Collision

Q: What is the difference between elastic and inelastic collisions?

In an elastic collision, both linear momentum and kinetic energy are conserved. In an inelastic collision, momentum is conserved but kinetic energy is not - some is lost to heat, sound, deformation, or vibrations. A perfectly inelastic collision (bodies stick together) has the maximum possible KE loss.

Q: Is momentum always conserved in a collision?

Yes, provided the net external impulse during the collision is negligible. Since collisions are typically very brief and internal collision forces are enormous compared to external forces like gravity, external impulse is essentially zero. So momentum of the (system of colliding bodies) is conserved in every practical collision.

Q: Why do two equal masses in an elastic 1D collision simply swap velocities?

Solving momentum and energy conservation for m_1 = m_2 gives v_1 = u_2 and v_2 = u_1 directly. Physically, this is the only solution consistent with both conservation laws. It is the reason Newton's cradle works: each ball's momentum and KE are transferred perfectly to the next.

Q: What does a coefficient of restitution of 0.8 physically mean?

After the collision, the two bodies separate at 80% of the relative speed with which they approached. So if a ball dropped from 1 m rebounds to 0.8^2 = 0.64 m, its coefficient with the floor is 0.8. The fraction of KE retained after one bounce is e^2 = 0.64.

Q: Why do equal masses colliding obliquely separate at 90° in an elastic collision?

Momentum conservation: \vec{u}_1 = \vec{v}_1 + \vec{v}_2. Squaring: u_1^2 = v_1^2 + v_2^2 + 2\vec{v}_1 \cdot \vec{v}_2. KE conservation (equal masses cancel): u_1^2 = v_1^2 + v_2^2. Comparing, \vec{v}_1 \cdot \vec{v}_2 = 0, so the two final velocities are perpendicular (unless one is zero, i.e. a head-on collision).

Q: Can the coefficient of restitution be greater than 1?

Not in ordinary mechanical collisions. e \gt 1 would violate energy conservation (the bodies would gain KE from nothing). It only occurs in explosive collisions (like a compressed spring released between bodies) where stored energy is added to the mechanical KE - such interactions are called super-elastic and are not covered by ordinary collision analysis.

PhysicsSystem Of Particles And Rotational MotionFor NEET aspirants

A collision is an event in which two or more bodies interact for a very short time, exchanging momentum through a large mutual force. During any collision, if no external force acts, the total linear momentum is always conserved. Whether kinetic energy is conserved distinguishes the two main types: an elastic collision conserves KE, while an inelastic collision does not (some KE converts to heat, sound, or deformation). The coefficient of restitution $e =$ (separation speed)/(approach speed) measures how elastic a collision is, ranging from $e = 1$ (perfectly elastic) to $e = 0$ (perfectly inelastic). Collisions can also be classified as head-on (1D) or oblique (2D).

Key Formulas - Quick Reference

Momentum conservation (always): $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
Elastic collision (1D): $v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1} + \frac{2 m _{2}}{m _{1} + m _{2}} u_{2}$
Elastic collision (1D): $v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1} + \frac{m _{2} - m _{1}}{m _{1} + m _{2}} u_{2}$
Elastic collision relative velocity: $u_{1} - u_{2} = - (v_{1} - v_{2})$
Coefficient of restitution: $e = \frac{v _{2} - v _{1}}{u _{1} - u _{2}} = \frac{separation speed}{approach speed}$
Perfectly inelastic (sticky) final speed: $v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$
Loss of KE (perfectly inelastic): $Δ K = \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (u_{1} - u_{2})^{2}$
General 1D (coefficient $e$ ): $v_{1} = \frac{m _{1} - e m _{2}}{m _{1} + m _{2}} u_{1} + \frac{( 1 + e ) m _{2}}{m _{1} + m _{2}} u_{2}$

1. What is a Collision?

A collision is any brief interaction between two (or more) bodies during which they exert large forces on each other for a short time. Physical contact is not required - the concept applies equally to two billiard balls hitting each other and to two charged particles scattering by electrostatic repulsion. What matters is:

During the collision, the internal forces are much larger than any external forces, so external forces contribute negligible impulse.
Therefore, the total linear momentum of the colliding system is conserved (in each direction) between just before and just after the collision.
The collision may or may not conserve kinetic energy.

Classification:

Elastic collision: Kinetic energy is conserved. Total KE before = Total KE after.
Inelastic collision: Kinetic energy is not conserved. Some KE is converted to internal energy (heat, sound, deformation).
Perfectly inelastic collision: The bodies stick together after impact and move with a common velocity. Maximum possible loss of KE.

Geometrically, collisions are:

Head-on (or 1D): Velocities before and after lie along the same straight line (the line joining the centres).
Oblique (or 2D): Velocities before and after are not collinear; the collision has a component transverse to the line of centres.

2. Elastic Collision in One Dimension

Consider two smooth spheres of masses $m_{1}$ and $m_{2}$ moving along the same line with initial velocities $u_{1}$ and $u_{2}$ (taking rightward as positive). After a head-on elastic collision, their velocities are $v_{1}$ and $v_{2}$ .

Figure 1: Head-on collision between two bodies. In an elastic collision, momentum and KE are both conserved; in an inelastic collision, only momentum is conserved.

Derivation of final velocities

Momentum conservation:

m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}

which we rewrite as

m_{1} (u_{1} - v_{1}) = m_{2} (v_{2} - u_{2}) (1)

Kinetic energy conservation:

\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}

which factors as

m_{1} (u_{1} - v_{1}) (u_{1} + v_{1}) = m_{2} (v_{2} - u_{2}) (v_{2} + u_{2}) (2)

Dividing (2) by (1) gives

u_{1} + v_{1} = v_{2} + u_{2} ⟹ u_{1} - u_{2} = - (v_{1} - v_{2}) (3)

Beautiful result: In a 1D elastic collision, the relative velocity of approach equals the relative velocity of separation. The bodies bounce off with the same relative speed as they came together with.

Solving (1) and (3) simultaneously:

v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1} + \frac{2 m _{2}}{m _{1} + m _{2}} u_{2}

v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1} + \frac{m _{2} - m _{1}}{m _{1} + m _{2}} u_{2}

Special cases (memorise)

Equal masses ( $m_{1} = m_{2}$ ): $v_{1} = u_{2}$ and $v_{2} = u_{1}$ . The two bodies simply exchange velocities. This is why one billiard ball striking a stationary equal one stops dead, and the target ball moves off with the original velocity.
Heavy hits light stationary ( $m_{1} ≫ m_{2}$ , $u_{2} = 0$ ): $v_{1} \approx u_{1}$ (heavy body barely slows), $v_{2} \approx 2 u_{1}$ (light body flies off at twice the incoming speed).
Light hits heavy stationary ( $m_{1} ≪ m_{2}$ , $u_{2} = 0$ ): $v_{1} \approx - u_{1}$ (light body bounces back with nearly the same speed), $v_{2} \approx 0$ (heavy body barely moves).

Solved Example 1

A ball of mass 0.4 kg moving at 3 m/s has a head-on, perfectly elastic collision with a 0.6 kg ball initially at rest. Find the velocities after collision.

Solution:

By momentum conservation: $(0.4) (3) + 0 = 0.4 v_{1} + 0.6 v_{2}$ , i.e. $v_{1} + 1.5 v_{2} = 3$ .

Elastic collision relative velocity: $v_{2} - v_{1} = u_{1} - u_{2} = 3$ .

Adding: $2.5 v_{2} = 6$ , so $v_{2} = 2.4$ m/s and $v_{1} = - 0.6$ m/s. The lighter ball bounces back with 0.6 m/s; the heavier ball moves forward at 2.4 m/s.

3. Inelastic Collision

In an inelastic collision, momentum is conserved but some kinetic energy is lost. If the two bodies stick together (perfectly inelastic collision), they share a common velocity after impact:

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}}

Figure 2: Perfectly inelastic collision. The two bodies stick and move together at v = (m₁u₁ + m₂u₂)/(m₁ + m₂). Momentum is conserved; KE loss is maximum.

The loss of kinetic energy is

Δ K = K_{i} - K_{f} = \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} - \frac{1}{2} (m_{1} + m_{2}) v^{2}

Substituting $v$ and simplifying gives the compact form

Δ K = \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (u_{1} - u_{2})^{2}

The quantity $\frac{m _{1} m _{2}}{m _{1} + m _{2}}$ is called the reduced mass. Notice that $Δ K$ depends only on the relative velocity, not on absolute velocities - a consequence of the fact that this loss is intrinsic to the collision and independent of the observer's frame.

Solved Example 2

A moving particle of mass

m

makes a head-on collision with a particle of mass

2 m

initially at rest. If the collision is perfectly elastic, show that the incoming particle loses

8/9

of its initial kinetic energy.

Solution:

Using the elastic formulas with $m_{1} = m$ , $m_{2} = 2 m$ , $u_{1} = u$ , $u_{2} = 0$ :

v_{1} = \frac{m - 2 m}{3 m} u = - \frac{u}{3}

Fractional KE loss of the incoming particle:

\frac{Δ K}{K _{i}} = 1 - (\frac{v _{1}}{u})^{2} = 1 - \frac{1}{9} = \frac{8}{9}

The incoming particle keeps $1/9$ of its KE and transfers $8/9$ to the target.

4. Coefficient of Restitution

For a head-on collision between two bodies, the coefficient of restitution $e$ is defined as

e = \frac{speed of separation}{speed of approach} = \frac{v _{2} - v _{1}}{u _{1} - u _{2}}

where both velocities are measured along the line of impact (the line joining the centres).

Range of $e$ :

0 \leq e \leq 1

$e = 1$ : perfectly elastic - KE fully conserved.
$0 < e < 1$ : partially elastic - some KE lost.
$e = 0$ : perfectly inelastic - bodies stick, maximum KE loss.

The coefficient depends on the material and geometry of the two bodies but is very nearly a constant for a given pair over a wide range of speeds.

General 1D collision formulas

Combining momentum conservation with $v_{2} - v_{1} = e (u_{1} - u_{2})$ gives

v_{1} = \frac{m _{1} - e m _{2}}{m _{1} + m _{2}} u_{1} + \frac{( 1 + e ) m _{2}}{m _{1} + m _{2}} u_{2}

v_{2} = \frac{( 1 + e ) m _{1}}{m _{1} + m _{2}} u_{1} + \frac{m _{2} - e m _{1}}{m _{1} + m _{2}} u_{2}

Setting $e = 1$ recovers the elastic formulas; setting $e = 0$ gives $v_{1} = v_{2} = (m_{1} u_{1} + m_{2} u_{2}) / (m_{1} + m_{2})$ , the perfectly inelastic result.

Ball bouncing on floor

A useful special case: a ball dropped from height $h_{0}$ onto a rigid floor rebounds to a height $h_{1}$ . Since the floor is effectively infinitely massive, $e = v_{after} / v_{before}$ , and $v \propto h$ , so

e = \frac{h _{1}}{h _{0}}, h_{n} = e^{2 n} h_{0}

Figure 3: A ball repeatedly bouncing on a rigid floor loses a factor e² of height per bounce. Since v ∝ √h, the coefficient of restitution is e = √(h₁/h₀).

After $n$ successive bounces the ball reaches height $h_{n} = e^{2 n} h_{0}$ , and the fractional KE remaining is $e^{2 n}$ .

Solved Example 3

A ball of mass

m

moving with speed

v

collides head-on with an identical stationary ball. After the collision the total KE is

3/4

of the original. Find the coefficient of restitution.

Solution:

Using the general 1D formulas with equal masses, $u_{1} = v$ , $u_{2} = 0$ :

v_{1} = \frac{( 1 - e ) v}{2}, v_{2} = \frac{( 1 + e ) v}{2}

Final KE $= \frac{1}{2} m v_{1}^{2} + \frac{1}{2} m v_{2}^{2} = \frac{m v ^{2}}{8} [(1 - e)^{2} + (1 + e)^{2}] = \frac{m v ^{2}}{4} (1 + e^{2})$ .

Setting this equal to $\frac{3}{4} \cdot \frac{1}{2} m v^{2} = \frac{3 m v ^{2}}{8}$ :

\frac{m v ^{2}}{4} (1 + e^{2}) = \frac{3 m v ^{2}}{8} ⟹ 1 + e^{2} = \frac{3}{2} ⟹ e = \frac{1}{2}

5. Oblique (2D) Collision

When the velocities before and after collision are not collinear, the collision is called oblique. To analyse an oblique collision, resolve velocities along two perpendicular directions:

Along the line of impact (the line joining the centres at the moment of contact): the collision force acts here, so momentum is conserved and the coefficient of restitution relates approach/separation speeds along this line.
Perpendicular to the line of impact: no force acts here (smooth surfaces), so each body's velocity component in this direction remains unchanged.

For an oblique collision between smooth spheres with the coefficient of restitution $e$ along the line of impact:

v_{1} cos θ_{1} - v_{2} cos θ_{2} = - e (u_{1} cos α_{1} - u_{2} cos α_{2})

where the angles are measured with respect to the line of impact. The perpendicular components stay unchanged: $v_{1} sin θ_{1} = u_{1} sin α_{1}$ and similarly for body 2.

Figure 4: Two spheres shown at the moment of contact, touching along the line of impact (dashed). After the collision each sphere's velocity makes an angle (θ₁, θ₂) with the line of impact — the components along the line change, the perpendicular components do not.

Elastic collision of equal masses, one initially at rest: If a moving ball elastically collides obliquely with an identical stationary ball, they separate at right angles (90°) to each other after impact. This is a classic JEE result and comes directly from combining momentum and energy conservation for equal masses.

Solved Example 4

A ball moving with speed

v

collides elastically and obliquely with an identical stationary ball. If the incoming ball is deflected by angle

θ

, find the speeds of both balls after collision.

Solution:

For equal masses in an elastic collision with the target at rest, the two balls separate at right angles. So the target ball moves at $(9 0^{\circ} - θ)$ to the incoming direction.

From momentum conservation in the two components, or equivalently from KE conservation: $v_{1} = v cos θ$ and $v_{2} = v sin θ$ .

Check: $v_{1}^{2} + v_{2}^{2} = v^{2} cos^{2} θ + v^{2} sin^{2} θ = v^{2}$ , so KE is conserved.

Common Mistakes to Avoid

Watch out

Assuming kinetic energy is conserved in every collision. It is only conserved in elastic collisions. Momentum is always conserved (given no external impulse), but energy is not.
Using the elastic-collision formulas $v_{1}, v_{2}$ for inelastic problems. If the collision has $e < 1$ , use the general formulas with the correct value of $e$ , or apply momentum conservation together with the definition of $e$ .
For oblique collisions, forgetting that only the components along the line of impact change. The perpendicular components of each body's velocity remain unchanged.
Computing the loss of KE for a perfectly inelastic collision by first finding both final velocities, when the compact formula $Δ K = \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (u_{1} - u_{2})^{2}$ is much faster.
Confusing the coefficient of restitution with a ratio of kinetic energies. $e$ is a ratio of speeds (approach vs separation), not energies. The KE ratio after one collision is $e^{2}$ only in special cases.
Applying momentum conservation across an entire event that includes an external force (e.g. a ball colliding with a wall attached to the Earth): only momentum of the ball changes; total momentum with the Earth is conserved but the Earth's velocity change is negligibly small.

Frequently Asked Questions

Q1. What is the difference between elastic and inelastic collisions?

In an elastic collision, both linear momentum and kinetic energy are conserved. In an inelastic collision, momentum is conserved but kinetic energy is not - some is lost to heat, sound, deformation, or vibrations. A perfectly inelastic collision (bodies stick together) has the maximum possible KE loss.

Q2. Is momentum always conserved in a collision?

Yes, provided the net external impulse during the collision is negligible. Since collisions are typically very brief and internal collision forces are enormous compared to external forces like gravity, external impulse is essentially zero. So momentum of the (system of colliding bodies) is conserved in every practical collision.

Q3. Why do two equal masses in an elastic 1D collision simply swap velocities?

Solving momentum and energy conservation for $m_{1} = m_{2}$ gives $v_{1} = u_{2}$ and $v_{2} = u_{1}$ directly. Physically, this is the only solution consistent with both conservation laws. It is the reason Newton's cradle works: each ball's momentum and KE are transferred perfectly to the next.

Q4. What does a coefficient of restitution of 0.8 physically mean?

After the collision, the two bodies separate at 80% of the relative speed with which they approached. So if a ball dropped from 1 m rebounds to $0. 8^{2} = 0.64$ m, its coefficient with the floor is 0.8. The fraction of KE retained after one bounce is $e^{2} = 0.64$ .

Q5. Why do equal masses colliding obliquely separate at 90° in an elastic collision?

Momentum conservation: $u_{1} = v_{1} + v_{2}$ . Squaring: $u_{1}^{2} = v_{1}^{2} + v_{2}^{2} + 2 v_{1} \cdot v_{2}$ . KE conservation (equal masses cancel): $u_{1}^{2} = v_{1}^{2} + v_{2}^{2}$ . Comparing, $v_{1} \cdot v_{2} = 0$ , so the two final velocities are perpendicular (unless one is zero, i.e. a head-on collision).

Q6. Can the coefficient of restitution be greater than 1?

Not in ordinary mechanical collisions. $e > 1$ would violate energy conservation (the bodies would gain KE from nothing). It only occurs in explosive collisions (like a compressed spring released between bodies) where stored energy is added to the mechanical KE - such interactions are called super-elastic and are not covered by ordinary collision analysis.

Q7. In a perfectly inelastic collision, where does the lost kinetic energy go?

It converts to internal energy of the joined system: heat (raised temperature), sound (compression waves in air), permanent deformation of the bodies, and internal vibrations. Total energy is still conserved, but the useful mechanical KE decreases. The maximum KE loss occurs precisely when both bodies end up moving with the same velocity.

Previous year questions on Collision

2 questions from past papers, each with a step-by-step solution.

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