Collisions and Momentum — Elastic, Inelastic, and Everything Between

Why Collisions Matter

Every car crash, billiard ball strike, and nuclear reaction is governed by the same principle: conservation of momentum. When two objects interact, their total momentum before equals their total momentum after — as long as no external force acts on the system.

This chapter is among the highest-weightage topics in JEE Main (3–4 questions most years) and appears in CBSE Class 11. The concepts seem abstract until you realise they predict the exact outcome of any collision — from everyday accidents to particle physics experiments.

Linear Momentum

Momentum ( $\vec{p}$ ) is the product of mass and velocity:

\vec{p} = m\vec{v}

Units: kg·m/s (or N·s — same unit, different name).

Momentum is a vector — it has direction. A 10 kg object at 5 m/s east has different momentum from the same object at 5 m/s west.

Why does momentum matter? Newton’s Second Law can be restated as: net force equals rate of change of momentum:

\vec{F}_{net} = \frac{d\vec{p}}{dt}

This more general form covers both constant-mass situations ( $F = ma$ ) and variable-mass problems like rockets.

Conservation of Linear Momentum

If the net external force on a system is zero:

\vec{p}_{total} = \text{constant} \implies \vec{p}_{initial} = \vec{p}_{final}

This is not “approximately true” — it’s an exact law, following directly from Newton’s Third Law.

When is momentum conserved?

During all collisions (the collision forces are internal to the system)
When no friction, gravity, or external forces act on the system
Even when kinetic energy is NOT conserved

Common misconception: Momentum is “always conserved.” It’s conserved when net external force = 0. If friction acts (e.g., a block sliding on a rough surface), momentum of the block alone is NOT conserved — the friction from the ground is an external force.

Types of Collisions

Elastic Collision

Both momentum AND kinetic energy are conserved.

This is an ideal case — no energy is “lost” to deformation, heat, or sound. In reality, no macroscopic collision is perfectly elastic, but it’s a good approximation for billiard balls, steel balls, and atomic/nuclear particles.

Equations for 1D elastic collision (masses $m_1$ , $m_2$ ; initial velocities $u_1$ , $u_2$ ; final velocities $v_1$ , $v_2$ ):

Conservation of momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

Conservation of KE: $\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Solving these simultaneously:

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

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

Special cases (with $u_2 = 0$ , i.e., second body at rest):

Scenario	$v_1$	$v_2$
$m_1 = m_2$	0 (stops completely)	$u_1$ (original speed)
$m_1 \gg m_2$	$\approx u_1$ (barely slows)	$\approx 2u_1$
$m_1 \ll m_2$	$\approx -u_1$ (bounces back)	$\approx 0$

The $m_1 = m_2$ case is JEE-favourite: when two equal masses collide elastically (second at rest), they exchange velocities. The first stops, the second moves with the first’s original velocity. This underlies Newton’s Cradle!

Perfectly Inelastic Collision

Momentum is conserved, maximum kinetic energy is lost. The objects stick together after collision and move with a common velocity.

m_1 u_1 + m_2 u_2 = (m_1 + m_2)v

v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}

Energy lost:

\Delta KE = \frac{1}{2}\frac{m_1 m_2}{m_1 + m_2}(u_1 - u_2)^2

This lost energy goes into deformation, heat, sound — all the “crumple” you see in a car crash.

Example: Bullet embedding in a wooden block. The bullet-block system moves together after the collision.

Inelastic Collision (General)

Most real collisions fall between perfectly elastic and perfectly inelastic. Kinetic energy is partially lost.

Coefficient of Restitution (e):

e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}} = \frac{v_2 - v_1}{u_1 - u_2}

$e = 1$ : perfectly elastic
$e = 0$ : perfectly inelastic (stick together)
$0 < e < 1$ : partially inelastic (real collisions)

A rubber ball bouncing has $e \approx 0.8$ ; a clay ball $e \approx 0$ .

Impulse

Impulse ( $J$ ) is the change in momentum:

J = \Delta p = F \cdot \Delta t

Impulse-momentum theorem: $F_{avg} \cdot \Delta t = m\Delta v$

Why this matters: Car airbags increase the time of collision ( $\Delta t$ ), reducing the average force ( $F$ ) on the occupant even though the change in momentum is the same.

J = \Delta \vec{p} = m\vec{v}_f - m\vec{v}_i = F_{avg} \cdot \Delta t

For constant force: $\vec{J} = \vec{F} \cdot t$

Units: N·s = kg·m/s

Centre of Mass in Collisions

The centre of mass (CM) of a system moves at constant velocity when no external force acts — even during a collision. This is a powerful tool:

\vec{v}_{CM} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2} = \text{constant}

In the CM frame, total momentum is zero. Elastic collisions in the CM frame are symmetric — both objects reverse their velocities in that frame.

Solved Examples

Easy — CBSE Class 9/11

Problem: A 2 kg ball moving at 4 m/s collides with a stationary 6 kg ball and they stick together. Find the final velocity.

v = \frac{m_1 u_1}{m_1 + m_2} = \frac{2 \times 4}{2 + 6} = \frac{8}{8} = 1 \text{ m/s}

Also find energy lost:

\Delta KE = \frac{1}{2}(2)(4^2) - \frac{1}{2}(8)(1^2) = 16 - 4 = 12 \text{ J}

Medium — JEE Main Level

Problem: A 3 kg ball moving at 6 m/s collides elastically with a 1 kg ball at rest. Find the final velocities.

Using elastic collision formulas with $u_2 = 0$ :

v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 = \frac{3-1}{3+1} \times 6 = \frac{2}{4} \times 6 = 3 \text{ m/s}

v_2 = \frac{2m_1}{m_1 + m_2} u_1 = \frac{2 \times 3}{4} \times 6 = \frac{6}{4} \times 6 = 9 \text{ m/s}

Verify KE: Initial = $\frac{1}{2}(3)(36) = 54$ J. Final = $\frac{1}{2}(3)(9) + \frac{1}{2}(1)(81) = 13.5 + 40.5 = 54$ J ✓

Hard — JEE Advanced Level

Problem: A particle of mass $m$ moving at velocity $v$ hits a stationary particle of mass $M$ . The coefficient of restitution is $e$ . Find the final velocities.

From momentum conservation: $mv = mv_1 + Mv_2$ …(i)

From restitution: $e = \frac{v_2 - v_1}{v - 0} \implies v_2 - v_1 = ev$ …(ii)

From (i) and (ii):

v_1 = \frac{m - eM}{m + M}v \quad \text{and} \quad v_2 = \frac{(1+e)m}{m+M}v

When $e = 1$ : recovers elastic collision formulas ✓ When $e = 0$ : $v_1 = v_2 = \frac{mv}{m+M}$ ✓ (perfectly inelastic)

Common Mistakes

Mistake 1: Using momentum conservation when external forces act. If a ball is thrown horizontally and hits a wall, momentum changes (the wall exerts an external force). You can’t say “momentum before = momentum after” for the ball alone.

Mistake 2: Assuming kinetic energy is conserved in all collisions. KE is conserved only in elastic collisions. For ANY other collision (inelastic, perfectly inelastic), some KE is lost. The total energy is still conserved (first law of thermodynamics) — it’s just converted to other forms.

Mistake 3: Forgetting direction in momentum calculations. Momentum is a vector. If two balls collide head-on, assign one direction as positive. For a 2 kg ball at +4 m/s and a 3 kg ball at -2 m/s, total momentum = $2(4) + 3(-2) = 8 - 6 = 2$ kg·m/s, not $8 + 6 = 14$ kg·m/s.

Mistake 4: Confusing elastic and inelastic for macroscopic objects. Most daily-life collisions (cars, balls) are inelastic — energy is lost. “Elastic” in the physical sense means NO energy loss — this is only approximately true for hard steel balls or atomic particles.

Practice Questions

Q1. A gun of mass 3 kg fires a bullet of mass 50 g at 300 m/s. What is the recoil velocity of the gun?

Total initial momentum = 0 (both at rest).

By conservation: $0 = (0.05)(300) + (3)v_{gun}$

$v_{gun} = -\frac{15}{3} = -5 \text{ m/s}$

The gun recoils at 5 m/s in the opposite direction.

Q2. A 1500 kg car moving at 20 m/s collides with a stationary 1000 kg car. They stick together. Find: (a) final velocity, (b) KE lost.

(a) $v = \frac{1500 \times 20}{1500 + 1000} = \frac{30000}{2500} = 12 \text{ m/s}$

(b) Initial KE = $\frac{1}{2}(1500)(400) = 300,000$ J

Final KE = $\frac{1}{2}(2500)(144) = 180,000$ J

KE lost = $120,000$ J = 120 kJ

Q3. In a 1D elastic collision, the first body (mass $m$ ) is moving and hits a stationary body (also mass $m$ ). What happens?

Using elastic formulas: $v_1 = \frac{m-m}{m+m}u = 0$ and $v_2 = \frac{2m}{2m}u = u$ .

The first body stops completely; the second moves with the original speed. Complete velocity exchange.

Q4. Calculate the coefficient of restitution for a ball dropped from 2 m height that bounces back to 1.25 m.

$e = \sqrt{\frac{h_2}{h_1}} = \sqrt{\frac{1.25}{2}} = \sqrt{0.625} \approx 0.79$

(Derived from: speed before impact $= \sqrt{2gh_1}$ ; speed after $= \sqrt{2gh_2}$ ; e = ratio of speeds)

Two-Dimensional Collisions

Most real collisions are not head-on — they occur at angles. In 2D, we apply conservation of momentum separately in the $x$ and $y$ directions.

Worked Example — Oblique Elastic Collision

A billiard ball A moving at 5 m/s strikes an identical stationary ball B. After collision, A moves at 30° to the original direction. Find the speeds and direction of both balls.

$mv = mv_A\cos 30° + mv_B\cos\theta$

$5 = v_A \times \frac{\sqrt{3}}{2} + v_B\cos\theta$ …(i)

$0 = mv_A\sin 30° - mv_B\sin\theta$

$v_A \times \frac{1}{2} = v_B\sin\theta$ …(ii)

For elastic collision between equal masses, there is a beautiful result: the two balls move at 90° to each other after collision. So $\theta = 60°$ .

From (ii): $v_A/2 = v_B\sin 60° = v_B\sqrt{3}/2 \implies v_A = v_B\sqrt{3}$

From (i): $v_B\sqrt{3} \times \sqrt{3}/2 + v_B\cos 60° = 5 \implies 3v_B/2 + v_B/2 = 5 \implies 2v_B = 5$

$v_B = 2.5$ m/s, $v_A = 2.5\sqrt{3} \approx 4.33$ m/s.

For equal-mass elastic collision where one body is initially at rest:

The two bodies always move at 90° to each other after collision
$v_A^2 + v_B^2 = u^2$ (KE conservation)
This 90° rule is the basis of many JEE problems on oblique collisions

JEE Main 2023 had a 2D collision problem involving a ball hitting a smooth wall at an angle. For perfectly elastic collision with a wall (infinite mass): the component of velocity perpendicular to the wall reverses; the component parallel to the wall remains unchanged. Angle of incidence = angle of reflection. This is the same as light reflecting off a mirror.

Q5. A 10 g bullet travelling at 400 m/s embeds in a 2 kg block on a frictionless surface. The block then slides up a frictionless incline. Find the maximum height reached.

Collision (perfectly inelastic): $v = \frac{0.01 \times 400}{0.01 + 2} = \frac{4}{2.01} \approx 1.99$ m/s.

After collision, KE converts to PE on the incline: $\frac{1}{2}(2.01)(1.99)^2 = (2.01)(10)h$

$h = \frac{(1.99)^2}{2 \times 10} = \frac{3.96}{20} \approx 0.198$ m $\approx$ 19.8 cm.

FAQs

Q: Is momentum always conserved? Momentum is conserved when the net external force on the system is zero. During most collision analyses, we treat the collision forces as internal to the two-body system, so momentum is conserved. However, if gravity or friction acts significantly during the collision, those are external forces and would change the momentum.

Q: Why do airbags save lives? Airbags increase the time duration of the collision from ~5 ms (hard impact) to ~25–30 ms. By impulse-momentum theorem: $F = \Delta p / \Delta t$ . The change in momentum (impulse) is the same, but the larger $\Delta t$ reduces the average force on the occupant by 5–6 times.

Q: What is Newton’s Cradle? Newton’s Cradle demonstrates elastic collisions between equal masses. When one ball swings and hits the row, that ball stops and the ball at the other end swings out (momentum and energy are “transmitted” through the row). The intermediate balls barely move because momentum and energy transfer occurs sequentially.

Q: Can momentum be negative? Yes. Momentum is a vector — it takes on the sign of the velocity. If you choose rightward as positive, a leftward-moving object has negative momentum. What’s physically meaningful is the change in momentum and the total momentum of a system.