Classical-Mechanics · Unit 19 · Video 6 · Interactive Practice

The Right-Angle Rule for Equal Masses, an Unequal-Mass Collision, and the CM Scattering Angle

IKey Formulas

Formula	Name	What it takes
$\vec{v}_{1,i} = \vec{v}_{1,f} + \vec{v}_{2,f}$	Momentum, as a vector	Equal masses: $m_1$ divides out of every term
$v_{1,i}^2 = v_{1,f}^2 + 2\,\vec{v}_{1,f}\cdot\vec{v}_{2,f} + v_{2,f}^2$	The same equation dotted with itself	$\vec{v}\cdot\vec{v} = v^2$ , and the dot product distributes
$\vec{v}_{1,f}\cdot\vec{v}_{2,f} = 0 \;\Longrightarrow\; \theta_{1,f} + \theta_{2,f} = 90\degree$	Right-angle rule	Equal masses, elastic, target at rest, neither speed zero
$\vec{v}_{cm} = \dfrac{m_1\vec{v}_{1,i}}{m_1 + m_2}$	Boost to the centre-of-mass frame	Target 2 at rest; constant, so $m_1\vec{v}^{\,\prime}_{1} + m_2\vec{v}^{\,\prime}_{2} = 0$ before and after

Key Insight: Elasticity with equal masses supplies $v_{1,i}^2 = v_{1,f}^2 + v_{2,f}^2$ . Subtract that from the squared momentum equation and only the cross term survives, so $\vec{v}_{1,f}\cdot\vec{v}_{2,f} = 0$ — a right angle at every scattering angle, with no trigonometry and no named angle.

IIVisualization 1 — One Dot Product, One Right Angle

Equal masses: momentum lets the tip of $\vec{v}_{1,f}$ sit anywhere, but energy pins it to one circle.

θ₁f (deg) 40 v₁f (m/s) 2.30

💡 The rule excludes the two degenerate hits — the head-on strike, where $v_{1,f} = 0$ , and the clean miss, where $v_{2,f} = 0$ — because a zero vector has no direction to be perpendicular to.

IIIVisualization 2 — Momentum Fixes the Speeds, Energy Is the Test

With $m_2 = m_1/3$ and a moving target, two momentum components leave the energy equation free to decide elasticity.

Step 1

Step 1 — The two components of momentum

m_1 v_{1,i} - \tfrac{m_1}{3}\, v_{2,i} \;=\; \tfrac{m_1}{3}\,\tfrac{\sqrt{2}}{2}\, v_{2,f} \qquad (1)

m_1 v_{1,f} \;=\; \tfrac{m_1}{3}\,\tfrac{\sqrt{2}}{2}\, v_{2,f} \qquad (2)

Nothing moves vertically before the collision, so in

(2)

the downward momentum of particle 1 must cancel the upward component of particle 2 — one equation, one unknown.

Step 2 — Solve $(2)$ for $v_{2,f}$

v_{2,f} = \frac{6}{\sqrt{2}}\, v_{1,f} = 3\sqrt{2}\, v_{1,f} = \frac{3\sqrt{2}}{2}\, v_{1,i} \approx 2.121\, v_{1,i}

Particle 2 leaves faster than particle 1 arrived: it is the lighter body, and it must carry off the whole upward momentum.

Step 3 — Substitute into $(1)$

\tfrac{m_1}{3}\cdot\tfrac{\sqrt{2}}{2}\cdot\tfrac{3\sqrt{2}}{2}\, v_{1,i} = \tfrac{1}{2}\, m_1 v_{1,i}

m_1 v_{1,i} - \tfrac{m_1}{3}\, v_{2,i} = \tfrac{1}{2}\, m_1 v_{1,i} \quad\Longrightarrow\quad v_{2,i} = \tfrac{3}{2}\, v_{1,i}

Step 4 — The kinetic energy before

K_i = \tfrac{1}{2} m_1 v_{1,i}^2 + \tfrac{1}{2}\cdot\tfrac{m_1}{3}\left(\tfrac{3}{2} v_{1,i}\right)^{\!2} = \tfrac{4}{8} m_1 v_{1,i}^2 + \tfrac{3}{8} m_1 v_{1,i}^2 = \tfrac{7}{8} m_1 v_{1,i}^2

Both unknown speeds came from momentum alone, so the energy equation was never spent — it is free to serve as a test.

Step 5 — The kinetic energy after, and the verdict

K_f = \tfrac{1}{2} m_1\left(\tfrac{v_{1,i}}{2}\right)^{\!2} + \tfrac{1}{2}\cdot\tfrac{m_1}{3}\left(\tfrac{3\sqrt{2}}{2} v_{1,i}\right)^{\!2} = \tfrac{1}{8} m_1 v_{1,i}^2 + \tfrac{3}{4} m_1 v_{1,i}^2 = \tfrac{7}{8} m_1 v_{1,i}^2

K_f = K_i

, so the collision is elastic — yet the outgoing directions are

135\degree

apart. The right-angle rule needs equal masses and a target at rest, and this collision has neither.

IVVisualization 3 — A Collision Seen from the Centre of Mass

Subtracting the constant $\vec{v}_{cm}$ from every velocity turns the collision back-to-back, in and out.

Frame lab Θcm (deg) 52

💡 The laboratory pair $(v_{1,f}, \theta_{1,f})$ and the single angle $\Theta_{cm}$ carry the same information; converting one into the other is the next video's work.

VQuiz Questions

Problem 1 · The Speed the Target Leaves With

Given: a particle of mass $m$ moving at $v_{1,i} = 4.00\ \text{m/s}$ strikes an identical particle at rest; the collision is elastic and glancing, and particle 1 leaves at $v_{1,f} = 3.00\ \text{m/s}$ — find $v_{2,f}$ .

✅ Correct! Equal masses and elasticity give

v_{1,i}^2 = v_{1,f}^2 + v_{2,f}^2

, so

v_{2,f} = \sqrt{16 - 9} = \sqrt{7} \approx 2.65\ \text{m/s}

— and the two outgoing velocities are

90\degree

apart.

❌ Speeds do not subtract.

\vec{v}_{1,i} = \vec{v}_{1,f} + \vec{v}_{2,f}

is a vector equation; the three vectors form a triangle, not a straight line, so

4.00 - 3.00

is not

v_{2,f}

❌ The hypotenuse is the wrong side.

v_{1,i}

is the long side of the right triangle:

v_{1,f}^2 + v_{2,f}^2 = v_{1,i}^2

, so subtract

9

from

16

rather than adding them.

❌ That is $v_{2,f}^2$ .

16 - 9 = 7

is the square of the speed, in

\text{m}^2/\text{s}^2

; take the square root.

❌ Not quite. Use the equal-mass elastic relation

v_{1,i}^2 = v_{1,f}^2 + v_{2,f}^2

Show solution

Momentum is conserved as a vector, and with $m_2 = m_1$ the common mass divides out:

\vec{v}_{1,i} = \vec{v}_{1,f} + \vec{v}_{2,f}

Dot each side with itself, using $\vec{v}\cdot\vec{v} = v^2$ :

v_{1,i}^2 = v_{1,f}^2 + 2\,\vec{v}_{1,f}\cdot\vec{v}_{2,f} + v_{2,f}^2

Elasticity with equal masses cancels $\tfrac{1}{2}m$ from every term of the energy equation:

v_{1,i}^2 = v_{1,f}^2 + v_{2,f}^2

Comparing the two lines kills the cross term, $\vec{v}_{1,f}\cdot\vec{v}_{2,f} = 0$ , and leaves a Pythagorean relation among the three speeds:

v_{2,f} = \sqrt{v_{1,i}^2 - v_{1,f}^2} = \sqrt{16.0 - 9.0} = \sqrt{7} \approx 2.65\ \text{m/s}

As a check, $\cos\theta_{1,f} = v_{1,f}/v_{1,i} = 0.75$ , so $\theta_{1,f} \approx 41.4\degree$ and $\theta_{2,f} \approx 48.6\degree$ — a sum of $90\degree$ .

Problem 2 · Why $135\degree$ and Not $90\degree$

Given: the collision with $m_2 = m_1/3$ and a moving target, whose outgoing velocities are $135\degree$ apart even though $K_f = K_i = \tfrac{7}{8} m_1 v_{1,i}^2$ — find the ingredient of the right-angle proof that this collision is missing.

✅ Correct! The proof needs both ingredients: equal masses so that

m_1

cancels out of the momentum and energy equations, and elasticity to supply the second equation. Here elasticity holds but the mass condition does not, so the cross term never has to vanish.

❌ It is elastic. The computation gave

K_f = K_i = \tfrac{7}{8} m_1 v_{1,i}^2

exactly, so energy conservation is available here — it simply is not enough on its own.

❌ Momentum is always conserved here. No external force acts, so both components are conserved — that is precisely how

v_{2,f}

and

v_{2,i}

were found. What fails is the simplified form

\vec{v}_{1,i} = \vec{v}_{1,f} + \vec{v}_{2,f}

❌ Equal masses do real work. They are what let

m_1

cancel from both equations so the two squared relations share the same terms. Drop them and the cross term

2\,\vec{v}_{1,f}\cdot\vec{v}_{2,f}

survives — as the

135\degree

here shows.

❌ Not quite. List the two facts the derivation actually used, then check which of them this collision still satisfies.

Show solution

The right-angle argument used exactly two inputs. First, momentum with equal masses and a target at rest:

m_1\vec{v}_{1,i} = m_1\vec{v}_{1,f} + m_1\vec{v}_{2,f} \quad\Longrightarrow\quad \vec{v}_{1,i} = \vec{v}_{1,f} + \vec{v}_{2,f}

Second, elasticity with the same equal masses:

\tfrac{1}{2}m v_{1,i}^2 = \tfrac{1}{2}m v_{1,f}^2 + \tfrac{1}{2}m v_{2,f}^2 \quad\Longrightarrow\quad v_{1,i}^2 = v_{1,f}^2 + v_{2,f}^2

Squaring the first and subtracting the second leaves $2\,\vec{v}_{1,f}\cdot\vec{v}_{2,f} = 0$ .

In this collision the second input holds — $K_f = K_i$ was verified by direct calculation — but the first does not: with $m_2 = m_1/3$ the masses do not cancel, and with $\vec{v}_{2,i} \neq \vec{0}$ the incoming momentum is not $m_1\vec{v}_{1,i}$ alone. So the vector equation never collapses to a triangle of the three velocities, the cross term is under no obligation to vanish, and the outgoing directions come out $135\degree$ apart.

Problem 3 · Two Components, Two Unknown Speeds

Given: $m_1$ moves in the $+x$ direction at $v_{1,i}$ and $m_2 = m_1/3$ moves in the $-x$ direction at an unknown $v_{2,i}$ ; afterwards particle 1 moves straight down at $v_{1,f} = v_{1,i}/2$ and particle 2 moves at $v_{2,f}$ , $45\degree$ above the $+x$ axis — find both unknown speeds.

What is $v_{2,f}$ ?

What is $v_{2,i}$ ?

✅ Correct! The

y

-equation carried only

v_{2,f}

, so it went first; with

v_{2,f}

known the

x

-equation gave

v_{2,i} = \tfrac{3}{2} v_{1,i}

. Momentum alone fixed both speeds.

❌ The mass factor is missing. Particle 2 has only

m_1/3

, so it needs three times the speed to balance particle 1's downward momentum:

v_{2,f} = 3\sqrt{2}\,v_{1,f}

, not the

\sqrt{2}\,v_{1,f} = \tfrac{\sqrt{2}}{2}\,v_{1,i}

you get by dropping it.

❌ That is $3\sqrt{2}\,v_{1,f}$ with $v_{1,f}$ mistaken for $v_{1,i}$ . The problem states

v_{1,f} = v_{1,i}/2

, so

v_{2,f} = 3\sqrt{2}\cdot\tfrac{v_{1,i}}{2} = \tfrac{3\sqrt{2}}{2} v_{1,i}

❌ Only the $y$ -component of $\vec{v}_{2,f}$ balances particle 1. Particle 2 leaves at

45\degree

, so the vertical share is

v_{2,f}\sin 45\degree = \tfrac{\sqrt{2}}{2} v_{2,f}

, not the whole speed.

❌ Check the $y$ -equation.

m_1 v_{1,f} = \tfrac{m_1}{3}\,\tfrac{\sqrt{2}}{2}\, v_{2,f}

, with

v_{1,f} = v_{1,i}/2

❌ One factor of 3 short. The left-over term is

\tfrac{m_1}{3} v_{2,i} = \tfrac{1}{2} m_1 v_{1,i}

; multiplying both sides by 3 gives

v_{2,i} = \tfrac{3}{2} v_{1,i}

❌ The right-hand side is not zero. After the collision particle 2 still carries

x

-momentum

\tfrac{m_1}{3} v_{2,f}\cos 45\degree = \tfrac{1}{2} m_1 v_{1,i}

, so only half of

m_1 v_{1,i}

is left for the

v_{2,i}

term.

❌ Sign slip. Particle 2 comes in along

-x

, so its momentum enters as

-\tfrac{m_1}{3} v_{2,i}

: the equation is

m_1 v_{1,i} - \tfrac{m_1}{3} v_{2,i} = \tfrac{1}{2} m_1 v_{1,i}

, which leaves

\tfrac{1}{2} m_1 v_{1,i}

on the right, not

\tfrac{3}{2} m_1 v_{1,i}

❌ Check the $x$ -equation.

m_1 v_{1,i} - \tfrac{m_1}{3} v_{2,i} = \tfrac{m_1}{3}\,\tfrac{\sqrt{2}}{2}\, v_{2,f}

, with

v_{2,f}

from the first part.

Show solution

Step 1 — $y$ -momentum. Nothing moves vertically before the collision, so the two vertical momenta afterwards must cancel:

m_1 v_{1,f} = \tfrac{m_1}{3}\,\tfrac{\sqrt{2}}{2}\, v_{2,f} \quad\Longrightarrow\quad v_{2,f} = \frac{6}{\sqrt{2}}\, v_{1,f} = 3\sqrt{2}\, v_{1,f}

With $v_{1,f} = v_{1,i}/2$ :

v_{2,f} = \frac{3\sqrt{2}}{2}\, v_{1,i} \approx 2.121\, v_{1,i}

Step 2 — $x$ -momentum. Particle 1 has no $x$ -component afterwards, so

m_1 v_{1,i} - \tfrac{m_1}{3}\, v_{2,i} = \tfrac{m_1}{3}\,\tfrac{\sqrt{2}}{2}\, v_{2,f} = \tfrac{m_1}{3}\cdot\tfrac{\sqrt{2}}{2}\cdot\tfrac{3\sqrt{2}}{2}\, v_{1,i} = \tfrac{1}{2}\, m_1 v_{1,i}

\tfrac{m_1}{3}\, v_{2,i} = \tfrac{1}{2}\, m_1 v_{1,i} \quad\Longrightarrow\quad v_{2,i} = \tfrac{3}{2}\, v_{1,i}

Check with energy. Both speeds came from momentum, so energy is an independent test:

K_i = \tfrac{1}{2} m_1 v_{1,i}^2 + \tfrac{1}{2}\cdot\tfrac{m_1}{3}\left(\tfrac{3}{2} v_{1,i}\right)^{\!2} = \tfrac{4}{8} m_1 v_{1,i}^2 + \tfrac{3}{8} m_1 v_{1,i}^2 = \tfrac{7}{8} m_1 v_{1,i}^2

K_f = \tfrac{1}{8} m_1 v_{1,i}^2 + \tfrac{3}{4} m_1 v_{1,i}^2 = \tfrac{7}{8} m_1 v_{1,i}^2

$K_f = K_i$ , so the collision is elastic — the conclusion of the calculation, not an assumption in it.

Problem 4 · Into the Centre-of-Mass Frame

Given: a particle of mass $m_1$ approaches with speed $v_{1,i}$ along $+x$ and strikes a target of mass $m_2 = 3m_1$ at rest — find the two incoming velocities in the centre-of-mass frame, taking $+x$ as positive.

What is $v^{\,\prime}_{1,i}$ ?

What is $v^{\,\prime}_{2,i}$ ?

✅ Correct! The check is the definition of the frame:

m_1\left(\tfrac{3}{4} v_{1,i}\right) + 3m_1\left(-\tfrac{1}{4} v_{1,i}\right) = 0

. The two particles approach on one line, back-to-back, the lighter one three times faster.

❌ That is $v_{cm}$ itself.

v_{cm} = \tfrac{m_1 v_{1,i}}{m_1 + 3m_1} = \tfrac{1}{4} v_{1,i}

is the boost you subtract, not the boosted velocity:

v^{\,\prime}_{1,i} = v_{1,i} - \tfrac{1}{4} v_{1,i}

❌ The denominator is the total mass.

\tfrac{1}{3} v_{1,i}

\tfrac{m_1}{m_2} v_{1,i}

v_{cm}

divides by

m_1 + m_2 = 4m_1

, not by

m_2 = 3m_1

, so the boost is

\tfrac{1}{4} v_{1,i}

and

v^{\,\prime}_{1,i} = v_{1,i} - v_{cm} = \tfrac{m_2}{m_1+m_2} v_{1,i} = \tfrac{3}{4} v_{1,i}

❌ The boost was not subtracted. Every velocity in the new frame is

\vec{v}^{\,\prime} = \vec{v} - \vec{v}_{cm}

, so particle 1 slows from

v_{1,i}

\tfrac{3}{4} v_{1,i}

❌ Check the boost.

v_{cm} = \tfrac{m_1 v_{1,i}}{m_1 + m_2}

, then

v^{\,\prime}_{1,i} = v_{1,i} - v_{cm}

❌ Only the laboratory sees it at rest. An observer riding at

v_{cm}

sees the target drifting backwards at

-v_{cm}

; if it were still at rest the total momentum in this frame could not be zero.

❌ The two factors are swapped. The heavier particle moves slower:

m_1 v^{\,\prime}_{1,i} + m_2 v^{\,\prime}_{2,i} = 0

forces speeds in inverse proportion to the masses, so the

3m_1

target gets

\tfrac{1}{4} v_{1,i}

and particle 1 gets

\tfrac{3}{4} v_{1,i}

❌ Too fast. The target's laboratory velocity is zero, so in the boosted frame it is exactly

0 - v_{cm} = -\tfrac{1}{4} v_{1,i}

❌ Check the second boost. Particle 2 is at rest in the laboratory, so

v^{\,\prime}_{2,i} = 0 - v_{cm}

Show solution

Only particle 1 is moving, so the centre-of-mass velocity is

v_{cm} = \frac{m_1 v_{1,i} + m_2 \cdot 0}{m_1 + m_2} = \frac{m_1 v_{1,i}}{4m_1} = \tfrac{1}{4}\, v_{1,i}

No external force acts, so $v_{cm}$ is constant and one boost serves before, during and after. Each velocity in the new frame is the laboratory velocity minus $v_{cm}$ :

v^{\,\prime}_{1,i} = v_{1,i} - \tfrac{1}{4} v_{1,i} = \tfrac{3}{4}\, v_{1,i} = \frac{m_2}{m_1 + m_2}\, v_{1,i}

v^{\,\prime}_{2,i} = 0 - \tfrac{1}{4} v_{1,i} = -\tfrac{1}{4}\, v_{1,i} = -\frac{m_1}{m_1 + m_2}\, v_{1,i}

The defining property of the frame follows at once:

m_1 v^{\,\prime}_{1,i} + m_2 v^{\,\prime}_{2,i} = \tfrac{3}{4} m_1 v_{1,i} - 3m_1 \cdot \tfrac{1}{4} v_{1,i} = 0

Zero total momentum puts the two incoming velocities on one line pointing in opposite directions, with the lighter particle faster. After the collision the total is still zero, so the outgoing pair is back-to-back as well — and the single angle between the two lines, $\Theta_{cm}$ , describes the whole collision.

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