Elastic collisions — 1D - solve for final velocities: Visual walkthrough

Step 1 — The picture we are trying to explain

WHAT. Two blocks slide on a smooth, straight, frictionless track. We agree that rightward = positive — this is a choice, but once we make it, every velocity is a signed number: a block moving left has a negative velocity.

Let me name everything, so no symbol appears later without meaning:

WHY. Before we can conserve anything, we must know exactly what "before" and "after" mean and which way is positive. Sign discipline is where most collision mistakes are born.

PICTURE. Two blocks, arrows showing signed velocities, the track and the positive direction marked.

Step 2 — Fact one: total momentum can't change

WHAT. Momentum of one block is its mass times its velocity, $m v$ — a signed number. The total momentum is the sum for both blocks. We claim this total is the same before and after the crash:

$block 1’s push, before m_{1} u_{1} + block 2’s push, before m_{2} u_{2} = block 1’s push, after m_{1} v_{1} + block 2’s push, after m_{2} v_{2}$

WHY this law? During the collision the blocks push on each other, but by Newton's third law those two forces are equal and opposite — they cancel in the total. With no outside horizontal force, the total momentum has nothing that could change it. See Conservation of Momentum.

PICTURE. A "momentum bar" for each block; the two bars stacked before equal the two bars stacked after, even though the split changes.

Step 3 — Fact two: total kinetic energy can't change either

WHAT. Kinetic energy (KE) is the "energy of motion," $\frac{1}{2} m v^{2}$ for one block. Notice the velocity is squared — so a leftward block ( $v < 0$ ) still has positive energy. "Elastic" is the special promise that the total KE is unchanged:

$total KE before \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = total KE after \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$

WHY squared, and why this law? Squaring is what makes energy blind to direction — energy is about how fast, not which way. We need a second equation because we have two unknowns ( $v_{1}$ and $v_{2}$ ); momentum alone leaves infinitely many possibilities. The elastic promise (no heat, no dent, no sound) is exactly the missing constraint. See Kinetic Energy.

PICTURE. Energy drawn as area of squares (side = speed); total tile-area before = total tile-area after.

Step 4 — Group the masses (momentum, rewritten)

WHAT. Take the momentum equation and shove all of block 1 to the left, all of block 2 to the right:

$m_{1} u_{1} - m_{1} v_{1} = m_{2} v_{2} - m_{2} u_{2} ⟹ momentum block 1 *lost* m_{1} (u_{1} - v_{1}) = momentum block 2 *gained* m_{2} (v_{2} - u_{2}) (A)$

WHY. This says something physical and beautiful: whatever momentum block 1 gives up, block 2 picks up. The two grouped chunks $m_{1} (u_{1} - v_{1})$ and $m_{2} (v_{2} - u_{2})$ are the stars of the whole trick — remember their shapes.

PICTURE. An arrow of momentum leaving block 1 and the identical arrow arriving at block 2.

Step 5 — Group the same way in energy (difference of squares)

WHAT. Do the identical grouping to the energy equation. First cancel every $\frac{1}{2}$ (multiply both sides by 2), then gather block 1 on the left, block 2 on the right:

$m_{1} (u_{1}^{2} - v_{1}^{2}) = m_{2} (v_{2}^{2} - u_{2}^{2})$

Now the key algebra tool: a difference of two squares always factors as $a^{2} - b^{2} = (a - b) (a + b)$ .

$same chunk as (A)! m_{1} (u_{1} - v_{1}) (u_{1} + v_{1}) = same chunk as (A)! m_{2} (v_{2} - u_{2}) (v_{2} + u_{2}) (B)$

PICTURE. The area rectangle $u^{2} - v^{2}$ sliced into the $(u - v)$ strip times the $(u + v)$ length.

Step 6 — Divide (B) by (A): the messiness evaporates

WHAT. Equation (B) is equation (A) with an extra factor $(u_{1} + v_{1})$ on the left and $(v_{2} + u_{2})$ on the right. Divide (B) by (A) — the identical chunks cancel top and bottom:

$\frac{m _{1} ( u _{1} - v _{1} ) ( u _{1} + v _{1} )}{m _{1} ( u _{1} - v _{1} )} = \frac{m _{2} ( v _{2} - u _{2} ) ( v _{2} + u _{2} )}{m _{2} ( v _{2} - u _{2} )} ⟹ u_{1} + v_{1} = v_{2} + u_{2}$

Rearrange so "before" quantities sit together and "after" together:

$u_{1} - u_{2} = v_{2} - v_{1}$

WHY divide? Dividing turns two quadratic equations into one linear one. Linear is easy; that's the payoff. (Guard note: we divided by $m_{1} (u_{1} - v_{1})$ , which is fine as long as $u_{1} \neq = v_{1}$ — i.e. block 1 actually changed velocity, i.e. a real collision happened. The no-collision case $v_{1} = u_{1}, v_{2} = u_{2}$ is handled separately in Step 8.)

PICTURE. Approach arrows before, separation arrows after — same length.

Step 7 — Two linear equations → the final formulas

WHAT. We now hold two linear equations:

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2} (1) v_{2} = v_{1} + (u_{1} - u_{2}) (rule)$

Substitute the rule into (1) and solve for $v_{1}$ , then back-substitute for $v_{2}$ :

WHY. One substitution kills one unknown, leaving a single equation in $v_{1}$ . This is the entire "algebra omitted for brevity" from the parent note, done in the open.

PICTURE. The two-line linear system as a pair of crossing lines meeting at one point $(v_{1}, v_{2})$ — a unique solution.

Step 8 — Every case, including the degenerate ones

WHAT. Feed the formulas special inputs and watch them behave. The right formula must survive all of these — never let the reader hit an unshown scenario.

Case	Inputs	Result	Reading
Equal masses, target still	$m_{1} = m_{2}$ , $u_{2} = 0$	$v_{1} = 0, v_{2} = u_{1}$	They swap — Newton's Cradle
Heavy hits light (still)	$m_{1} ≫ m_{2}$ , $u_{2} = 0$	$v_{1} \approx u_{1}, v_{2} \approx 2 u_{1}$	Light flies off at double speed
Light hits heavy wall	$m_{1} ≪ m_{2}$ , $u_{2} = 0$	$v_{1} \approx - u_{1}, v_{2} \approx 0$	Ball bounces back
No collision (miss)	$u_{1} = u_{2}$	$v_{1} = u_{1}, v_{2} = u_{2}$	Nothing happens — approach speed 0

WHY show the degenerate case? When $u_{1} = u_{2}$ the approach speed is zero, so nothing changes — and notice this is exactly the case Step 6 warned about ( $u_{1} = v_{1}$ ). The formulas still return the right answer here, so nothing breaks; we just can't run the "divide by (A)" shortcut to derive it.

PICTURE. Four mini-panels, one per case, arrows before/after.

The one-picture summary

Here is the entire journey compressed: two conservation laws in, grouped, divided, and out come the two formulas — with the special cases hanging off the side.

Recall Feynman: tell the whole walkthrough to a friend

We had two bouncy carts on a smooth track and wanted their speeds after they crash. Two rules never break: the total push (mass × velocity, added up) stays the same, and the total motion-energy (mass × velocity², halved, added up) stays the same. I wrote both rules, then shoved cart 1 to one side and cart 2 to the other in each rule. The push-rule became "push lost by cart 1 = push gained by cart 2." The energy-rule, after factoring a difference of squares, grew the exact same cart-chunks with an extra piece stuck on. So I divided one rule by the other, the ugly chunks cancelled, and what fell out was gorgeous: they separate at the same speed they approached. That's one clean straight-line equation. Pair it with the push-rule, do one substitution, and out pop the two final-velocity formulas. To sanity-check, I plug in equal weights (they trade speeds), a heavy cart hitting a light one (light one shoots off at double speed), and a light cart hitting a wall (it bounces straight back). All four cases behave — so the formula is trustworthy everywhere.

Recall Quick self-test

Approach speed if $u_{1} = 5, u_{2} = - 2$ ? ::: $u_{1} - u_{2} = 5 - (- 2) = 7$ , so they separate at $7$ too. Which algebra tool converts the energy equation into the same chunks as momentum? ::: Difference of squares, $u^{2} - v^{2} = (u - v) (u + v)$ . After dividing (B) by (A), what single equation appears? ::: $u_{1} - u_{2} = v_{2} - v_{1}$ (the relative-velocity rule). Why is the relative-velocity rule preferred over raw energy conservation? ::: It's linear — no squares — so two linear equations solve instantly.

Connections

Conservation of Momentum — the first pillar (Step 2).
Kinetic Energy — the elastic promise (Step 3).
Coefficient of Restitution — Step 6's rule is the $e = 1$ case.
Center of Mass Frame — the slick re-derivation behind Step 8's "double speed."
Newton's Cradle — the equal-mass swap of Step 8.
Inelastic Collisions — 1D — what happens when the KE promise is broken.