The bob A of a pendulum released from $30^\circ$ to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15. ![Figure 5.15 — pendulum bob A, released from 30° to the vertical, swings down and strikes bob B of equal mass at rest on a table.](/figures/work-energy-and-power/pendulum-bob-a-hits-bob-b-collision-fig-5-15.svg) How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

> [!TLDR] > Using conservation of mechanical energy for the pendulum swing and conservation of momentum + kinetic energy for the elastic collision, bob A comes to rest after hitting bob B, so it rises to zero height — it simply stops. The key insight here is that the collision is **elastic** and both bobs have **equal mass**. When a moving object of mass $m$ collides elastically with a stationary object of the same mass, the moving object **transfers all its velocity** to the stationary one and itself comes to rest. That's a standard result from elastic collision theory, and it's the heart of this problem. Let's walk through it carefully. --- 1. **Find the speed of bob A just before collision** Bob A is released from rest at an angle of $30^\circ$ to the vertical. As it swings down, gravitational potential energy converts to kinetic energy. The height from which it falls is the vertical drop from its release point to the lowest point. If the pendulum string has length $L$, then at $30^\circ$ the bob is at a height $h = L(1 - \cos 30^\circ)$ above the lowest point. Using conservation of mechanical energy: $$ m g h = \frac{1}{2} m v^2 $$ $$ v = \sqrt{2 g h} = \sqrt{2 g L (1 - \cos 30^\circ)} $$ Since $\cos 30^\circ = \frac{\sqrt{3}}{2}$, we get $1 - \cos 30^\circ = 1 - \frac{\sqrt{3}}{2}$. But we don't actually need the numeric value — the exact speed isn't required for the final answer. 2. **The elastic collision between equal masses** Bob A (mass $m$, velocity $v$ just before impact) hits stationary bob B (mass $m$). The collision is elastic, so both momentum and kinetic energy are conserved. Let $v_A'$ and $v_B'$ be velocities after collision. Conservation of momentum: $$ m v + m(0) = m v_A' + m v_B' \quad \Rightarrow \quad v = v_A' + v_B' $$ Conservation of kinetic energy: $$ \frac{1}{2} m v^2 + 0 = \frac{1}{2} m (v_A')^2 + \frac{1}{2} m (v_B')^2 \quad \Rightarrow \quad v^2 = (v_A')^2 + (v_B')^2 $$ Solving these two

The bob A of a pendulum released from 30^° to the vertical hits another bob B of the same mass at rest on a ta

Exercises · 5.17

30^\circ

Figure 5.15 — pendulum bob A, released from 30° to the vertical, swings down and strikes bob B of equal mass at rest on a table. — Figure 5.15

How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

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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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Using conservation of mechanical energy for the pendulum swing and conservation of momentum + kinetic energy for the elastic collision, bob A comes to rest after hitting bob B, so it rises to zero height — it simply stops.

The key insight here is that the collision is elastic and both bobs have equal mass. When a moving object of mass $m$ collides elastically with a stationary object of the same mass, the moving object transfers all its velocity to the stationary one and itself comes to rest. That's a standard result from elastic collision theory, and it's the heart of this problem.

Let's walk through it carefully.

Find the speed of bob A just before collision

Bob A is released from rest at an angle of $30^\circ$ to the vertical. As it swings down, gravitational potential energy converts to kinetic energy. The height from which it falls is the vertical drop from its release point to the lowest point.

If the pendulum string has length $L$ , then at $30^\circ$ the bob is at a height $h = L(1 - \cos 30^\circ)$ above the lowest point.

Using conservation of mechanical energy:

m g h = \frac{1}{2} m v^2

v = \sqrt{2 g h} = \sqrt{2 g L (1 - \cos 30^\circ)}

Since $\cos 30^\circ = \frac{\sqrt{3}}{2}$ , we get $1 - \cos 30^\circ = 1 - \frac{\sqrt{3}}{2}$ . But we don't actually need the numeric value — the exact speed isn't required for the final answer.

The elastic collision between equal masses

Bob A (mass $m$ , velocity $v$ just before impact) hits stationary bob B (mass $m$ ). The collision is elastic, so both momentum and kinetic energy are conserved.

Let $v_A'$ and $v_B'$ be velocities after collision. Conservation of momentum:

m v + m(0) = m v_A' + m v_B' \quad \Rightarrow \quad v = v_A' + v_B'

Conservation of kinetic energy:

\frac{1}{2} m v^2 + 0 = \frac{1}{2} m (v_A')^2 + \frac{1}{2} m (v_B')^2 \quad \Rightarrow \quad v^2 = (v_A')^2 + (v_B')^2

Solving these two equations: substitute $v_B' = v - v_A'$ into the energy equation:

v^2 = (v_A')^2 + (v - v_A')^2

v^2 = (v_A')^2 + v^2 - 2 v v_A' + (v_A')^2

0 = 2(v_A')^2 - 2 v v_A'

2 v_A' (v_A' - v) = 0

So either $v_A' = 0$ or $v_A' = v$ . The $v_A' = v$ solution would mean $v_B' = 0$ (no collision happened), which is physically impossible here. Therefore: …

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Q.The bob A of a pendulum released from $30^\circ$ to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15.

Concept understanding — Elastic Collision

Elastic Collision – From Intuition to Precision

The precise statement

What this means physically

A useful derived result

Concept: Energy Conservation on the Swing + Elastic Collision Between Equal Masses

Q.The bob A of a pendulum released from 30∘30^\circ30∘ to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15.

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Q.The bob A of a pendulum released from $30^\circ$ to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15.

Unlock everything free for 14 days