Two identical simple pendulums, having bobs of equal mass and strings of equal length, hang side by side from a common support so that in their rest positions the two bobs just touch each other. One bob (call it A) is pulled aside by a small angle of $10^\circ$ and released, so that it swings down and collides elastically, head-on, with the other bob (B). Let $T$ be the period of each pendulum. (a) Describe the subsequent motion of the two bobs. (b) Describe the graph showing how the energy of either one of the pendulums varies with time over the interval $0 \le t \le 2T$.

> [!TLDR] > Because the bobs are equal masses colliding elastically, each collision transfers all the motion from one bob to the other. So the bobs take turns: A swings down and stops on hitting B, B swings out to $10^\circ$ and returns to hand the motion back to A, and so on. The energy of either pendulum is therefore a rectangular (on–off) pulse train: it holds the value $E$ while that bob swings and is zero while the other one does. ### (a) Motion of the two bobs An elastic head-on collision between **equal masses** exchanges their velocities. Starting from rest at the mean position, the motion is carried from one bob to the other each time: 1. A is released from $10^\circ$, swings down, and at the mean position strikes B. A **stops**; B moves off with A's speed. 2. B swings up to $10^\circ$ on its side, swings back, and at the mean position strikes A. B **stops**; A moves off. 3. A swings up to $10^\circ$ on its side, returns, and strikes B again — and the cycle repeats. At any instant **only one bob is moving** while the other is momentarily at rest at the contact point; they alternately oscillate, each reaching a maximum angle of $10^\circ$. ### (b) Energy of one pendulum versus time Let $E$ be the (constant) mechanical energy of a swing to $10^\circ$. Taking A released at $t=0$: - A reaches the mean position (and collides) at $t=T/4$, so A holds energy $E$ during $[0, T/4]$, then drops to $0$. - B then holds $E$ during $[T/4, 3T/4]$ (it swings out and back, one half-period), then hands it back. …

Two identical simple pendulums, having bobs of equal mass and strings of equal length, hang side by side from

NCERT Exemplar · Q39

10^\circ

(a) Describe the subsequent motion of the two bobs.

(b) Describe the graph showing how the energy of either one of the pendulums varies with time over the interval

0 \le t \le 2T

Puducherry CbseShort· 3mImportance★★★★★est

89% · 74/83 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

Because the bobs are equal masses colliding elastically, each collision transfers all the motion from one bob to the other. So the bobs take turns: A swings down and stops on hitting B, B swings out to $10^\circ$ and returns to hand the motion back to A, and so on. The energy of either pendulum is therefore a rectangular (on–off) pulse train: it holds the value $E$ while that bob swings and is zero while the other one does.

(a) Motion of the two bobs

An elastic head-on collision between equal masses exchanges their velocities. Starting from rest at the mean position, the motion is carried from one bob to the other each time:

A is released from $10^\circ$ , swings down, and at the mean position strikes B. A stops; B moves off with A's speed.
B swings up to $10^\circ$ on its side, swings back, and at the mean position strikes A. B stops; A moves off.
A swings up to $10^\circ$ on its side, returns, and strikes B again — and the cycle repeats.

At any instant only one bob is moving while the other is momentarily at rest at the contact point; they alternately oscillate, each reaching a maximum angle of $10^\circ$ .

(b) Energy of one pendulum versus time

Let $E$ be the (constant) mechanical energy of a swing to $10^\circ$ . Taking A released at $t=0$ :

A reaches the mean position (and collides) at $t=T/4$ , so A holds energy $E$ during $[0, T/4]$ , then drops to $0$ .
B then holds $E$ during $[T/4, 3T/4]$ (it swings out and back, one half-period), then hands it back. …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ38 — A raindrop of mass 1.00 g falling from a height of 1 km hits the groun…→Next questionQ40 — Suppose the average mass of raindrops is 3.0 × 10⁻⁵ kg and their avera…

(a) Motion of the two bobs

(b) Energy of one pendulum versus time

🎓Unlock everything free for 14 days

Unlock everything free for 14 days