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Collisions
Elastic vs Inelastic Collisions
A collision involves an impact between bodies where momentum is always conserved, but Kinetic Energy may or may not be conserved depending on the collision type.
$$\text{Elastic: } K.E._{initial} = K.E._{final}$$
Kinetic energy is only conserved in perfectly elastic collisions, while total energy and momentum are always conserved in all collision types.
$K.E.$=Kinetic energy(Joules (J))
$e = 1$ (perfectly elastic)
→KE fully conserved. Bodies bounce off with no energy loss.
$e = 0$ (perfectly inelastic)
→Bodies stick together. Maximum KE is lost to heat, sound, deformation.
$0 < e < 1$ (real collisions)
→Partial KE loss. Most everyday collisions fall here.
Momentum Always Conserved: Total momentum of an isolated system remains constant regardless of collision type: $m_1 v_1 + m_2 v_2 = m_1 v'_1 + m_2 v'_2$.
Energy Transformation: In inelastic collisions, KE transforms into internal energy, sound, or permanent deformation — but total energy is still conserved.
1D Elastic Collision Fundamentals
For two non-rotating bodies in a Head-on collision, the sum of their initial momenta equals the sum of their final momenta. Combined with KE conservation, this gives us enough equations to solve for both final velocities.
$$m_1 v_1 + m_2 v_2 = m_1 v'_1 + m_2 v'_2$$
Conservation of linear momentum for two colliding masses in one dimension.
$m_1, m_2$=Masses of objects 1 and 2(kg)
$v_1, v_2$=Initial velocities (before collision)(m/s)
$v'_1, v'_2$=Final velocities (after collision)(m/s)
Sign Convention: Choose a positive direction (usually right). Velocities in the opposite direction are negative.
Mass Independence: Momentum conservation applies regardless of whether the collision is elastic or inelastic.
Dimensional Check: Both sides of $m_1 v_1 + m_2 v_2 = m_1 v'_1 + m_2 v'_2$ have dimensions of $[M][L][T]^{-1}$ (kg·m/s). If your answer has different units, something went wrong.
Derivation Path: By rearranging momentum conservation as $m_1(v_1 - v'_1) = m_2(v'_2 - v_2)$ and KE conservation similarly, dividing one by the other gives the relative velocity relation.
The Relative Velocity Theorem
In a perfectly elastic collision, the Relative velocity of approach equals the Relative velocity of separation in magnitude, but reversed in direction. This is a powerful shortcut that avoids the KE conservation equation entirely.
$$v_1 - v_2 = -(v'_1 - v'_2) = v'_2 - v'_1$$
The closing speed before impact equals the separation speed after impact in elastic collisions.
$v_1 - v_2$=Relative velocity of approach(m/s)
$v'_2 - v'_1$=Relative velocity of separation(m/s)
Shortcut Power: Instead of using the quadratic KE equation, combine momentum conservation with this linear relation — much simpler algebra.
Only for Elastic: This exact equality only holds when $e = 1$. For partially inelastic collisions, the separation speed is reduced by the Coefficient of restitution: $v'_2 - v'_1 = e(v_1 - v_2)$.
Quick Verification: After solving a collision problem, check that the approach speed equals the separation speed. If it doesn't, either the collision isn't elastic or there's an error.
General Velocity Equations
Combining momentum conservation and the relative velocity relation yields closed-form expressions for each body's final velocity in terms of the masses and initial velocities.
$$v'_1 = \frac{m_1 - m_2}{m_1 + m_2} v_1 + \frac{2m_2}{m_1 + m_2} v_2$$
Final velocity of body 1 after a 1D elastic collision.
$\frac{m_1 - m_2}{m_1 + m_2}$=Mass ratio coefficient for body 1's initial velocity
$\frac{2m_2}{m_1 + m_2}$=Transfer coefficient from body 2's initial velocity
Unit Shortcut: Since masses appear as ratios $\frac{m_1 - m_2}{m_1 + m_2}$, you can use any consistent mass unit (grams, kg) — they cancel.
Symmetry: The formula for $v'_2$ has the same structure but with subscripts swapped: $v'_2 = \frac{2m_1}{m_1 + m_2} v_1 + \frac{m_2 - m_1}{m_1 + m_2} v_2$.
Dimensional Sanity: The output $v'_1$ has the same dimensions as $v_1$ and $v_2$ (m/s) since the mass ratios are dimensionless. If you accidentally mix mass units (e.g., one in grams, another in kg), the ratios are wrong.
The second velocity formula gives the final velocity of body 2, derived identically by solving the same pair of equations.
$$v'_2 = \frac{2m_1}{m_1 + m_2} v_1 + \frac{m_2 - m_1}{m_1 + m_2} v_2$$
Final velocity of body 2 after a 1D elastic collision.
$\frac{2m_1}{m_1 + m_2}$=Transfer coefficient from body 1's initial velocity
$\frac{m_2 - m_1}{m_1 + m_2}$=Mass ratio coefficient for body 2's initial velocity
When $v_2 = 0$: Simplifies to $v'_2 = \frac{2m_1}{m_1 + m_2} v_1$ — the target object always moves in the direction of the incoming object.
Verification: After calculating both $v'_1$ and $v'_2$, verify momentum conservation: $m_1 v'_1 + m_2 v'_2$ should equal $m_1 v_1 + m_2 v_2$.
Special Cases in Elastic Collisions
When two bodies of equal mass collide elastically, a remarkably simple result emerges: they exchange velocities completely.
$$m_1 = m_2 \implies v'_1 = v_2 \text{ and } v'_2 = v_1$$
Complete Swap: Each body takes the other's velocity. If one was stationary, the moving one stops and the stationary one takes off.
Newton's Cradle: This principle explains why in Newton's cradle, only the end ball swings out when the opposite end ball strikes.
Billiard Balls: A cue ball hitting a stationary target ball of equal mass stops dead — all motion transfers to the target.
When a light body collides with a much heavier stationary body, the light body bounces back with nearly unchanged speed while the heavy body barely moves.
$$m_2 \gg m_1, \; v_2 = 0 \implies v'_1 \approx -v_1, \; v'_2 \approx 0$$
Limiting case of the general elastic collision formula when the target is overwhelmingly massive.
$\frac{m_1 - m_2}{m_1 + m_2}$=Approaches $-1$ when $m_2 \gg m_1$
$\frac{2m_1}{m_1 + m_2}$=Approaches $0$ when $m_2 \gg m_1$
Wall Analogy: A ball bouncing off a wall is the extreme case — the wall (infinite mass) doesn't move at all.
Squash Player: The squash ball reverses direction off the massive front wall while the wall stays perfectly still.
Speed Preserved: Only the direction changes; the magnitude of velocity stays approximately the same.
Limiting Case: As $m_2 / m_1 \to \infty$, the result converges to exactly $v'_1 = -v_1$ and $v'_2 = 0$.
When a massive body collides with a light stationary body, the massive body continues nearly unchanged while the light body flies off at approximately twice the massive body's speed.
$$m_1 \gg m_2, \; v_2 = 0 \implies v'_1 \approx v_1, \; v'_2 \approx 2v_1$$
Limiting case of the general elastic collision formula when the projectile is overwhelmingly massive.
$\frac{m_1 - m_2}{m_1 + m_2}$=Approaches $+1$ when $m_1 \gg m_2$
$\frac{2m_1}{m_1 + m_2}$=Approaches $2$ when $m_1 \gg m_2$
Double Speed: The light object leaves at $\approx 2v_1$ — twice the heavy body's speed, not equal to it.
Truck vs Pebble: A heavy truck barely notices hitting a small stone, but the stone flies off at twice the truck's speed.
Limiting Case Check: As $m_1 / m_2 \to \infty$, the result approaches exactly $v'_2 = 2v_1$.
Coefficient of Restitution
The coefficient of restitution ($e$) measures the "bounciness" of a collision. It is the ratio of the relative velocity of separation to the relative velocity of approach.
$$e = \frac{v'_2 - v'_1}{v_1 - v_2}$$
Measures what fraction of the approach speed is retained as separation speed after collision.
$e$=Coefficient of restitution
$v'_2 - v'_1$=Relative velocity of separation(m/s)
$v_1 - v_2$=Relative velocity of approach(m/s)
$e = 1$
→Perfectly elastic — approach speed = separation speed. No KE lost.
$e = 0$
→Perfectly inelastic — bodies stick together, separation speed = 0. Maximum KE lost.
$0 < e < 1$
→Partially inelastic — some KE lost. All real-world collisions fall here.
Dimensionless: $e$ has no units — it is a pure ratio.
Material Property: $e$ depends on the materials of the colliding objects. A steel ball on marble has $e \approx 0.95$; clay on clay has $e \approx 0$.
Height Connection: For a ball dropped from height $h_1$ bouncing to height $h_2$: $e = \sqrt{\frac{h_2}{h_1}}$, since velocity at ground level $\propto \sqrt{h}$.
Proportionality: $e \propto$ separation speed for fixed approach speed — higher $e$ means faster separation.
Perfectly Inelastic Collisions
In a perfectly inelastic collision, the two bodies stick together after impact and move as one combined mass. This is the collision type that loses the maximum kinetic energy.
$$v_f = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$$
The combined velocity after a perfectly inelastic collision, derived from momentum conservation alone.
$v_f$=Common final velocity of the combined mass(m/s)
$m_1 + m_2$=Total combined mass after sticking(kg)
$v_2 = 0$ (stationary target)
→Simplifies to $v_f = \frac{m_1}{m_1 + m_2} v_1$ — always slower than the incoming object.
$m_1 = m_2$ and $v_2 = 0$
→$v_f = \frac{v_1}{2}$ — they move together at half the initial speed.
Maximum KE Loss: Among all collision types with the same initial conditions, perfectly inelastic loses the most kinetic energy.
Only Momentum: Since bodies stick together ($e = 0$), you cannot use KE conservation. Use momentum conservation only.
Real Examples: Car crashes (crumple zones), catching a ball (hand + ball move together), bullet embedding in a block.
Limiting Case: As $m_2 \to 0$, $v_f \to v_1$ (picking up a negligible mass barely affects you). As $m_2 \to \infty$, $v_f \to 0$ (slamming into an immovable wall — you stop).
The kinetic energy lost in a perfectly inelastic collision can be calculated directly from the masses and initial velocities without finding the final velocity first.
$$\Delta K.E. = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (v_1 - v_2)^2$$
The kinetic energy converted to heat, sound, and deformation in a perfectly inelastic collision.
$\Delta K.E.$=Kinetic energy lost (always positive)(Joules (J))
$\frac{m_1 m_2}{m_1 + m_2}$=Reduced mass of the system(kg)
$(v_1 - v_2)^2$=Square of relative approach speed(m²/s²)
Proportionality: $\Delta K.E. \propto (v_1 - v_2)^2$ — doubling the relative speed quadruples the energy loss.
Fraction Lost: When $v_2 = 0$, the fraction of KE lost is $\frac{m_2}{m_1 + m_2}$. A heavier target means more energy lost.
Equal Masses: When $m_1 = m_2$ and one is stationary, exactly half the KE is lost.
Limiting Case: As $m_1 \gg m_2$ (heavy hits light), fraction lost $\to 0$ — barely any energy wasted. As $m_2 \gg m_1$ (light hits heavy), fraction lost $\to 1$ — almost all KE absorbed.