W_friction

📐 Worked Example 2 — Conservation of Energy on a Smooth Surface

A particle of mass 2 kg slides from rest down a smooth curved surface from a height of 5 m. Find its speed at the bottom.

No friction → mechanical energy conserved.
Loss in GPE = Gain in KE:
mgh = ½mv² → gh = ½v² → v² = 2gh = 2 × 9.8 × 5 = 98

v = √98 = 7√2 ≈ 9.90 m/s
Note: the mass cancels — speed is independent of mass for smooth surfaces.

📐 Worked Example 3 — Energy with Friction

A 4 kg block slides 10 m down a rough incline at 30°. μ = 0.2. It starts from rest. Find the speed at the bottom.

Height lost: h = 10 sin30° = 5 m. GPE lost = 4 × 9.8 × 5 = 196 J
R = 4 × 9.8 × cos30° = 33.9 N. Friction = 0.2 × 33.9 = 6.79 N
W_friction = 6.79 × 10 = 67.9 J

Energy equation: GPE lost = KE gained + W_friction
196 = ½ × 4 × v² + 67.9
2v² = 128.1 → v = √64.05 = 8.00 m/s

📐 Worked Example 4 — Work-Energy Theorem

A car of mass 1500 kg accelerates from 10 m/s to 30 m/s over a distance of 200 m on a level road. Resistance is 600 N. Find the driving force.

Change in KE = ½ × 1500 × 30² − ½ × 1500 × 10²
= ½ × 1500 × (900−100) = 750 × 800 = 600 000 J

Net work done = (F − 600) × 200 = 600 000
F − 600 = 3000 → F = 3600 N

3. Power M1

Power: The rate of doing work — how quickly energy is transferred. Measured in Watts (W) where 1 W = 1 J/s.

Power Formulae

Power = Work done / time = W/t

P = Fv (force × velocity, when force is constant)

P = F · v (dot product when force and velocity are vectors)

Maximum speed: when driving force = total resistance (acceleration = 0)

At maximum speed: P = F_resistance × v_max

📐 Worked Example 5 — Power and Maximum Speed

A car has engine power 40 kW and experiences resistance of 500 N on a level road. (a) Find the maximum speed on the level. (b) Find the maximum speed up a slope of sin⁻¹(1/15) if the car has mass 1200 kg. (c) Find the acceleration when travelling at 20 m/s on the level.

(a) At max speed, F_drive = F_resist = 500 N
v_max = P/F = 40000/500 = 80 m/s

(b) On slope, total resistance = 500 + mg sinθ = 500 + 1200 × 9.8 × (1/15) = 500 + 784 = 1284 N
v_max = 40000/1284 = 31.2 m/s

(c) At v=20 m/s: F_drive = P/v = 40000/20 = 2000 N
Net force = 2000 − 500 = 1500 N
a = F/m = 1500/1200 = 1.25 m/s²

📐 Worked Example 6 — Power with Variable Speed

A pump raises water from a depth of 8 m at a rate of 300 kg per minute. What power is required?

Mass per second = 300/60 = 5 kg/s
Weight lifted per second = 5 × 9.8 = 49 N

Work done per second = Force × distance per second = 49 × 8 = 392 J/s
Power = 392 W

4. Momentum and Impulse M1

Momentum: The product of mass and velocity — p = mv. A vector quantity measured in kg m/s or N s. The total momentum of a closed system is conserved in the absence of external forces.

Momentum and Impulse Formulae

Momentum: p = mv (vector)

Impulse: J = Ft = change in momentum = mv − mu

Impulse-Momentum Theorem: Ft = m(v−u)

Conservation of Momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (closed system, no external forces)

For variable force: Impulse = ∫F dt

📐 Worked Example 7 — Impulse and Momentum

A ball of mass 0.5 kg moving at 8 m/s hits a wall and rebounds at 6 m/s. (a) Find the impulse from the wall. (b) If the contact time is 0.02 s, find the average force from the wall.

Take initial direction as positive. u = +8, v = −6 (rebounding).
Impulse = m(v−u) = 0.5 × (−6−8) = 0.5 × (−14) = −7 N s
The wall exerts an impulse of 7 N s in the direction opposing original motion.

Average force = Impulse/time = 7/0.02 = 350 N

📐 Worked Example 8 — Conservation of Momentum (Direct Collision)

Particle A (mass 3 kg, speed 4 m/s) collides directly with particle B (mass 2 kg, speed 1 m/s in the same direction). After collision A has speed 2 m/s. Find B's speed after collision. Verify momentum is conserved.

Before: total momentum = 3×4 + 2×1 = 12+2 = 14 kg m/s

After: 3×2 + 2×v_B = 14
6 + 2v_B = 14 → v_B = 4 m/s

Check: After momentum = 3×2 + 2×4 = 6+8 = 14 ✓
Also check: v_B = 4 > v_A = 2 ✓ (B moves faster — physically sensible, A cannot catch B).

5. Newton's Law of Restitution M1

Coefficient of Restitution (e): A measure of the elasticity of a collision. Ranges from 0 (perfectly inelastic — objects stick together) to 1 (perfectly elastic — no kinetic energy lost).

Newton's Law of Restitution

e = speed of separation / speed of approach

e = (v_B − v_A) / (u_A − u_B) (taking positive direction consistently)

0 ≤ e ≤ 1

e = 0: perfectly inelastic (coalesce) e = 1: perfectly elastic

For bounce on fixed surface: e = speed after / speed before

📐 Worked Example 9 — Collision with Restitution

Particle A (mass 2 kg, velocity 5 m/s) collides directly with particle B (mass 3 kg, at rest). The coefficient of restitution is e = 0.4. Find the velocities after collision and the energy lost.

Conservation of momentum:
2×5 + 3×0 = 2v_A + 3v_B
2v_A + 3v_B = 10 ...(1)

Newton's law of restitution:
e = (v_B − v_A)/(u_A − u_B) → 0.4 = (v_B − v_A)/(5−0)
v_B − v_A = 2 ...(2)

From (2): v_B = v_A + 2. Sub into (1): 2v_A + 3(v_A+2) = 10
5v_A = 4 → v_A = 0.8 m/s, v_B = 2.8 m/s

KE before = ½×2×25 = 25 J. KE after = ½×2×0.64 + ½×3×7.84 = 0.64+11.76 = 12.4 J
Energy lost = 25−12.4 = 12.6 J

📐 Worked Example 10 — Successive Bounces

A ball is dropped from height 4 m onto a fixed floor. The coefficient of restitution with the floor is e = 0.6. Find the height reached after the first two bounces and the total distance travelled before coming to rest.

Speed just before 1st impact: v₀ = √(2gh) = √(2×9.8×4) = √78.4 = 8.854 m/s
Speed just after 1st impact: v₁ = e×v₀ = 0.6×8.854 = 5.313 m/s
Height after 1st bounce: h₁ = v₁²/(2g) = 28.22/19.6 = 1.44 m

After 2nd bounce: h₂ = e²×h₁ = 0.36×1.44 = 0.518 m
In general: hₙ = e^(2n)×h₀ = (0.6)^(2n)×4

Total distance = h₀ + 2h₁ + 2h₂ + ... = 4 + 2×4×(e²+e⁴+...) = 4 + 8e²/(1−e²)
= 4 + 8×0.36/(1−0.36) = 4 + 2.88/0.64 = 4 + 4.5 = 8.5 m

6. Elastic Strings and Springs (Hooke's Law) M1

Hooke's Law and Elastic Potential Energy

Tension in elastic string/spring: T = ke = λe/l₀

k = spring constant (N/m) λ = modulus of elasticity (N) l₀ = natural length e = extension

Elastic Potential Energy: EPE = ½ke² = λe²/(2l₀)

For elastic string: tension only when extended (T=0 when slack)

For spring: tension when extended, thrust when compressed

📐 Worked Example 11 — Elastic String Energy

A particle of mass 0.5 kg is attached to one end of an elastic string of natural length 2 m and modulus of elasticity 20 N. The other end is fixed to a ceiling. The particle hangs in equilibrium. Find: (a) the extension (b) the EPE stored.

(a) At equilibrium: T = mg → λe/l₀ = mg
20e/2 = 0.5×9.8 → 10e = 4.9 → e = 0.49 m

(b) EPE = λe²/(2l₀) = 20×0.49²/(2×2) = 20×0.2401/4 = 1.20 J

📝 Exam Practice Questions

Q1 [4 marks] — A particle of mass 3 kg is projected up a rough inclined plane (angle 25°, μ=0.35) with initial speed 12 m/s. Using energy methods, find: (a) the distance travelled before coming to rest (b) whether the particle slides back down.

R=3×9.8×cos25°=26.6N. Friction=0.35×26.6=9.32N (opposing motion up, then down).
(a) KE lost=GPE gained+W_friction:
½×3×144=3×9.8×d×sin25°+9.32d
216=(12.43+9.32)d=21.75d → d=9.93 m
(b) Force down slope=mg sin25°=12.43N. Max friction (opposing slide back)=9.32N.
Net force down slope=12.43−9.32=3.11N>0 → particle slides back down.

Q2 [4 marks] — A pump of efficiency 75% raises water from a depth of 12 m at a rate of 500 litres per minute. Find the power input to the pump. (1 litre of water has mass 1 kg.)

Mass per second = 500/60 = 8.33 kg/s
Useful power output = mgh/t = 8.33×9.8×12 = 979 W
Input power = useful power/efficiency = 979/0.75 = 1305 W ≈ 1.31 kW

Q3 [4 marks] — Particles A (mass 4 kg, velocity 6î m/s) and B (mass 2 kg, velocity −3î m/s) collide. After collision they coalesce. Find their common velocity and the energy lost in the collision.

Conservation of momentum: 4×6+2×(−3)=(4+2)v
24−6=6v → v=18/6=3 m/s (in original direction of A)
KE before=½×4×36+½×2×9=72+9=81J
KE after=½×6×9=27J
Energy lost=81−27=54J

Exam Tip: When particles coalesce (stick together), it is a perfectly inelastic collision (e=0). The combined object moves with a single velocity found from conservation of momentum. Maximum energy is always lost in a perfectly inelastic collision.

Q4 [5 marks] — Particle A (2 kg, 8 m/s) collides with stationary particle B (4 kg). After collision, A moves at 2 m/s in the same direction. (a) Find v_B. (b) Find e. (c) Find the energy lost. (d) Show the collision is physically valid.

(a) Momentum: 2×8+0=2×2+4v_B → 16=4+4v_B → v_B=3 m/s
(b) e=(v_B−v_A)/(u_A−u_B)=(3−2)/(8−0)=1/8=0.125
(c) KE before=½×2×64=64J. KE after=½×2×4+½×4×9=4+18=22J.
Energy lost=42J
(d) v_B=3>v_A=2 ✓ (B moves faster than A — separation occurs, not overlap).
0≤e=0.125≤1 ✓. Both velocities in same direction ✓. Physically valid.

Q5 [4 marks] — A car (mass 1000 kg, engine power 25 kW, resistance 400 N) tows a trailer (mass 500 kg, resistance 200 N) on a level road. Find: (a) the maximum speed (b) the tension in the tow bar at half the maximum speed (c) the acceleration at half the maximum speed.

Total resistance = 400+200 = 600N
(a) v_max = P/F_total = 25000/600 = 41.7 m/s
(b) At v=20.8 m/s: F_drive=25000/20.8=1202N
Net force on system=1202−600=602N. a=602/1500=0.401 m/s²
For trailer only: T−200=500×0.401 → T=400N
(c) a=0.401 m/s²

Q6 [4 marks] — A particle of mass 0.3 kg falls from rest and hits the ground after falling 5 m. It rebounds to a height of 3 m. Find: (a) the speed just before impact (b) the speed just after impact (c) the coefficient of restitution (d) the energy lost on impact.

(a) v²=2×9.8×5=98 → v=9.90 m/s
(b) v_after²=2×9.8×3=58.8 → v_after=7.67 m/s
(c) e=v_after/v_before=7.67/9.90=0.775
Alternatively: e=√(h_rebound/h_drop)=√(3/5)=√0.6≈0.775 ✓
(d) KE before=½×0.3×98=14.7J. KE after=½×0.3×58.8=8.82J.
Energy lost=5.88J

Exam Tip: For a ball bouncing on a fixed surface, the elegant shortcut is e = √(h_after/h_before). This comes from combining v = √(2gh) for each bounce — very useful in multi-bounce problems.

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