Perfectly inelastic collisions — maximum KE loss: Visual walkthrough

We use only these ideas, and we build each one:

mass — how much "stuff" a lump has (a number, in kilograms);
velocity — how fast and which way something moves (an arrow: length = speed, direction = way it points);
momentum — mass times velocity, the "push" a moving lump carries;
kinetic energy — the "motion-energy" stored in a moving lump.

Everything else we derive.

Step 1 — Draw the two lumps before they touch

WHAT. Two clay blobs slide along one straight line. The left one has mass $m_{1}$ and moves with velocity $u_{1}$ ; the right one has mass $m_{2}$ and velocity $u_{2}$ . A velocity is drawn as an arrow: longer arrow = faster, and the arrow points the way the lump travels. Pointing right we call $+$ , pointing left we call $-$ .

WHY. Before we can talk about what is conserved, we must fix a positive direction so that every speed becomes a signed number. Without signs, a head-on crash and a rear-end bump look the same — and they are not.

PICTURE. Below, the pale-yellow arrow ( $u_{1}$ ) is longer than the blue arrow ( $u_{2}$ ), so the left blob is catching up.

Figure — Perfectly inelastic collisions — maximum KE loss

Step 2 — Give each lump a momentum arrow

WHAT. Momentum is $p = m u$ : take the velocity arrow and stretch it by the mass. A heavy slow lump can carry the same push as a light fast one.

WHY. The single fact that survives every collision is that total momentum does not change (nothing outside pushes the pair — see Conservation of Linear Momentum). So momentum, not velocity, is the quantity we track through the crash. We switch to momentum arrows because those are the arrows that will still add up the same after impact.

PICTURE. Each blob now carries a thick momentum arrow $p_{1} = m_{1} u_{1}$ and $p_{2} = m_{2} u_{2}$ . Their tip-to-tail sum (pink) is the total momentum $p_{tot}$ .

Step 3 — They stick: one lump, one velocity

WHAT. After the crash the blobs are glued into a single lump of mass $m_{1} + m_{2}$ moving at one shared velocity $v$ . This "one velocity for both" is the whole meaning of perfectly inelastic: the front and back are the same object now, so they cannot move at different speeds.

WHY. Because the pieces are glued, we no longer have two unknown final velocities — we have exactly one, $v$ . That single unknown is what momentum conservation will pin down.

PICTURE. Two arrows collapse into one combined arrow $v$ on a bigger blob.

Step 4 — Set the pushes equal, and solve for $v$

WHAT. Total momentum before = total momentum after. The "before" push is $m_{1} u_{1} + m_{2} u_{2}$ ; the "after" push is one lump of mass $m_{1} + m_{2}$ moving at $v$ , i.e. $(m_{1} + m_{2}) v$ . Setting them equal and dividing gives $v$ .

WHY. This is the only equation we are allowed to write. Kinetic energy is off-limits here because some of it secretly leaks into heat during the squish — we don't yet know how much. Momentum leaks nowhere, so it is the safe bookkeeping quantity.

$push before m_{1} u_{1} + m_{2} u_{2} = glued mass (m_{1} + m_{2}) shared velocity v$

Divide both sides by $m_{1} + m_{2}$ :

$v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$

PICTURE. The pink total-momentum arrow from Step 2 is reused unchanged on the big blob; only the mass under it grew, so the arrow-per-mass (the velocity) shrinks. That shrink is $v$ .

Step 5 — Split every velocity into "COM drift" + "relative wiggle"

WHAT. Sit on the centre of mass and watch. From there you see each blob's velocity as two pieces: the common drift $v$ (which the COM shares), plus a leftover relative motion. The relative velocity is $u_{rel} = u_{1} - u_{2}$ — how fast lump 1 approaches lump 2.

WHY. We split like this because momentum conservation freezes the drift piece ( $v$ can't change) but says nothing protecting the relative piece. So all the energy that can be destroyed lives entirely in the relative wiggle. Separating the two pieces is what makes the energy loss obvious.

PICTURE. Each blob's arrow is drawn as (grey drift $v$ ) + (coloured relative part). After sticking, the coloured relative parts vanish; only the grey drift remains.

Step 6 — Measure the energy: before vs after

WHAT. Kinetic energy is $\frac{1}{2} m u^{2}$ for one lump. Add both lumps for "before," and use the single glued lump for "after."

WHY. Now that $v$ is known, KE is the only thing left to compute. We subtract to get the loss.

$K E_{i} = \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2}, K E_{f} = \frac{1}{2} (m_{1} + m_{2}) v^{2}$

PICTURE. KE grows as the square of an arrow's length — so we draw each energy as the area of a square built on its velocity arrow. The two "before" squares are large; the single "after" square (built on the short $v$ ) is small. The missing area is the loss.

Step 7 — Substitute $v$ and watch it collapse to reduced mass

WHAT. Put $v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ into $K E_{f}$ and subtract from $K E_{i}$ . After the dust settles, a beautiful thing happens: the $u_{1}$ and $u_{2}$ only survive as $(u_{1} - u_{2})^{2}$ , and the masses collapse into the single combination $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$ .

WHY. This is the payoff of Step 5. The algebra is just confirming what the split already told us: the lost energy is precisely the kinetic energy of the relative motion.

$Δ K E = K E_{i} - K E_{f} = \frac{1}{2} μ = reduced mass \frac{m _{1} m _{2}}{m _{1} + m _{2}} u_{rel}^{2} (u_{1} - u_{2})^{2} = \frac{1}{2} μ u_{rel}^{2}$

PICTURE. The "missing square" area from Step 6 is redrawn as one clean square of side $u_{rel}$ scaled by $μ$ — the relative-motion energy, now standing alone.

Step 8 — Why this is the MAXIMUM loss (the valley picture)

WHAT. Among all final states that keep total momentum fixed, sticking sits at the very bottom of the kinetic-energy valley.

WHY. In the COM frame the total momentum is zero, so the two final momenta must be equal and opposite: $+ q$ and $- q$ for some number $q$ . The final KE in that frame is $\frac{q ^{2}}{2 m _{1}} + \frac{q ^{2}}{2 m _{2}}$ — a bowl-shaped curve in $q$ that is smallest at $q = 0$ . And $q = 0$ means both lumps at rest in the COM frame = both moving together = sticking. You cannot go lower, because KE can never be negative.

PICTURE. A parabola of final KE versus the leftover relative momentum $q$ ; its minimum is marked exactly at $q = 0$ , the sticking point.

Step 9 — The three edge cases, drawn

WHAT & PICTURE. Three special inputs, each a mini-scene on the board:

Equal masses, one at rest ( $m_{1} = m_{2} = m$ , $u_{2} = 0$ ): $v = u /2$ , and exactly half the KE is lost. The valley floor is halfway up.
Tiny into huge (bullet into block): $v$ is nearly zero, so almost all KE is lost — but not quite, because $p_{tot} \neq = 0$ .
Equal and opposite momenta ( $m_{1} u_{1} = - m_{2} u_{2}$ ): $p_{tot} = 0$ , so $v = 0$ and 100% of KE is destroyed — the valley itself sits at zero.

WHY. These cover every regime: partial loss, near-total loss, and total loss. Nothing else can happen, because $v$ and $Δ K E$ depend only on the signed inputs you feed in.

The one-picture summary

Everything on one board: the two momentum arrows add to $p_{tot}$ (unchanged), the glued lump inherits $v = p_{tot} / (m_{1} + m_{2})$ , and the energy that disappears is the relative-motion square $\frac{1}{2} μ u_{rel}^{2}$ — the depth of the KE valley.

Recall Feynman: the whole walkthrough in plain words

Picture two clay blobs sliding at each other. Give each one an arrow for how it moves, then a fatter arrow for its "push" (mass times velocity). Glue them: now it's one bigger blob, and the only rule the universe enforces is that the total push can't change. So the fat push-arrows just add up and sit on the heavier blob — and because the blob is heavier, its speed comes out smaller. That smaller shared speed is the answer $v$ . Now for the energy: motion-energy grows like the square of the arrow, so watch the areas. Split each blob's motion into a shared drift (which survives) and a leftover "closing-in" wiggle (which doesn't). Sticking erases every bit of the wiggle — and it turns out the erased energy is exactly $\frac{1}{2}$ times a special combined mass $μ$ times the closing speed squared. Why is that the most you can lose? Because if the blobs tried to keep any wiggle, they'd have to move apart again — glue says no. In the balance-point frame, the lowest energy you can reach is "both at rest," and that's precisely what sticking gives. If they were charging at each other with equal-and-opposite pushes, the blob just stops dead and all the motion-energy is gone.

Recall

After sticking, what single quantity is unchanged from before? ::: The total momentum $m_{1} u_{1} + m_{2} u_{2}$ . Why can't we use KE conservation to find $v$ ? ::: KE partly leaks to heat/sound/deformation, so it is unknown until $v$ is found from momentum. The lost KE equals the kinetic energy of what motion? ::: The relative motion, $\frac{1}{2} μ u_{rel}^{2}$ . When is the loss 100%? ::: When total momentum is zero (equal-and-opposite pushes), so $v = 0$ .

Connections

Parent topic — the master results this page draws out.
Conservation of Linear Momentum — the one law used in Steps 2–4.
Centre of mass motion — $v$ is the COM velocity (Step 4).
Reduced mass — the $μ$ that appears in Step 7.
Coefficient of restitution — sticking is the $e = 0$ end; Elastic collisions is $e = 1$ .
Ballistic pendulum — the bullet-into-block edge case (Step 9).
Work–Energy theorem — the missing KE equals the work done in deforming the clay.