Classical-Mechanics · Unit 19 · Video 7 · Interactive Practice

Scattering in the Center-of-Mass Frame: Translating Between CM and Lab Angles

IKey Formulas

Formula	Name	What you need
$\vec v_{1,f} = \vec v\,'_{1,f} + \vec v_{cm}$	Add the boost back	The rotated CM velocity and $\vec v_{cm}$
$\tan\theta_{1,f} = \dfrac{\sin\Theta_{cm}}{\cos\Theta_{cm} + m_1/m_2}$	Laboratory scattering angle	$\Theta_{cm}$ and the mass ratio
$v_{1,f} = \dfrac{\sqrt{m_2^2 + 2m_1m_2\cos\Theta_{cm} + m_1^2}}{m_1 + m_2}\,v_{1,i}$	Final laboratory speed	$\Theta_{cm}$ , the masses, $v_{1,i}$
$\dfrac{\Delta K_1}{K_{1,i}} = \dfrac{2m_1m_2\left(\cos\Theta_{cm} - 1\right)}{(m_1 + m_2)^2}$	Fraction of $K_1$ lost	$\Theta_{cm}$ and the masses

Key Insight: Zero total momentum plus conserved energy force $v'_{1,f} = v'_{1,i} = \frac{m_2}{m_1+m_2}v_{1,i}$ and $v'_{2,f} = v'_{2,i} = v_{cm}$ , so an elastic collision is a pure rotation of the relative velocity through $\Theta_{cm}$ , with the one-dimensional reversal as the case $\Theta_{cm} = 180\degree$ . The same triangle run backward returns the CM angle from a detector reading: $\tan\Theta_{cm} = \frac{v_{1,f}\sin\theta_{1,f}}{v_{1,f}\cos\theta_{1,f} - v_{cm}}$ .

IIVisualization 1 — The Boost-Back Triangle

The collision only rotates $\vec v\,'_{1,f}$ ; adding $\vec v_{cm}$ back tips the arrow forward to the laboratory angle.

Mass pair (incident speed 4) Θcm 60°

IIIVisualization 2 — From Triangle to Formula

Two components, one division: the unknown $v_{1,f}$ cancels and only the mass ratio survives.

Step 1

Step 1 — Take components

v_{1,f}\cos\theta_{1,f} = v'_{1,f}\cos\Theta_{cm} + v_{cm}

v_{1,f}\sin\theta_{1,f} = v'_{1,f}\sin\Theta_{cm}

The boost has no perpendicular part.

Step 2 — Divide the second by the first

\tan\theta_{1,f} = \frac{v'_{1,f}\sin\Theta_{cm}}{v'_{1,f}\cos\Theta_{cm} + v_{cm}}

The unknown

v_{1,f}

cancels.

Step 3 — Substitute the CM speeds

v'_{1,f} = \frac{m_2 v_{1,i}}{m_1 + m_2}, \qquad v_{cm} = \frac{m_1 v_{1,i}}{m_1 + m_2}

\tan\theta_{1,f} = \frac{m_2\sin\Theta_{cm}}{m_1 + m_2\cos\Theta_{cm}}

Step 4 — Divide through by $m_2$

\tan\theta_{1,f} = \frac{\sin\Theta_{cm}}{\cos\Theta_{cm} + m_1/m_2}\ \checkmark

m_1 = m_2

gives

\theta_{1,f} = \tfrac{1}{2}\Theta_{cm}

;

\ m_2 \gg m_1

gives

\theta_{1,f} \to \Theta_{cm}

Step 5 — The printed formula fails the check

\frac{m_2\sin\Theta_{cm}}{\cos\Theta_{cm} - m_1/m_2} = \frac{3\sin 60\degree}{\cos 60\degree - \tfrac{1}{3}} = 15.59

\Rightarrow\ \theta = 86.3\degree > \Theta_{cm}

A forward boost cannot enlarge the angle.

💡 Erratum: the textbook prints this conversion as $\tan\theta_{1,i} = \frac{m_2\sin\Theta_{cm}}{\cos\Theta_{cm} - m_1/m_2}$ — a final angle labelled $i$ , the boost subtracted instead of added, and a stray $m_2$ left in the numerator after the mass ratio was divided out. The triangle above gives the correct form.

IVVisualization 3 — Energy Handed to the Target

How much of $K_1$ can a single elastic collision hand to a target at rest?

Mass pair Your guess: largest fraction of K₁ lost 0.50

VQuiz Questions

Problem 1 · Speeds in the CM Frame

Given: $m_1 = 1$ kg moving at $v_{1,i} = 4$ m/s scatters elastically off $m_2 = 3$ kg at rest — find particle 1's speed in the CM frame after the collision, $v'_{1,f}$ , and the CM speed $v_{cm}$ .

✅ Correct! Both are fixed by

v_{1,i}

and the masses alone —

\Theta_{cm}

never enters. And

v_{cm} = v'_{2,f}

: particle 2 leaves the CM frame at the speed of the frame itself.

❌ The masses are swapped. Particle 1's CM speed carries the other mass:

v'_{1,f} = \frac{m_2}{m_1+m_2}v_{1,i}

, while

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

❌ That is the laboratory speed. Boosting into the CM frame subtracts

v_{cm}

v'_{1,i} = v_{1,i} - v_{cm} = 4 - 1 = 3

m/s, and the collision leaves that speed unchanged.

❌ Not quite. Use

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

and

v'_{1,f} = v'_{1,i} = \frac{m_2 v_{1,i}}{m_1+m_2}

with

m_1 + m_2 = 4

kg.

Show solution

The frame moves with

v_{cm} = \frac{m_1 v_{1,i}}{m_1 + m_2} = \frac{(1)(4)}{4} = 1\ \text{m/s}

so particle 1 arrives in that frame at $v'_{1,i} = v_{1,i} - v_{cm} = 3$ m/s, which is also

v'_{1,i} = \frac{m_2\,v_{1,i}}{m_1 + m_2} = \frac{(3)(4)}{4} = 3\ \text{m/s}

Zero total momentum plus conserved kinetic energy force $v'_{1,f} = v'_{1,i}$ , so

v'_{1,f} = 3\ \text{m/s}, \qquad v_{cm} = 1\ \text{m/s}

The collision changes direction only.

Problem 2 · The Laboratory Angle (Sign Trap)

Given: the same pair — $m_1 = 1$ kg, $m_2 = 3$ kg at rest, $v_{1,i} = 4$ m/s — scattering through $\Theta_{cm} = 60\degree$ . Find the laboratory scattering angle $\theta_{1,f}$ .

✅ Correct!

\tan\theta_{1,f} = \frac{\sin 60\degree}{\cos 60\degree + 1/3} = 1.039

, and the forward boost has pulled the arrow in from

60\degree

46.1\degree

❌ That is the equal-mass answer.

\theta_{1,f} = \tfrac{1}{2}\Theta_{cm}

holds only when

m_1 = m_2

, because then

\frac{\sin\Theta}{\cos\Theta + 1} = \tan\frac{\Theta}{2}

. Here

m_1/m_2 = 1/3

❌ That is the heavy-target limit.

\theta_{1,f} \to \Theta_{cm}

only as

m_1/m_2 \to 0

, when

v_{cm}

is negligible. With

m_1/m_2 = 1/3

the boost still swings the arrow forward.

❌ That is the textbook's printed formula. It uses

\frac{m_2\sin\Theta_{cm}}{\cos\Theta_{cm} - m_1/m_2}

— the boost subtracted, and a stray

m_2

. Adding a forward boost can only decrease the angle, so

\theta_{1,f} < \Theta_{cm} = 60\degree

❌ Not quite. Build the triangle: the along component is

v'_{1,f}\cos\Theta_{cm} + v_{cm}

and the perpendicular one is

v'_{1,f}\sin\Theta_{cm}

Show solution

With $v'_{1,f} = 3$ m/s and $v_{cm} = 1$ m/s, the components of $\vec v_{1,f} = \vec v\,'_{1,f} + \vec v_{cm}$ are

v_{1,f}\cos\theta_{1,f} = 3\cos 60\degree + 1 = 2.500, \qquad v_{1,f}\sin\theta_{1,f} = 3\sin 60\degree = 2.598

Dividing cancels $v_{1,f}$ :

\tan\theta_{1,f} = \frac{2.598}{2.500} = 1.039 \quad\Rightarrow\quad \theta_{1,f} = 46.1\degree

The general form follows by substituting both CM speeds and dividing by $m_2$ :

\tan\theta_{1,f} = \frac{\sin\Theta_{cm}}{\cos\Theta_{cm} + m_1/m_2} = \frac{\sin 60\degree}{\cos 60\degree + \tfrac{1}{3}} = \frac{0.866}{0.833} = 1.039\ \checkmark

Both routes agree, and $\theta_{1,f} < \Theta_{cm}$ as a forward boost requires.

Problem 3 · Final Speed and Energy Lost

Given: still $m_1 = 1$ kg, $m_2 = 3$ kg at rest, $v_{1,i} = 4$ m/s, $\Theta_{cm} = 60\degree$ — find the final laboratory speed $v_{1,f}$ and the fractional change in particle 1's kinetic energy.

What is the final laboratory speed?

What fraction of $K_1$ changes?

✅ Correct!

v_{1,f}^2/v_{1,i}^2 = 13/16

, so particle 1 keeps

13/16

of its kinetic energy and hands

3/16

to the target.

❌ Check the speed. The triangle's legs are

2.500

and

2.598

; the hypotenuse is

\sqrt{2.5^2 + 2.598^2}

, not either leg on its own.

❌ Check the energy.

K \propto v^2

, so the change is

\frac{v_{1,f}^2 - v_{1,i}^2}{v_{1,i}^2}

— subtract 1 from the speed-squared ratio, and the result must be negative for a target at rest.

Show solution

Step 1 — final speed. Dot the triangle with itself:

v_{1,f}^2 = v'^{\,2}_{1,f} + 2v'_{1,f}v_{cm}\cos\Theta_{cm} + v_{cm}^2 = \frac{m_2^2 + 2m_1m_2\cos\Theta_{cm} + m_1^2}{(m_1+m_2)^2}\,v_{1,i}^2

v_{1,f} = \frac{\sqrt{9 + 2(1)(3)(\tfrac{1}{2}) + 1}}{4}(4) = \sqrt{13} \approx 3.61\ \text{m/s}

The triangle agrees: $\sqrt{2.5^2 + 2.598^2} = \sqrt{6.25 + 6.75} = \sqrt{13}$ ✓

Step 2 — energy. Since $K \propto v^2$ ,

\frac{\Delta K_1}{K_{1,i}} = \frac{v_{1,f}^2 - v_{1,i}^2}{v_{1,i}^2} = \frac{13 - 16}{16} = -\frac{3}{16}

and the general formula gives the same thing:

\frac{2m_1m_2\left(\cos\Theta_{cm} - 1\right)}{(m_1+m_2)^2} = \frac{2(1)(3)\left(-\tfrac{1}{2}\right)}{16} = -\frac{3}{16}\ \checkmark

The cosine never exceeds 1, so this fraction is never positive: particle 1's loss is the target's gain.

Problem 4 · Running the Conversion Backward

Given: a detector measures $\theta_{1,f} = 46.1\degree$ and $v_{1,f} = \sqrt{13}$ m/s for the same pair ( $v_{cm} = 1$ m/s) — find the center-of-mass scattering angle $\Theta_{cm}$ that produced it.

✅ Correct!

\tan\Theta_{cm} = \frac{2.598}{2.500 - 1} = \sqrt{3}

, so the round trip returns exactly the angle we started from.

❌ That is the printed sign again. With

+v_{cm}

the denominator is

3.5

, the ratio

0.742

, and the angle

36.6\degree

— the round trip fails. Removing the boost means subtracting it:

v_{1,f}\cos\theta_{1,f} - v_{cm}

❌ The two angles are not equal.

\theta_{1,f} = \Theta_{cm}

only in the heavy-target limit

m_1/m_2 \to 0

; here

v_{cm}

is a third of

v'_{1,f}

❌ Doubling only works for equal masses.

\Theta_{cm} = 2\theta_{1,f}

requires

m_1 = m_2

; with

m_1/m_2 = 1/3

you must remove the boost explicitly.

❌ Not quite. Rearrange the along-equation as

v_{1,f}\cos\theta_{1,f} - v_{cm} = v'_{1,f}\cos\Theta_{cm}

, then divide the perpendicular equation by it.

Show solution

Move $v_{cm}$ to the left of the along-equation and keep the perpendicular one as it is:

v_{1,f}\cos\theta_{1,f} - v_{cm} = v'_{1,f}\cos\Theta_{cm}, \qquad v_{1,f}\sin\theta_{1,f} = v'_{1,f}\sin\Theta_{cm}

Dividing cancels $v'_{1,f}$ this time:

\tan\Theta_{cm} = \frac{v_{1,f}\sin\theta_{1,f}}{v_{1,f}\cos\theta_{1,f} - v_{cm}}

With $v_{1,f}\sin\theta_{1,f} = 2.598$ , $\ v_{1,f}\cos\theta_{1,f} = 2.500$ and $v_{cm} = 1$ :

\tan\Theta_{cm} = \frac{2.598}{1.500} = 1.732 = \sqrt{3} \quad\Rightarrow\quad \Theta_{cm} = 60\degree\ \checkmark

Fixing any one of $v_{1,f}$ , $v_{2,f}$ , $\theta_{1,f}$ , $\theta_{2,f}$ fixes the other three, so the pair this formula needs is one measurement's worth of information.

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