Collisions in one dimension

A cricket ball smacks the middle of a bat, a railway wagon slams into a stationary one and locks on, two marbles click together on a straight groove — in every one of these the motion, before and after, lies along a single straight line. That single fact turns collision physics from a messy vector problem into pure algebra: two unknowns, at most two equations, and a definite numerical answer every time. The one law that never fails, no matter how violent or gentle the impact, no matter what crumples or what bounces, is conservation of momentum. Nothing about the internal mess of a collision — the crunch of metal, the elastic twang of a rubber ball, the heat of friction — can change the total momentum of the two colliding bodies, because none of that mess involves any external force. Kinetic energy is far more fickle: sometimes it is fully conserved (an elastic collision, the idealised limit of billiard balls and gas molecules), sometimes it is fully surrendered (a perfectly inelastic collision, where the bodies stick and move as one), and most real collisions sit somewhere between the two. This article builds the complete one-dimensional toolkit CBSE Class 11 expects: the general algebra for a head-on elastic collision, the three special cases every board paper loves to test, and the single momentum equation that handles every inelastic case. Two worked examples anchor the formulas to real numbers — a rebounding ball and a pair of coupling railway wagons — so that by the end you can set up and solve any 1-D collision problem confidently, and explain in words why momentum survives when kinetic energy does not.

Key takeaways

Momentum is conserved in every collision, elastic or not: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ , because the colliding pair feels no net external force during impact.
Kinetic energy is conserved only in an elastic collision — a second, independent equation that, combined with momentum conservation, pins down both final velocities uniquely.
The general elastic 1-D formulas give $v_1$ and $v_2$ directly in terms of $m_1, m_2, u_1, u_2$ — memorise the algebra, don't just memorise the result.
Three special cases are exam favourites: equal masses (velocities swap), a light body bouncing off a heavy stationary one (rebounds at nearly the same speed), and a heavy body ploughing into a light stationary one (light body shoots off at about twice the heavy body's speed).
A perfectly inelastic collision needs only momentum conservation — the bodies move off together at one common velocity $v = \dfrac{m_1u_1+m_2u_2}{m_1+m_2}$ , and the kinetic energy "lost" is computed as $\Delta K = K_i - K_f$ .
Coefficient of restitution is out of the CBSE 2025-26 syllabus as a formal derived quantity — know only that it qualitatively measures "how elastic" a collision is, from 1 (perfectly elastic) to 0 (perfectly inelastic).
Key idea: momentum conservation alone always solves an inelastic collision, while an elastic collision needs the extra kinetic-energy equation, and their combination for a head-on collision gives the exact algebraic formulas for both final velocities.

The core insight

In any 1-D collision total momentum $m_1u_1+m_2u_2 = m_1v_1+m_2v_2$ is always conserved, and adding kinetic-energy conservation (elastic case only) turns that single equation into the two you need to solve for both final velocities exactly.

Where it fits in the syllabus

This topic sits inside the CBSE Class 11 unit Work, Energy and Power, as the direct sequel to distinguishing elastic from inelastic collisions in general. The prescribed scope is explicit: elastic and inelastic collisions in one and two dimensions, with momentum conservation as the unifying tool — this article covers the one-dimensional (head-on) case in full algebraic detail, the foundation the two-dimensional case builds on. Two things are deliberately kept out, honestly: the coefficient of restitution is not treated as a formal derived quantity here (only mentioned qualitatively, as the syllabus intends), and collision cross-sections — a scattering-theory idea from university physics — do not appear at all.

The physical picture

Picture two bodies of masses $m_1$ and $m_2$ , both constrained to a single straight line, moving with initial velocities $u_1$ and $u_2$ (one fixed direction taken as positive, so a body moving the other way has a negative velocity). They collide head-on and separate — or stick — with final velocities $v_1$ and $v_2$ along that same line. Whatever else happens during the brief violent contact, the external forces on the two-body system (gravity, the ground's support) do not change during the collision, and the forces the bodies exert on each other are internal and cancel by Newton's third law. That guarantees momentum conservation holds for every 1-D collision, without exception:

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

where: $m_1, m_2$ are the two masses (kg), $u_1, u_2$ their velocities just before impact (m/s, signed), and $v_1, v_2$ their velocities just after (m/s, signed). Read it physically: the total push of the system is unchanged by however the two bodies choose to redistribute it between themselves. If the collision is elastic, the bodies additionally conserve kinetic energy — no permanent deformation, no heat, no sound carries energy away:

\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2

where: each term $\tfrac12 mv^2$ is the kinetic energy of one body; the equation says the sum is identical before and after, even though individual bodies may speed up or slow down. If the collision is perfectly inelastic, the two bodies do not separate at all — they move off locked together at one common velocity $v$ , and only the single momentum equation applies (kinetic energy is not conserved; some is lost to deformation, heat and sound). Most real collisions — a car crumpling on impact, a bat striking a ball with some sting lost to vibration — fall between these two idealised limits, and are solved using momentum conservation together with whatever fraction of kinetic energy the problem states is retained.

Figure

Two carts on a straight track shown before and after a head on collision, with velocity arrows and labels for the momentum conservation equation

Elastic collision formulas

Derivation AThe general elastic 1-D collision formulas

The goal is to turn the two conservation laws — momentum and kinetic energy — into standalone formulas for $v_1$ and $v_2$ that can be applied directly to any head-on elastic collision, without re-deriving them from scratch every time. Two unknowns, two independent equations: the algebra is entirely within reach of factorising and rearranging linear and quadratic expressions, no calculus required.

Step 1. Write both conservation laws with the unknowns ( $v_1, v_2$ ) collected on one side. Momentum gives $m_1(u_1-v_1) = m_2(v_2-u_2)$ , obtained by moving the $m_1$ terms together and the $m_2$ terms together. Kinetic energy gives $m_1(u_1^2-v_1^2) = m_2(v_2^2-u_2^2)$ in exactly the same way, after multiplying through by 2 to clear the halves.

Step 2. Factorise the kinetic-energy equation using the difference-of-squares identity. The left side becomes $m_1(u_1-v_1)(u_1+v_1)$ and the right side $m_2(v_2-u_2)(v_2+u_2)$ , so the equation reads $m_1(u_1-v_1)(u_1+v_1) = m_2(v_2-u_2)(v_2+u_2)$ .

Step 3. Divide the Step 2 equation by the Step 1 equation. Provided $u_1 \neq v_1$ (a genuine collision actually changes body 1's velocity), the common factors $m_1(u_1-v_1)$ on the left and $m_2(v_2-u_2)$ on the right cancel completely, leaving the strikingly simple relation $u_1 + v_1 = v_2 + u_2$ , or equivalently $u_1 - u_2 = -(v_1 - v_2)$ . In words: the relative velocity of approach before impact equals the relative velocity of separation after impact, reversed in sign — the elastic collision is a perfect "bounce" in relative motion.

Step 4. Use this relation to eliminate $v_2$ . From Step 3, $v_2 = u_1 + v_1 - u_2$ . Substitute this into the momentum equation $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ and collect every term containing $v_1$ on one side: $m_1u_1 + m_2u_2 - m_2u_1 + m_2u_2 = m_1v_1 + m_2v_1$ , which simplifies to $v_1(m_1+m_2) = (m_1-m_2)u_1 + 2m_2u_2$ .

Step 5. Divide by $(m_1+m_2)$ to isolate $v_1$ , and use the same substitution (now eliminating $v_1$ instead) to get the matching expression for $v_2$ :

v_1 = \frac{(m_1-m_2)u_1 + 2m_2u_2}{m_1+m_2}, \qquad v_2 = \frac{(m_2-m_1)u_2 + 2m_1u_1}{m_1+m_2}

Each is a complete, ready-to-use formula: plug in the two masses and two initial velocities (with correct signs) and read off both final velocities directly, with no need to solve simultaneous equations from scratch again.

Three special cases, all with the second body initially at rest ( $u_2 = 0$ ), turn up repeatedly in board and competitive papers:

Equal masses ( $m_1 = m_2$ ): the formulas collapse to $v_1 = 0$ and $v_2 = u_1$ — the incoming body stops dead and the target moves off with exactly the incoming body's original velocity. The two bodies have "exchanged" velocities.
A very light body strikes a very heavy, stationary body ( $m_1 \ll m_2$ ): $v_1 \approx -u_1$ (it rebounds almost straight back at nearly its original speed) and $v_2 \approx 0$ (the heavy body barely moves) — exactly like a ball bouncing elastically off a wall.
A very heavy body strikes a very light, stationary body ( $m_1 \gg m_2$ ): $v_1 \approx u_1$ (the heavy body is barely slowed) and $v_2 \approx 2u_1$ (the light body shoots off at roughly twice the heavy body's incoming speed).

For a perfectly inelastic collision the two bodies move off together at one common velocity, found from momentum conservation alone (Step 3's relative-velocity trick does not apply here, since the bodies do not separate):

v = \frac{m_1u_1 + m_2u_2}{m_1+m_2}

The kinetic energy lost, $\Delta K = K_i - K_f$ , is computed as two separate numbers — initial KE using $u_1, u_2$ and final KE using the single common $v$ for both masses — and then subtracted; it is never zero except in the trivial case of no relative approach at all.

Worked examples

Board level - elastic collision of a ball off a heavier stationary ball

A ball of mass $1\ \text{kg}$ moving at $6\ \text{m/s}$ collides elastically head-on with a stationary ball of mass $3\ \text{kg}$ . Find the velocities of both balls after the collision, and verify momentum conservation.

Step 1. Identify the values before touching any formula: $m_1 = 1\ \text{kg}$ , $u_1 = 6\ \text{m/s}$ , $m_2 = 3\ \text{kg}$ , $u_2 = 0$ (stationary target). This is an elastic collision, so the general formulas from the derivation apply directly.

Step 2. Substitute into the $v_1$ formula, with $u_2=0$ dropping the second term entirely: $v_1 = \dfrac{(m_1-m_2)u_1}{m_1+m_2} = \dfrac{(1-3)(6)}{1+3} = \dfrac{-12}{4} = -3\ \text{m/s}$ .

Step 3. Substitute into the $v_2$ formula, similarly with $u_2 = 0$ dropping its term: $v_2 = \dfrac{2m_1u_1}{m_1+m_2} = \dfrac{2(1)(6)}{4} = \dfrac{12}{4} = 3\ \text{m/s}$ .

Step 4. Interpret the signs. $v_1 = -3\ \text{m/s}$ is negative, meaning the lighter $1\ \text{kg}$ ball rebounds backward, opposite to its original direction, at $3\ \text{m/s}$ . The heavier $3\ \text{kg}$ ball moves forward at $3\ \text{m/s}$ — this is the light-body-hits-heavy-body special case in action, close to but not exactly the $m_1 \ll m_2$ limit since the mass ratio here is only $1{:}3$ .

Step 5. Verify with momentum conservation as a check, which every collision answer should carry. Before: $1(6) + 3(0) = 6\ \text{kg}\,\cdot\,\text{m/s}$ . After: $1(-3) + 3(3) = -3 + 9 = 6\ \text{kg}\,\cdot\,\text{m/s}$ . The two final velocities are $v_1 = -3\ \text{m/s}$ (rebounds) and $v_2 = 3\ \text{m/s}$ (moves forward), and momentum matches exactly, confirming the algebra.

Competitive level - railway wagons coupling perfectly inelastically

A railway wagon of mass $8000\ \text{kg}$ moving at $2\ \text{m/s}$ collides with a stationary wagon of mass $12{,}000\ \text{kg}$ and the two couple together (perfectly inelastic collision). Find the common velocity after coupling and the kinetic energy lost in the impact.

Step 1. Recognise the collision type from the wording — "couple together" means perfectly inelastic, so only momentum conservation applies (no kinetic-energy equation is available here). Values: $m_1 = 8000\ \text{kg}$ , $u_1 = 2\ \text{m/s}$ , $m_2 = 12{,}000\ \text{kg}$ , $u_2 = 0$ .

Step 2. Apply the common-velocity formula: $v = \dfrac{m_1u_1+m_2u_2}{m_1+m_2} = \dfrac{8000(2) + 12000(0)}{8000+12000} = \dfrac{16000}{20000} = 0.8\ \text{m/s}$ .

Step 3. Compute the initial kinetic energy, using only the moving wagon (the stationary one contributes zero): $K_i = \tfrac12(8000)(2)^2 = \tfrac12(8000)(4) = 16{,}000\ \text{J}$ .

Step 4. Compute the final kinetic energy, now using the combined mass moving at the single common velocity $v$ : $K_f = \tfrac12(20000)(0.8)^2 = \tfrac12(20000)(0.64) = 6400\ \text{J}$ .

Step 5. Subtract to get the loss: $\Delta K = K_i - K_f = 16000 - 6400 = 9600\ \text{J}$ . The coupled wagons move off together at $0.8\ \text{m/s}$ , with $9600\ \text{J}$ of kinetic energy lost — dissipated as heat, sound and permanent deformation in the coupling mechanism, roughly $60\%$ of the original kinetic energy. Timing note: in a competitive-exam setting this whole problem should take under ninety seconds — the common-velocity formula is one substitution, and both kinetic energies are single-line calculations. The trap that costs time is reaching for the elastic-collision $v_1, v_2$ formulas out of habit; the word "couple" is the signal to switch immediately to the one-equation inelastic route, saving the algebra entirely.

Common traps

Mixing up which conservation laws apply. Applying the elastic-collision KE-conservation equation to a collision explicitly described as inelastic (or skipping it for a collision that actually is elastic) is the single most common scoring error — always identify the collision type from the wording first.
Sign errors from an inconsistent positive direction. Fix one direction as positive before writing a single equation; a body moving opposite to that direction must enter every formula as a negative number, which matters most when two bodies approach each other from opposite sides.
Treating the special cases as general rules. "Equal masses swap velocities" and "light body bounces off heavy body" are consequences of the general formulas under specific mass and initial-velocity conditions ( $u_2=0$ ) — they do not apply when both bodies are already moving, or when masses are only moderately different.
Reaching for the two-equation elastic system when momentum alone suffices, or the reverse — using only momentum for a problem that explicitly states the collision is elastic and expects both final velocities uniquely determined.
Forgetting the momentum check. Re-substituting your final $v_1, v_2$ (or $v$ ) back into the momentum equation costs ten seconds and catches almost every arithmetic slip before it reaches the answer sheet.

You should be able to…

State and apply momentum conservation for any 1-D collision, elastic or not, with a consistently signed positive direction.
Derive the general elastic-collision formulas for $v_1$ and $v_2$ from the two conservation laws, rather than only recalling the final results.
Recognise and correctly apply the equal-mass, light-hits-heavy and heavy-hits-light special cases without mistaking them for general rules.
Solve for the common velocity in a perfectly inelastic 1-D collision using momentum conservation alone.
Compute kinetic energy lost in an inelastic collision and use momentum conservation as an independent check on any collision answer.
Identify from a problem's wording whether it is elastic, perfectly inelastic, or a general inelastic collision, and select the correct equations accordingly.

Ask the tutor: Want to see how these same conservation laws extend to a two-dimensional (oblique) collision, where momentum must be split into perpendicular components before you can apply any of this algebra? Ask the AI tutor to walk through a 2-D elastic collision example.

Why does momentum conservation always work but kinetic energy conservation sometimes fails?

Momentum conservation follows from Newton's third law: whatever force body 1 exerts on body 2 during the collision, body 2 exerts an exactly equal and opposite force back on body 1, for exactly the same tiny duration. Those internal forces always cancel in the total momentum sum, regardless of what physically happens at the point of contact — crumpling metal, a bouncing spring, anything. Kinetic energy has no such guarantee: some of it can convert into heat, sound and permanent deformation during the impact, and that conversion is real and irreversible unless the materials are perfectly elastic (like idealised billiard balls or gas molecules). So momentum is a bookkeeping law about forces; kinetic energy conservation is an extra physical property of the specific collision, true only when nothing dissipates energy.

How do I quickly tell whether a collision problem is elastic, inelastic, or perfectly inelastic just from the wording?

Look for a few standard phrases. "Perfectly inelastic," "stick together," "couple," "embed into" or "get stuck in" all signal a common final velocity found from momentum alone. "Elastic" or "collision in which kinetic energy is conserved" signals you should use the full two-equation system (or the ready-made $v_1, v_2$ formulas). If the problem gives no such label but instead states a fraction of kinetic energy retained, or gives you both final velocities as unknowns with extra data like a coefficient of restitution, it is a general inelastic collision, and you fall back on momentum conservation plus whatever extra information the problem supplies — never assume KE conservation unless told.

Is the coefficient of restitution examinable this year, since I have seen it in some reference books?

Not as a formal quantity for CBSE 2025-26 Class 11 — it is explicitly kept out of the current prescription as a derived law you would need to compute or manipulate algebraically. You should still know, qualitatively, that it is a number between 0 and 1 describing how much of the relative approach speed survives as relative separation speed after impact (1 means perfectly elastic, 0 means perfectly inelastic), because it helps build intuition for why real collisions sit between the two idealised extremes covered in this article. But expect board and competitive questions at this level to stick to the elastic and perfectly-inelastic formulas developed here, not restitution-coefficient numerics.

In the wagon-coupling example, where did the 9600 J of lost kinetic energy actually go?

It did not vanish — energy is always conserved overall, just not as kinetic energy of the two wagons anymore. During the brief, violent coupling, the mechanical linkage deforms permanently (plastic deformation absorbs energy), the metal parts heat up slightly from friction and impact, and some energy radiates away as the loud clang you would hear standing near the tracks. All three are forms of energy the wagons' straight-line motion can no longer reclaim, which is exactly why a perfectly inelastic collision, by definition, always loses the maximum kinetic energy possible for a given momentum — sticking together is the most "wasteful" outcome consistent with momentum conservation.