Two equal masses $m_1$ and $m_2$ are moving along the same straight line with velocities $5\ \mathrm{m\,s^{-1}}$ and $9\ \mathrm{m\,s^{-1}}$ respectively. If the collision is elastic, then calculate the velocities after the collision of $m_1$ and $m_2$, respectively (a) $4\ \mathrm{m\,s^{-1}}$ and $10\ \mathrm{m\,s^{-1}}$ (b) $10\ \mathrm{m\,s^{-1}}$ and $0\ \mathrm{m\,s^{-1}}$ (c) $9\ \mathrm{m\,s^{-1}}$ and $5\ \mathrm{m\,s^{-1}}$ (d) $5\ \mathrm{m\,s^{-1}}$ and $1\ \mathrm{m\,s^{-1}}$

Two equal masses m₁ and m₂ are moving along the same straight line with velocities 5mathrmms⁻¹ and 9mathrmms⁻¹

I. Multiple Choice Questions · Q12

m_1

(a)

4\ \mathrm{m\,s^{-1}}

and

10\ \mathrm{m\,s^{-1}}

(b)

10\ \mathrm{m\,s^{-1}}

and

0\ \mathrm{m\,s^{-1}}

(c)

9\ \mathrm{m\,s^{-1}}

and

5\ \mathrm{m\,s^{-1}}

(d)

5\ \mathrm{m\,s^{-1}}

and

1\ \mathrm{m\,s^{-1}}

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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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Step 1. The two masses are equal, $m_1=m_2$ , and the collision is elastic.

Step 2. For equal masses in a one-dimensional elastic collision, the standard result derived from simultaneously conserving momentum and kinetic energy is that the two bodies exchange velocities: $v_1=u_2$ and $v_2=u_1$ .

Step 3. Here $u_1=5\ \mathrm{m\,s^{-1}}$ (for $m_1$ ) and $u_2=9\ \mathrm{m\,s^{-1}}$ (for $m_2$ ), so after collision $m_1$ takes on velocity $v_1=9\ \mathrm{m\,s^{-1}}$ and $m_2$ takes on velocity $v_2=5\ \mathrm{m\,s^{-1}}$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. A ball of mass 2 kg moving with a speed of 10 m/s collides elastically with another ball of mass 3 kg at rest. What will be the speed of the 3 kg ball after collision ?$
(a) 0 m/s
(b) 2 m/s
(c) − 2 m/s
(d) 8 m/s
›Reveal solutionSolution

Elastic collision: v2 = 2 m1 u/(m1 + m2) = 8 m/s. Answer (D).

For a one-dimensional elastic collision where mass m2 is initially at rest, the velocity of the struck body is:

v2 = 2 m1 u /(m1 + m2).

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. For perfect elastic collision (A) e = 1 (B) e > 1 (C) e < 1 (D) e = 0$
›Reveal solutionSolution

A perfectly elastic collision has coefficient of restitution e = 1.

The coefficient of restitution is defined as:

$e = \dfrac{\text{relative velocity of separation}}{\text{relative velocity of approach}}$

For a perfectly elastic collision, both momentum and kinetic energy are conserved, which mathematically requires the relative speed of separation to exactly equal the relative speed of approach, giving $e = 1$ . For an inelastic collision …

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CBSE 2020Set ANNUAL1 mark
$Q. Two identical solid balls, one of ivory and the other of wet-clay, are dropped from the same height on the floor. Which one will rise to a greater height after striking the floor and why?$
›Reveal solutionSolution

Ivory is a hard, elastic material, so its collision with the floor is close to perfectly elastic and it rebounds to nearly the same height it was dropped from; wet clay is soft and deforms permanently on impact (a highly inelastic collision), losing almost all of its kinetic energy, so it barely rises after striking the floor.

When a ball is dropped and strikes the floor, the height h' to which it rebounds depends on how much of its kinetic energy just before impact is retained just after impact (rather than being lost to permanent deformation, heat, and sound during the collision) — this is measured by the coefficient of restitution, e, of the collision (e close to 1 = nearly elastic, e close to 0 = highly inelastic).

Ivory is a hard, relatively rigid and elastic material. When an ivory ball strikes the floor, it undergoes very little permanent deformation — it compresses briefly on impact and then springs back to its original shape, converting most of the kinetic energy back into the ball's motion. So the collision is close to perfectly elastic (e close to 1), and the ball rebounds to nearly the height it was dropped from.

…

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CBSE 2019Set ANNUAL1 mark
$Q. The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.$
›Reveal solutionSolution
Elastic collision + equal masses + target initially at rest ⇒ velocities are exchanged, so the striking bob (A) stops completely and rises to zero height.

For a head-on elastic collision between two bodies of equal mass $m$ , where the second body ( $B$ ) is initially at rest, momentum and kinetic-energy conservation together give the standard result that the velocities are exchanged: body A (the one that was moving) ends up at rest, and body B moves off with A's original velocity.

Let A be released from $30°$ to the vertical. By energy conservation on the way down (before the collision), A arrives at the bottom (table level) with some speed $v$ , having converted its initial gravitational PE fully into KE.

At the collision: since $m_A = m_B = m$ and it is elastic, the exchange-of-velocity result gives:

A's velocity after collision $= 0$

B's velocity after collision $= v$ (A's velocity just before collision) …
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CBSE 2018Set ANNUAL1 markMCQ
$Q. In Elastic Collision:$
(a) Both momentum and K.E are conserved
(b) Both momentum and K.E are not conserved
(c) Momentum is conserved and K.E is not conserved
(d) Momentum is not conserved and K.E is conserved
›Reveal solutionSolution

Elastic collision = momentum conserved AND kinetic energy conserved; this is what distinguishes it from an inelastic collision.

In any collision (elastic or inelastic), as long as no external force acts on the system, linear momentum is always conserved — this follows directly from Newton's third law (the colliding bodies exert equal and opposite forces on each other for equal times, so their net impulse on the system is zero).

…

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