Perfectly inelastic collisions — maximum KE loss: Foundations

Before you can read the parent note Perfectly inelastic collisions, every squiggle it uses must mean something to you. Below, we build each symbol from nothing — plain words first, then a picture, then why the topic needs it. Read top to bottom; each brick sits on the one before.

1. Mass — $m$

The picture. Think of a shopping trolley. An empty trolley (small $m$ ) is easy to shove into motion or to stop. A trolley full of bricks (big $m$ ) resists both. Same push, different response — that resistance is mass.

Why the topic needs it. The whole chapter is about two bodies, so we label them $m_{1}$ and $m_{2}$ . The subscripts $1$ and $2$ are just name tags — "object one" and "object two" — nothing mathematical. When they stick, we add them: $m_{1} + m_{2}$ is the mass of the combined blob.

2. Velocity — $u$ , $v$ , and the crucial minus sign

Why two letters? Pure convenience: $u$ = "before" (u comes before v in the alphabet), $v$ = "after". So $u_{1}$ is object 1's velocity before impact; $v$ is the shared velocity after they stick.

The picture — direction is a sign. On a straight line, pick one direction as $+$ (say, rightward). Anything moving left gets a minus sign. A ball at $- 6$ m/s is moving left at $6$ m/s.

Figure — Perfectly inelastic collisions — maximum KE loss

3. Momentum — $p = m v$

Why multiply, not add? A truck at walking pace and a bullet are each hard to stop, but for opposite reasons — huge $m$ vs huge $v$ . Multiplying blends both into one number. A big- $m$ slow object and a small- $m$ fast object can have the same momentum.

The picture. Represent each object's momentum as an arrow: its length is $∣ m v ∣$ (how big) and its direction is which way it points. To combine two objects, lay their arrows tip-to-tail and read the total.

4. Kinetic energy — $K E = \frac{1}{2} m v^{2}$

Why the square, and why the $\frac{1}{2}$ ? The square comes from the Work–Energy theorem: to stop a moving object you must remove work, and the maths of "force over stopping distance" delivers a $v^{2}$ . Doubling the speed quadruples the KE. The $\frac{1}{2}$ is the constant that theorem hands us — accept it for now.

The picture — two axes tell two different stories.

5. Change and loss — $Δ$ and $Δ K E$

The picture. If you had $27$ J of motion-energy before and $0$ J after, the $Δ K E = 27$ J didn't vanish from the universe — it turned into heat, sound, and the permanent squish of the clay. $Δ$ is just the subtraction that measures that leak.

Subscripts $i$ and $f$ simply mean initial (before) and final (after). No maths — labels.

6. The reduced mass — $μ$

Why invent it? The KE-loss formula collapses into the tidy $Δ K E = \frac{1}{2} μ u_{r e l}^{2}$ — the same shape as ordinary $\frac{1}{2} m v^{2}$ , but for the relative motion of the two bodies. It packages the two-body problem into a one-body picture. You meet it again in Reduced mass.

7. Relative velocity — $u_{r e l} = u_{1} - u_{2}$

The picture. Ride on object 2. Object 1 rushes toward you at $u_{1} - u_{2}$ . When they stick, that closing motion is exactly what gets destroyed — after sticking, neither moves relative to the other. That destroyed relative-motion energy is the $Δ K E$ .

Why the topic needs it. $Δ K E$ depends only on this relative velocity, squared. It also links straight to the Coefficient of restitution (which is $0$ here — no bounce-apart at all).

8. Centre of mass — $v_{c m}$

The punchline of the whole topic: the shared final velocity $v$ is identical to $v_{c m}$ . Sticking together simply means both bodies come to rest relative to the centre of mass, killing all relative-motion KE. See Centre of mass motion.

How the foundations feed the topic

Read it as: mass and signed velocity build momentum and kinetic energy; conserved momentum gives the shared final velocity (= the COM velocity); reduced mass and relative velocity give the KE that dies — and together they explain why the loss is maximal.

Equipment checklist

Test yourself — cover the right side. If any answer is fuzzy, reread that section before the parent note.

What does

m

measure, in one phrase?

The amount of stuff / resistance to a change in motion (kg).

Why must velocity carry a

+

-

sign?

It is speed with direction; a leftward motion is negative on a rightward-positive line.

What is momentum and its formula?

"How hard to stop"

= p = m v

; it keeps the sign of

v

Why is momentum, not energy, conserved in a crash?

Internal collision forces are equal and opposite, so total momentum is unchanged with no external force; energy can leak to heat.

Write the kinetic-energy formula and say why

v

is squared.

K E = \frac{1}{2} m v^{2}

; the

v^{2}

comes from the work–energy theorem and makes KE grow as a curve, always positive.

What does

Δ

mean?

"The change in" — here

Δ K E = K E_{i} - K E_{f}

, the energy lost.

Define reduced mass and give its value for equal masses

m

μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}

; equals

m /2

for equal masses.

What is

u_{r e l}

and what happens to it when bodies stick?

u_{r e l} = u_{1} - u_{2}

, the closing speed; sticking destroys it entirely.

What is the shared final velocity equal to?

The centre-of-mass velocity

v_{c m} = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}

In one sentence, why is straight-line momentum vs curved energy the key?

You can keep the momentum-sum fixed while lowering the energy-sum, and sticking hits the bottom of that dip — maximum KE loss.

Connections

Conservation of Linear Momentum — the surviving law built from $p = m v$ .
Work–Energy theorem — where the $\frac{1}{2} m v^{2}$ and its $v^{2}$ come from.
Reduced mass — the $μ$ packaging introduced here.
Centre of mass motion — the shared $v$ is $v_{c m}$ .
Coefficient of restitution — sticking means $e = 0$ .
Elastic collisions — the opposite extreme where KE fully survives.
Ballistic pendulum — a classic application of these foundations.