Collisions — 1D vs 2D Approach

Most students learn 1D collisions cleanly and then crash on 2D problems. The reason: in 2D, momentum has to be conserved component-wise, but a single energy equation handles all directions at once. Mixing this up costs marks every JEE/NEET cycle.

This guide builds the toolkit step by step. By the end, you’ll know exactly which equations to write for 1D, which for 2D, and how to handle elastic vs inelastic in both.

Key Terms & Definitions

Collision — a brief interaction between two bodies during which they exert large forces on each other for a short time. External forces are negligible compared to internal forces during the contact, so momentum is conserved.

Elastic collision — kinetic energy is conserved (in addition to momentum). Examples: ideal billiard balls, atomic collisions in gases.

Inelastic collision — kinetic energy is not conserved (some lost to heat, sound, deformation). Momentum is still conserved.

Perfectly inelastic — the bodies stick together after the collision. Maximum KE loss possible (subject to momentum conservation).

Coefficient of restitution $e$ — ratio of relative speed of separation to relative speed of approach (for 1D). $e = 1$ for elastic, $e = 0$ for perfectly inelastic, $0 < e < 1$ for partially inelastic.

Methods & Concepts

1D collisions — the easy case

Two bodies along the same line. Let masses be $m_1, m_2$ with initial velocities $u_1, u_2$ and final velocities $v_1, v_2$ .

Momentum conservation (always): $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

For elastic (KE conservation): $\tfrac{1}{2}m_1 u_1^2 + \tfrac{1}{2}m_2 u_2^2 = \tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2$

Solving the two together gives the famous result:

$v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1 + \frac{2m_2}{m_1 + m_2}u_2$

$v_2 = \frac{2m_1}{m_1 + m_2}u_1 + \frac{m_2 - m_1}{m_1 + m_2}u_2$

For perfectly inelastic (common $v$ ): $v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$

For partially inelastic with given $e$ : $e = \frac{v_2 - v_1}{u_1 - u_2}$

Solve this with momentum conservation — two equations, two unknowns.

2D collisions — write components

In 2D, you choose coordinate axes (often along and perpendicular to the initial velocity). Then:

$\sum p_x \text{ before} = \sum p_x \text{ after}$

$\sum p_y \text{ before} = \sum p_y \text{ after}$

For elastic 2D collisions, also use KE conservation (single scalar equation).

A common 2D scenario: $m_1$ moving with $u_1$ collides with stationary $m_2$ . After collision, $m_1$ moves at angle $\theta_1$ with speed $v_1$ ; $m_2$ moves at angle $\theta_2$ (on the other side of the original line).

Momentum along original line: $m_1 u_1 = m_1 v_1 \cos\theta_1 + m_2 v_2 \cos\theta_2$

Momentum perpendicular: $0 = m_1 v_1 \sin\theta_1 - m_2 v_2 \sin\theta_2$

(Sign convention: one body up, other down — hence the minus.)

Elastic only: $\tfrac{1}{2}m_1 u_1^2 = \tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2$

Three equations, four unknowns ( $v_1, v_2, \theta_1, \theta_2$ ) — so one of them must be given to solve completely.

Type	Momentum	KE	Constraint
Elastic 1D	conserved	conserved	$e = 1$
Inelastic 1D	conserved	not	$0 \leq e < 1$
Perfectly inelastic 1D	conserved	not	$v_1' = v_2'$ , $e = 0$
Elastic 2D	conserved (each axis)	conserved	$e = 1$ along impact line

Special case: equal mass elastic 1D

When $m_1 = m_2$ , the elastic collision formulas simplify dramatically: $v_1 = u_2$ and $v_2 = u_1$ . The bodies swap velocities. This is what happens with billiard balls of the same mass.

Special case: heavy hits light at rest

If $m_1 \gg m_2$ , $u_2 = 0$ : $v_1 \approx u_1$ (heavy ball barely slows), $v_2 \approx 2u_1$ (light ball flies off at twice the incoming speed). Think basketball + tennis ball trick.

Solved Examples

Example 1 (CBSE) — Perfectly inelastic 1D

Question. A bullet of mass $20$ g hits a wooden block of mass $4$ kg at rest with speed $200$ m/s and embeds in it. Find the final speed.

Solution. $v = (0.02 \times 200)/(0.02 + 4) = 4/4.02 \approx 0.995$ m/s.

Example 2 (JEE Main) — Elastic 1D

Question. A $2$ kg ball moving at $4$ m/s strikes a stationary $1$ kg ball elastically head-on. Find the velocities after collision.

Solution. $v_1 = (2-1)/(2+1) \cdot 4 = 4/3$ m/s. $v_2 = 2(2)/(2+1) \cdot 4 = 16/3$ m/s. Check: total momentum before $= 8$ kg·m/s; after $= 2(4/3) + 1(16/3) = 24/3 = 8$ . ✓

Example 3 (JEE Advanced) — 2D elastic

Question. A particle of mass $m$ moving at $u$ hits a stationary particle of equal mass $m$ . After elastic collision, they move at angles $\theta_1$ and $\theta_2$ with the original direction. Show $\theta_1 + \theta_2 = 90°$ .

Solution. For equal-mass elastic collision in 2D with one stationary, momentum conservation and KE conservation together force the two outgoing velocities to be perpendicular. Cleanly seen by switching to centre-of-mass frame, where the velocities are equal and opposite — they remain perpendicular when transformed back.

This is the “cue ball trick” of carrom/billiards.

Exam-Specific Tips

JEE Main weightage. Collisions appears every year, usually as one 4-mark question (numerical, often 1D elastic) plus one conceptual MCQ. 2D collisions show up roughly once every 2 years.

NEET weightage. 1D collisions only, mostly perfectly inelastic. 2D is rare in NEET.

JEE Advanced. Loves the “show two outgoing velocities are perpendicular” result. Also asks for energy lost in inelastic collisions in disguise (e.g., “fraction of original KE retained”).

For “find energy lost in perfectly inelastic,” use the reduced-mass formula: $\Delta KE = \tfrac{1}{2}\mu(u_1 - u_2)^2$ where $\mu = m_1 m_2/(m_1 + m_2)$ . Avoids subtracting two big numbers.

Common Mistakes to Avoid

Using KE conservation in inelastic collisions. Don’t. KE is lost in inelastic. Use momentum and the inelastic constraint (common $v$ or $e$ ).
Adding momenta as scalars in 2D. Momentum is a vector — split into components. Adding magnitudes gives wrong results.
Forgetting that the angle in 2D collisions is measured from the impact line. Two angle conventions in textbooks: from the line of initial motion (most common in JEE) and from the impact normal. Read the question.
Using $e = 1$ for “perfectly inelastic.” $e = 0$ for perfectly inelastic. $e = 1$ is elastic. They’re swapped in many students’ heads.
Forgetting to check momentum conservation as a sanity check. After solving, plug $v_1, v_2$ back into the momentum equation. Catches arithmetic slips.

Practice Questions

Q1. A $5$ kg ball at $6$ m/s collides head-on elastically with a $5$ kg ball at rest. Final velocities?

Equal mass elastic: velocities swap. So first ball stops, second ball moves at $6$ m/s.

Q2. A $3$ kg ball at $4$ m/s strikes a $2$ kg ball at $-2$ m/s (head-on). $e = 0.5$ . Find final velocities.

Momentum: $3(4) + 2(-2) = 8 = 3v_1 + 2v_2$ . Restitution: $v_2 - v_1 = 0.5(4 - (-2)) = 3$ . From the second, $v_2 = v_1 + 3$ . Sub: $3v_1 + 2(v_1 + 3) = 8 \Rightarrow 5v_1 = 2 \Rightarrow v_1 = 0.4$ m/s, $v_2 = 3.4$ m/s.

Q3. Energy lost in a perfectly inelastic head-on collision between $m_1 = 4$ kg at $v$ and $m_2 = 1$ kg at rest?

$\Delta KE = \tfrac{1}{2}\mu v^2$ with $\mu = 4/5$ . So $\Delta KE = (1/2)(4/5)v^2 = 0.4 v^2$ J. Original KE $= 2v^2$ . Fraction lost $= 0.2 = 1/5$ .

Q4. A $1$ kg ball moving at $3$ m/s strikes a $1$ kg ball at rest. After elastic 2D collision, the first ball moves at $30°$ above horizontal. Find both speeds and the second ball’s angle.

Equal mass elastic 2D with one at rest: angles sum to $90°$ , so second ball moves at $60°$ below. KE: $9 = v_1^2 + v_2^2$ . Momentum x: $3 = v_1 \cos 30° + v_2 \cos 60° = 0.866 v_1 + 0.5 v_2$ . Momentum y: $0 = v_1 \sin 30° - v_2 \sin 60°$ giving $v_1 = \sqrt{3} v_2$ . Sub into KE: $3v_2^2 + v_2^2 = 9$ , $v_2 = 1.5$ m/s, $v_1 = 1.5\sqrt{3} \approx 2.6$ m/s.

Q5. Why do equal mass elastic collisions in 2D give perpendicular outgoing velocities?

Momentum: $\vec{u} = \vec{v}_1 + \vec{v}_2$ (since $m$ cancels). KE: $u^2 = v_1^2 + v_2^2$ . Square the momentum: $u^2 = v_1^2 + v_2^2 + 2\vec{v}_1\cdot\vec{v}_2$ . Subtract: $\vec{v}_1\cdot\vec{v}_2 = 0$ , i.e., perpendicular.

Q6. A $2$ kg block at $5$ m/s collides perfectly inelastically with a $3$ kg block at rest. Final speed?

$v = (2 \times 5)/(2+3) = 2$ m/s.

Q7. Define coefficient of restitution.

$e = (v_2 - v_1)/(u_1 - u_2)$ — ratio of relative velocity of separation to relative velocity of approach along the line of impact.

Q8. A bullet embeds in a hanging block (ballistic pendulum). Is this elastic, inelastic, or perfectly inelastic?

Perfectly inelastic — they move together after impact. Used to measure bullet speed by measuring pendulum’s swing height.

FAQs

Is momentum always conserved in collisions? Yes, as long as no significant external force acts during the brief collision time. Internal forces between the bodies cancel by Newton’s third law.

When should I switch to centre-of-mass frame? For 2D elastic collisions where you want clean angle/velocity relationships. The CM frame makes equal-mass and elastic conditions especially neat.

Why is KE not conserved in inelastic collisions? Some KE converts to heat, sound, internal vibration, or permanent deformation. Total energy is conserved, but mechanical KE is not.

What does $e > 1$ mean? It would mean the bodies separate faster than they approached — only possible if some internal energy is released (explosion). Real collisions have $0 \leq e \leq 1$ .

Can we have a 2D inelastic collision? Yes. Momentum conservation in each direction, plus an inelastic constraint (often common velocity in perpendicular case). KE not conserved.

How does the line of impact matter? In 2D, $e$ is defined along the line of impact (line joining the centres at contact), not along the original motion. For oblique collisions of spheres, this matters.