Work, Energy, Power, Centre of Mass and Collisions

This chapter gives you the two most powerful shortcuts in mechanics. Energy conservation lets you connect a starting state to a finishing state without ever touching the messy path in between, and momentum conservation lets you solve a collision without knowing anything at all about the enormous forces that acted during it. Both are shortcuts you must earn, though, because each has a precise condition: energy is conserved only when no non-conservative force does work, and momentum is conserved only when no net external force acts. Knowing exactly when each applies is worth more marks than any formula in it.

Work — and the Cases Where It Is Zero

Work is the dot product W = F⃗ · s⃗ = Fs cos θ, where θ is the angle between the force and the displacement. For a varying force, W = ∫F⃗ · ds⃗, which is the area under a force–displacement graph. Because it is a dot product, work is a scalar and it carries a sign, and the sign is where most marks are lost.

When work is zero, and why
SituationReason
Carrying a bag horizontallyThe upward force is perpendicular to the horizontal displacement, so cos 90° = 0
Centripetal force in circular motionAlways perpendicular to velocity — which is exactly why speed stays constant
Pushing a wall that does not moveDisplacement is zero, so W = 0 however tired you get
Normal force on a body sliding along a surfacePerpendicular to the motion at every instant
Tension in a string during circular motionAlong the radius, motion along the tangent
⚠️ Friction does not always do negative work
It is repeated everywhere that friction opposes motion and therefore does negative work. That is true for the body that is sliding, but consider a block placed on a moving conveyor belt. Friction is the only horizontal force on the block and it acts forwards, accelerating the block up to belt speed — so friction does positive work on the block. The same applies to the friction that drives a car: without it the wheels would spin and the car would not move at all. The reliable statement is that friction opposes relative sliding at the surface of contact, and whether the work it does on a particular body is positive or negative depends on which body you are looking at and which way it moves.

The Two Theorems, and When Energy Is Conserved

The work–energy theorem says W_net = ΔKE = ½mv² − ½mu², and it is always true — for any force, conservative or not, constant or varying. The conservation of mechanical energy, KE + PE = constant, is a much narrower statement: it holds only when every force doing work is conservative.

Conservative versus non-conservative forces
PropertyConservative (gravity, spring, electrostatic)Non-conservative (friction, viscosity, air drag)
Work depends onOnly the endpoints, never the pathThe path taken — a longer path dissipates more
Work around a closed loopExactly zeroNon-zero and always negative
Potential energy defined?Yes — F = −dU/dxNo
🧠 Use energy when the path is complicated
A bead slides down a smooth curved wire of arbitrary shape from height h. Finding the final speed with forces would need the shape of the curve, the varying normal force, and an integration. With energy it is one line: the normal force does no work because it is always perpendicular to the motion, so ½mv² = mgh and v = √(2gh) — independent of the shape entirely, and the same as free fall from that height. That is the whole art of this chapter: whenever a question gives you a strange path, two heights and asks for a speed, the answer is energy conservation and the shape is a red herring. Note the condition though — the moment the wire is rough, friction does path-dependent work and you need the length of the path after all.
🎯 Power, and why a car cannot accelerate hard at high speed
Power is the rate of doing work, P = dW/dt = F⃗ · v⃗. Written that way it explains something everyday: an engine delivering constant power has F = P/v, so as speed rises the available force falls, and acceleration falls with it. That is why a car pulls away briskly from rest and struggles to gain the last 20 km/h, and it is why maximum speed is reached when the driving force has dropped to equal the drag. The same relation is why a truck climbing a hill changes to a low gear: at constant engine power, less speed buys more force. Note the units — 1 horsepower ≈ 746 W, and 1 kWh = 3.6 × 10⁶ J, a commercial unit of energy rather than power.

Centre of Mass and Collisions

The centre of mass is the point that moves as though all the mass were concentrated there and all external forces acted on it: x_cm = Σmᵢxᵢ / Σmᵢ. Its power comes from one consequence — if the net external force is zero, the centre of mass has constant velocity, no matter what the parts of the system do to each other.

🎯 Why an exploding shell's centre of mass keeps its parabola
A shell in flight explodes into a dozen fragments. The explosion forces are enormous — but they are internal, and internal forces cancel in pairs by Newton's third law. So the only external force is still gravity, and the centre of mass of all the fragments continues along exactly the original parabola as though nothing had happened. This makes a famously hard-looking question easy: if a shell explodes at the top of its path into two equal pieces and one piece falls straight down, the other must move so that the centre of mass still lands where the whole shell would have. Since the centre of mass lands at range R and one half is at R/2, the other half is at 3R/2. No knowledge of the explosion is needed at all.
Collisions — what is conserved in each
TypeMomentumKinetic energye
Perfectly elasticConservedConserved1
InelasticConservedNot conserved — some becomes heat and sound0 < e < 1
Perfectly inelasticConservedMaximum possible loss; bodies move together0

The coefficient of restitution is e = (relative velocity of separation) / (relative velocity of approach). Momentum is conserved in every collision, elastic or not, because the collision forces are internal — that is the fact worth relying on, and kinetic energy is the thing that may or may not survive.

🧠 Elastic collision of equal masses — the result to memorise
In a one-dimensional elastic collision between equal masses where one is at rest, the velocities are simply exchanged: the moving one stops dead and the stationary one leaves with the whole original velocity. Every carrom shot and every break in billiards is this result. The general formulae are v₁ = ((m₁−m₂)/(m₁+m₂))u₁ and v₂ = (2m₁/(m₁+m₂))u₁, and it is worth reading off the two limits rather than memorising them: a light ball hitting a very heavy wall bounces back with almost the same speed (m₂ ≫ m₁ makes v₁ ≈ −u₁), while a heavy ball hitting a very light one barely slows and sends the light one off at nearly twice its own speed. That factor of two surprises students and is a favourite question.
⚠️ Do not use energy conservation on an inelastic collision
The standard trap runs like this: a bullet embeds in a block, and the block then swings up. Students apply energy conservation across the whole process and get the wrong answer, because the embedding is perfectly inelastic and destroys a large fraction of the kinetic energy. The correct method is in two distinct stages. Stage 1, the collision: use momentum conservation only, mv = (m+M)V. Stage 2, the swing after it: now no non-conservative force acts, so use energy conservation, ½(m+M)V² = (m+M)gh. Mixing the two — using energy in stage 1 — is one of the most reliably punished errors in mechanics, and the giveaway that stage 1 is inelastic is always the phrase "embeds", "sticks to" or "they move together".

Key takeaways

  • W = Fs cos θ, a scalar with a sign; work is zero whenever force is perpendicular to displacement.
  • Friction can do positive work — on a block carried forward by a conveyor belt, it is the driving force.
  • The work–energy theorem W_net = ΔKE is ALWAYS true; mechanical-energy conservation needs all forces conservative.
  • Sliding down any smooth curve from height h gives v = √(2gh), independent of the shape.
  • P = F·v, so at constant power the available force falls as speed rises — hence low gears on a hill.
  • Internal forces cancel, so an exploding shell's centre of mass continues on its original parabola.
  • Momentum is conserved in EVERY collision; kinetic energy only in an elastic one (e = 1).
  • Equal masses in a 1-D elastic collision exchange velocities. Bullet-in-block: momentum first, then energy.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A man carries a suitcase horizontally at constant speed across a room. The work done by him on the suitcase is:

    1. mgd, where d is the distance walked
    2. Zero, because the force is perpendicular to the displacement
    3. Negative, because he gets tired
    4. Equal to his kinetic energy
    Show answer

    Answer: B — Zero, because the force is perpendicular to the displacement

    He supports the suitcase with an upward force while the displacement is horizontal, so θ = 90° and cos 90° = 0. Feeling tired is a fact about his metabolism, not about mechanical work — the same reason pushing an immovable wall does zero work despite the effort.

  2. A bead slides from rest down a smooth wire of arbitrary shape through a vertical drop h. Its final speed is:

    1. √(gh)
    2. √(2gh), regardless of the shape of the wire
    3. 2gh
    4. It cannot be found without the shape of the wire
    Show answer

    Answer: B — √(2gh), regardless of the shape of the wire

    The normal force is always perpendicular to the motion and so does no work, leaving gravity as the only force doing work. Energy conservation then gives ½mv² = mgh and v = √(2gh) — the same as free fall from that height, with the shape irrelevant. If the wire were rough this would fail, because friction does path-dependent work.

  3. A bullet of mass m and speed v embeds itself in a block of mass M hanging at rest, and the pair swings up to height h. To relate v and h you must:

    1. Apply energy conservation across the whole process
    2. Apply momentum conservation to the collision, then energy conservation to the swing
    3. Apply momentum conservation across the whole process
    4. Apply energy conservation to the collision, then momentum to the swing
    Show answer

    Answer: B — Apply momentum conservation to the collision, then energy conservation to the swing

    The embedding is perfectly inelastic and destroys a large part of the kinetic energy, so energy conservation is invalid there — but momentum is conserved because the collision forces are internal. After the collision no non-conservative force acts, so energy conservation is valid for the swing. Using energy in stage one is among the most reliably punished errors in mechanics, and the words "embeds itself" are the giveaway.

  4. A shell moving on a parabolic path explodes into several fragments in mid-air. The centre of mass of the fragments:

    1. Stops immediately, since the shell no longer exists
    2. Continues on the original parabolic path
    3. Moves in a straight line after the explosion
    4. Falls vertically downward
    Show answer

    Answer: B — Continues on the original parabolic path

    The explosion forces are internal and cancel in pairs by Newton's third law, so gravity remains the only external force and the centre of mass keeps its original parabola. This is what makes such questions tractable: if one of two equal fragments falls straight down at range R/2, the other must be at 3R/2 so that the centre of mass still lands at R.

  5. A ball of mass m moving at u collides head-on and elastically with a stationary ball of the same mass. After the collision:

    1. Both move forward at u/2
    2. The first stops and the second moves at u
    3. The first rebounds at u and the second stays at rest
    4. Both move forward at u
    Show answer

    Answer: B — The first stops and the second moves at u

    For equal masses in a one-dimensional elastic collision the velocities are exchanged, so the incoming ball stops dead and the target leaves with the full speed u. Option 1 describes a perfectly inelastic collision, which would conserve momentum but not kinetic energy, and option 4 violates momentum conservation outright.

  6. Which quantity is conserved in ALL collisions, elastic and inelastic alike?

    1. Kinetic energy
    2. Linear momentum
    3. Both kinetic energy and momentum
    4. Neither, unless the surface is frictionless
    Show answer

    Answer: B — Linear momentum

    The collision forces are internal to the two-body system, so momentum is always conserved. Kinetic energy is conserved only when the collision is perfectly elastic (e = 1); in an inelastic collision some of it becomes heat, sound and permanent deformation.

  7. A car engine delivers constant power. As the car speeds up, its acceleration:

    1. Stays constant
    2. Decreases, because F = P/v
    3. Increases, because power is constant
    4. Depends only on the mass of the car
    Show answer

    Answer: B — Decreases, because F = P/v

    Since P = Fv, constant power means the available driving force falls as 1/v, and with it the acceleration. This is why a car pulls away briskly from rest but struggles for the last increment of speed, and why maximum speed is reached when the driving force has fallen to equal the drag.

  8. A block is gently placed on a conveyor belt already moving to the right. The friction on the block:

    1. Acts leftward and does negative work
    2. Acts rightward and does positive work on the block
    3. Is zero, since the block is not being pushed
    4. Acts rightward but does no work
    Show answer

    Answer: B — Acts rightward and does positive work on the block

    The belt moves right relative to the block, so friction on the block acts rightward and accelerates it up to belt speed — force and displacement are in the same direction, so the work is positive. This is the standard counter-example to the belief that friction always opposes motion and always does negative work; the reliable statement is that it opposes relative sliding at the contact.