Classical-Mechanics · Unit 19 · Video 5 · Interactive Practice

Fix One Angle: Solving the Two-Dimensional Elastic Collision

IKey Formulas

Formula	Name	What you need
$0 = (1+\alpha)v_{1,f}^2 - 2\alpha v_{1,i}v_{1,f}\cos\theta_{1,f} - (1-\alpha)v_{1,i}^2$	Quadratic for the scattered speed, with $\alpha \equiv m_1/m_2$	Both momentum components and elastic energy, with $\theta_{2,f}$ squared away
$v_{1,f} = \dfrac{\alpha v_{1,i}\cos\theta_{1,f} \pm v_{1,i}\sqrt{\alpha^2\cos^2\theta_{1,f} + 1 - \alpha^2}}{1+\alpha}$	Scattered speed	$v_{1,i}$ , $\theta_{1,f}$ and $\alpha$ ; the $+$ root alone when $\alpha \le 1$
$\tan\theta_{2,f} = \dfrac{v_{1,f}\sin\theta_{1,f}}{v_{1,i} - v_{1,f}\cos\theta_{1,f}}$	Recoil angle	$v_{1,f}$ from the quadratic — no mass appears in this relation
$v_{2,f} = \dfrac{\alpha v_{1,f}\sin\theta_{1,f}}{\sin\theta_{2,f}}$	Recoil speed	$\theta_{2,f}$ , and the mass ratio returns

Key Insight: Two momentum components and one energy equation are three equations for four unknowns, so a two-dimensional elastic collision keeps one free parameter — the laws cannot tell a head-on hit from a glancing blow. Supply $\theta_{1,f}$ and everything downstream is forced; for $m_1 = m_2$ the quadratic collapses to $v_{1,f} = v_{1,i}\cos\theta_{1,f}$ and the two objects always leave at $90\degree$ .

IIVisualization 1 — One Angle Chosen, Three Answers Forced

The conservation laws refuse to pick the scattering angle; pick it yourself and they fix all the rest.

Scattering angle θ1f 30°

IIIVisualization 2 — Which Root, and How Far It Can Scatter

The mass ratio decides which root of the quadratic is physical — and whether any root exists at all.

Mass ratio α = m1/m2 1.00 Scattering angle θ1f 30°

The ceiling on $\theta_{1,f}$ belongs to the laboratory frame alone: viewed from the center-of-mass frame every scattering angle stays open however heavy the projectile, which is where the next videos begin.

IVVisualization 3 — The Equal-Mass Example, Step by Step

Equal masses collapse the quadratic to a cosine, and send the two objects off at right angles.

Step 1

The right angle is elastic-only: any kinetic energy lost to deformation makes $\vec{v}_{1,f}\cdot\vec{v}_{2,f} > 0$ , and the angle sum drops below $90\degree$ .

VQuiz Questions

Problem 1 · Equal Masses, Angle Given

Given: an object of mass $m$ moving at $v_{1,i} = 4.0\ \text{m/s}$ collides elastically with a second object of the same mass at rest, and leaves at $\theta_{1,f} = 60\degree$ — find its final speed $v_{1,f}$ and the recoil angle $\theta_{2,f}$ .

What is $v_{1,f}$ ?

What is $\theta_{2,f}$ ?

✅ Correct! With

\alpha = 1

the quadratic gives

v_{1,f} = v_{1,i}\cos\theta_{1,f} = 2.0\ \text{m/s}

, and the equal-mass right angle leaves

\theta_{2,f} = 90\degree - 60\degree = 30\degree

❌ That is the target's speed.

v_{1,i}\sin 60\degree = 3.46\ \text{m/s}

v_{2,f}

; the projectile keeps the cosine,

v_{1,f} = v_{1,i}\cos\theta_{1,f}

❌ The speed cannot survive intact. If

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

the projectile keeps all the kinetic energy, leaving none for a target that has certainly started moving.

❌ Check the cosine. Equal masses collapse the quadratic to

v_{1,f} = v_{1,i}\cos\theta_{1,f} = (4.0\ \text{m/s})\cos 60\degree

❌ Check the angle sum. For equal masses

\theta_{1,f} + \theta_{2,f} = 90\degree

, whatever the scattering angle.

Show solution

Step 1 — the quadratic for equal masses. With $\alpha = m_1/m_2 = 1$ the term $(1-\alpha^2)v_{1,i}^2$ under the root vanishes, so the root equals the first term and the two solutions are

v_{1,f} = \frac{v_{1,i}\cos\theta_{1,f} \pm v_{1,i}\cos\theta_{1,f}}{2} = v_{1,i}\cos\theta_{1,f} \quad\text{or}\quad 0

The zero root is the head-on hit, in which the projectile stops dead; a scattering angle of $60\degree$ rules it out.

v_{1,f} = (4.0\ \text{m/s})\cos 60\degree = 2.0\ \text{m/s}

Step 2 — the recoil angle from the tangent relation.

\tan\theta_{2,f} = \frac{v_{1,f}\sin\theta_{1,f}}{v_{1,i} - v_{1,f}\cos\theta_{1,f}} = \frac{(2.0)(0.866)}{4.0 - (2.0)(0.500)} = \frac{1.732}{3.0} = 0.577

\theta_{2,f} = \tan^{-1}(0.577) = 30\degree

Step 3 — check with energy. The recoil speed is $v_{2,f} = v_{1,i}\sin\theta_{1,f} = 3.46\ \text{m/s}$ , and

v_{1,f}^2 + v_{2,f}^2 = 4.0 + 12.0 = 16.0 = v_{1,i}^2 \ \checkmark

The angles sum to $90\degree$ , as they must for equal masses.

Problem 2 · The Right Angle Is Not a Law

Given: a proton of mass $m_1 = m$ strikes a stationary helium nucleus of mass $m_2 = 4m$ elastically and scatters at $\theta_{1,f} = 30\degree$ — find the recoil angle $\theta_{2,f}$ of the helium nucleus.

✅ Correct! With

\alpha = \tfrac{1}{4}

the light proton keeps almost all its speed,

v_{1,f} = 0.967\,v_{1,i}

, and the heavy nucleus is knocked off at

71.4\degree

— an angle sum of

101.4\degree

, not

90\degree

❌ That is the equal-mass answer.

\theta_{1,f} + \theta_{2,f} = 90\degree

holds only when

m_1 = m_2

; here

\alpha = \tfrac{1}{4}

and a light projectile always pushes the sum above

90\degree

❌ The ratio is inverted. A largest scattering angle exists only for

\alpha > 1

. Here

\alpha = m_1/m_2 = \tfrac{1}{4}

, the projectile is the light one, and every angle up to

180\degree

is open to it.

❌ Not quite. Run the recipe in order: the quadratic gives

v_{1,f}

, then the tangent relation gives

\theta_{2,f}

— the

90\degree

shortcut is not available here.

Show solution

Step 1 — the quadratic. With $\alpha = m_1/m_2 = 0.25$ and $\theta_{1,f} = 30\degree$ , work in units of $v_{1,i}$ :

\alpha^2\cos^2\theta_{1,f} + 1 - \alpha^2 = (0.0625)(0.750) + 0.9375 = 0.9844

\frac{v_{1,f}}{v_{1,i}} = \frac{(0.25)(0.866) + \sqrt{0.9844}}{1.25} = \frac{0.2165 + 0.9922}{1.25} = 0.9669

Since $\alpha < 1$ only the $+$ root is physical — the $-$ root gives a negative speed.

Step 2 — the tangent relation.

\tan\theta_{2,f} = \frac{(0.9669)(0.500)}{1 - (0.9669)(0.866)} = \frac{0.4835}{0.1626} = 2.973

\theta_{2,f} = \tan^{-1}(2.973) = 71.4\degree

Step 3 — check with energy. The recoil speed follows from the sine equation,

v_{2,f} = \frac{\alpha v_{1,f}\sin\theta_{1,f}}{\sin\theta_{2,f}} = \frac{(0.25)(0.9669)(0.500)}{0.9478}\,v_{1,i} = 0.1275\,v_{1,i}

v_{1,f}^2 + \frac{v_{2,f}^2}{\alpha} = 0.9350 + \frac{0.01626}{0.25} = 0.9350 + 0.0650 = 1.000\,v_{1,i}^2 \ \checkmark

A light projectile bouncing off a heavy target barely slows down, and the angle sum $30\degree + 71.4\degree = 101.4\degree$ exceeds a right angle. Only $m_1 = m_2$ gives exactly $90\degree$ .

Problem 3 · How Far a Heavy Projectile Can Be Deflected

Given: an object of mass $m_1 = 3m_2$ collides elastically with a stationary object of mass $m_2$ — find the largest scattering angle $\theta_{1,f}$ the projectile can possibly have.

✅ Correct! The discriminant vanishes at

\sin\theta_{1,f} = 1/\alpha = \tfrac{1}{3}

, giving

\theta_{1,f}^{\max} = 19.47\degree

; beyond it the root is imaginary and no elastic collision produces that deflection.

❌ Wrong combination of masses.

\sin^{-1}\!\big(m_2/(m_1+m_2)\big) = \sin^{-1}(0.25) = 14.5\degree

. The discriminant condition is

\sin\theta_{1,f} \le m_2/m_1 = 1/\alpha

❌ That is $\cos^{-1}(1/3)$ . Solving

\alpha^2\cos^2\theta_{1,f} + 1 - \alpha^2 \ge 0

gives

\sin^2\theta_{1,f} \le 1/\alpha^2

— the condition constrains the sine, not the cosine.

❌ That holds only for $\alpha \le 1$ . When

\alpha > 1

the quantity

(1-\alpha^2)v_{1,i}^2

under the root is negative and eventually overwhelms the

\cos^2\theta_{1,f}

term, killing the solution altogether.

Show solution

Step 1 — demand a real root. The scattered speed is real only when the quantity under the square root is non-negative:

\alpha^2 v_{1,i}^2\cos^2\theta_{1,f} + (1-\alpha^2)v_{1,i}^2 \ge 0 \;\Longrightarrow\; \cos^2\theta_{1,f} \ge 1 - \frac{1}{\alpha^2}

Step 2 — turn it into a sine. Using $\cos^2\theta = 1 - \sin^2\theta$ ,

\sin^2\theta_{1,f} \le \frac{1}{\alpha^2} \;\Longrightarrow\; \sin\theta_{1,f} \le \frac{1}{\alpha} = \frac{m_2}{m_1}

Step 3 — evaluate. With $\alpha = 3$ ,

\theta_{1,f}^{\max} = \sin^{-1}\!\left(\frac{1}{3}\right) = 19.47\degree

For $\alpha < 1$ the right-hand side exceeds $1$ and the constraint is empty — a light projectile can be turned through any angle, right back along its original path. At $\alpha = 1$ the discriminant never dies either, but $v_{1,f} = v_{1,i}\cos\theta_{1,f}$ does: equal masses stop at $90\degree$ . A heavy projectile cannot even get that far — at $\theta_{1,f}^{\max}$ the two roots merge, and past it there is no elastic collision to be had.

Problem 4 · Sharing the Energy at a Chosen Angle

Given: two equal masses collide elastically and the projectile scatters at $\theta_{1,f} = 40\degree$ — find the fraction of the incident kinetic energy carried away by the target.

✅ Correct! For equal masses

v_{2,f} = v_{1,i}\sin\theta_{1,f}

, so the target's share is

\sin^2 40\degree = 0.41

— and the projectile's

\cos^2 40\degree = 0.59

completes the ledger.

❌ That is the projectile's share.

\cos^2 40\degree = 0.59

is the energy object 1 keeps; the target gets what is left,

\sin^2 40\degree

❌ That is a speed ratio, not an energy ratio.

\sin 40\degree = 0.64

equals

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

; kinetic energy goes as the square of the speed.

❌ Not quite. Get

v_{2,f}

first — for equal masses the recipe collapses to

v_{2,f} = v_{1,i}\sin\theta_{1,f}

— then square the ratio.

Show solution

Step 1 — the two final speeds. Equal masses give $v_{1,f} = v_{1,i}\cos\theta_{1,f}$ and $\theta_{2,f} = 90\degree - \theta_{1,f}$ , so the sine equation reads

v_{2,f} = \frac{\alpha v_{1,f}\sin\theta_{1,f}}{\sin\theta_{2,f}} = \frac{v_{1,i}\cos\theta_{1,f}\sin\theta_{1,f}}{\cos\theta_{1,f}} = v_{1,i}\sin\theta_{1,f}

Step 2 — the energy fraction. With equal masses the mass factors cancel:

\frac{K_{2,f}}{K_i} = \frac{\tfrac{1}{2}mv_{2,f}^2}{\tfrac{1}{2}mv_{1,i}^2} = \sin^2\theta_{1,f} = \sin^2 40\degree = (0.643)^2 = 0.41

Step 3 — check the books. The projectile keeps $\cos^2 40\degree = 0.59$ , and

\cos^2\theta_{1,f} + \sin^2\theta_{1,f} = 1 \ \checkmark

The same identity that eliminated $\theta_{2,f}$ from the momentum equations reappears as the energy ledger: a glancing blow ( $\theta_{1,f}$ small) hands over almost nothing, and a $90\degree$ deflection would hand over everything.

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