Two identical ball bearings (label them 2 and 3) are in contact with each other and rest on a smooth, frictionless table, lined up in a straight row. A third identical ball bearing (label 1) moves along the same line with speed $V$ and strikes ball 2 head-on. The collision is perfectly elastic. Which of the following is a possible result after the collision? (a) Ball 1 stops and balls 2 and 3 move off together, each with speed $V/2$. (b) Balls 1 and 2 remain at rest and ball 3 moves off with speed $V$. (c) All three balls move off together with a common speed $V/3$. (d) All three balls move off separately with speeds $V$, $V/2$ and $V/3$ respectively.

Two identical ball bearings (label them 2 and 3) are in contact with each other and rest on a smooth, friction

NCERT Exemplar · Q9

V

(a) Ball 1 stops and balls 2 and 3 move off together, each with speed

V/2

(b) Balls 1 and 2 remain at rest and ball 3 moves off with speed

V

V/3

(d) All three balls move off separately with speeds

V

V/2

and

V/3

respectively.

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Concept understanding — Elastic Collision

Elastic Collision – From Intuition to Precision

Imagine two billiard balls on a table. They smash into each other, bounce apart, and keep moving. Now imagine two lumps of clay that collide and stick together — they stop moving as separate objects. The first case feels like nothing is lost; the second clearly loses motion. That feeling is the seed of the idea.

An elastic collision is the idealised version of that first case: a collision where no kinetic energy is turned into heat, sound, or permanent deformation. All the motion-energy that went in comes out again as motion-energy of the same total amount. The objects bounce perfectly.

The precise statement

For any collision between two bodies (call them $A$ and $B$ ), two things are always true if no external force acts:

Total linear momentum is conserved — this is a law of physics, always.
Total kinetic energy may or may not be conserved — that depends on the nature of the collision.

An elastic collision is defined by the second condition: both total momentum and total kinetic energy are conserved.

Let the masses be $m_1$ , $m_2$ and the velocities before collision be $u_1$ , $u_2$ ; after collision, $v_1$ , $v_2$ . Then:

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

These two equations together define an elastic collision.

What this means physically

In an elastic collision, the objects do not get dented, heated, or stuck. They exchange energy and momentum purely through reversible deformation — like two perfect springs that compress and then fully recover. Real collisions are never perfectly elastic (some energy always leaks into sound or heat), but many are close enough: billiard balls, steel ball bearings, gas molecules.

Watch out

A common mistake is to think "elastic" means the objects themselves are elastic (like a rubber band). That's not the point. The collision is elastic — the total kinetic energy is the same before and after. A rubber ball hitting a wall can be nearly elastic; a lump of clay hitting a wall is not.

A useful derived result

From the two conservation equations, you can derive a neat relation for one-dimensional collisions:

u_1 - u_2 = -(v_1 - v_2)

That is, the relative speed of approach equals the relative speed of separation. This is often easier to use than the full energy equation in problems. …

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In a perfectly elastic collision of identical balls, both momentum and kinetic energy must be conserved. Only the outcome where a single ball (ball 3) moves off with the incoming speed $V$ satisfies both. Correct option (B).

Conservation laws

Before: momentum $p_i = mV$ , kinetic energy $KE_i = \tfrac{1}{2}mV^2$ . For an elastic collision both must be conserved.

(B) ball 3 with $V$ : $p = mV$ ✓ and $KE = \tfrac{1}{2}mV^2$ ✓. Both satisfied.
(A) balls 2,3 at $V/2$ : $p = 2m(V/2)=mV$ ✓ but $KE = 2\cdot\tfrac12 m(V/2)^2 = \tfrac14 mV^2 \ne \tfrac12 mV^2$ ✗. …

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Conservation laws

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