A bob of mass m, suspended by a string of length l1, is given a minimum velocity required to complet

Q: Wait, why is the velocity at the top $\sqrt{gl_1}$ and not $\sqrt{5gl_1}$?

$\sqrt{5gl_1}$ is the minimum velocity required at the *bottom* of the circle to make it all the way around. By the time the bob reaches the *top*, it has lost kinetic energy (converted to potential energy). The minimum speed it needs at the very top to keep the string taut (Tension $\ge 0$) is $\sqrt{gl_1}$.

Q: Why do the bobs exactly swap velocities?

In a perfectly elastic head-on collision between two objects of the *same mass*, the equations of conservation of momentum and conservation of kinetic energy dictate that they exchange their velocities. Since Bob 2 was at rest, Bob 1 comes to a complete stop, and Bob 2 takes off with Bob 1's exact speed.

Q: Why can't we just use energy conservation for the whole system from start to finish?

You can use energy conservation for the individual bobs during their circular motion, but during the collision, you must use momentum and collision rules. While kinetic energy is conserved *during* the elastic collision, the energy is transferred from one body to another. Breaking it into three distinct phases (Bob 1 up, Collision, Bob 2 around) is the safest and most logical approach.

Q: What if the masses were different?

If the masses were different, they wouldn't simply swap velocities. You would have to use the full 1D elastic collision formulas: $v_2' = \frac{2m_1}{m_1+m_2}v_1 + \frac{m_2-m_1}{m_1+m_2}v_2$. The problem specifically uses identical masses to create this elegant shortcut!

Q: Does the tension in the string affect the collision?

At the exact moment of collision, the strings are vertical. The collision forces are horizontal (tangential to the path). Since the strings are ideal and inextensible, they don't interfere with the horizontal exchange of momentum. The tension only acts as a centripetal force to keep them in circular paths before and after the impact.

Question

A bob of mass $m$, suspended by a string of length $l_1$, is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass $m$ suspended by a string of length $l_2$, which is initially at rest. Both the strings are massless and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio $l_1 / l_2$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a fascinating mechanical ballet. We have a bob of mass $m$ suspended by a string of length $l_1$. It is given a precise kick at the bottom—just enough to send it soaring through a complete vertical circle.

But the real magic happens at the very top of its trajectory. Waiting there, perfectly still, is a second bob of identical mass $m$, suspended by its own string of length $l_2$. The two bobs are destined to collide.

### The Master Equation: Bob 1's Ascent

Let's rewind to Bob 1's journey. To barely complete a vertical circle without the string going slack, the tension at the highest point must be exactly zero.

Using the centripetal force equation at the top, $mg = \frac{mv_1^2}{l_1}$, we can solve for the critical velocity.

This gives us the minimum speed of Bob 1 at the highest point:
$$v_1 = \sqrt{gl_1}$$

### The Perfect Swap: Elastic Collision

Right at this apex, Bob 1 crashes into Bob 2. The problem states this is a perfectly elastic collision.

Here is where a beautiful principle of physics comes into play: when two objects of identical mass undergo a perfectly elastic head-on collision, they completely exchange their velocities.

Since Bob 2 was initially at rest, Bob 1 transfers all its momentum and kinetic energy to Bob 2. Bob 1 comes to a dead stop, and Bob 2 is launched forward with the exact speed Bob 1 just had.

Therefore, the initial velocity of Bob 2 immediately after the collision is:
$$v_2 = \sqrt{gl_1}$$

### Bob 2's Awakening

Now, the spotlight shifts entirely to Bob 2. It has just received a massive horizontal kick and is sitting at the lowest point of its own string of length $l_2$.

The problem dictates that this newly acquired speed is exactly the minimum required for Bob 2 to complete its own vertical circle.

We know from the dynamics of vertical circular motion that the minimum speed required at the bottom to complete a full loop is $\sqrt{5gR}$. For Bob 2, this means its starting velocity must be:
$$v_2 = \sqrt{5gl_2}$$

### Final Calculation

We now have two distinct expressions for the exact same physical quantity—the velocity of Bob 2 right after the collision.

We can set these two expressions equal to each other:
$$\sqrt{gl_1} = \sqrt{5gl_2}$$

To solve this, we simply square both sides to eliminate the square roots:
$$gl_1 = 5gl_2$$

The acceleration due to gravity, $g$, elegantly cancels out from both sides, leaving us with a simple linear relationship:
$$l_1 = 5l_2$$

Finally, rearranging this to find the requested ratio, we get:
$$\frac{l_1}{l_2} = 5$$

**The final answer is 5.**

Enter Numerical Value:

Visualized Solution

The Setup: Bob 1

Velocity at the Highest Point

The Second Bob

Elastic Collision

Bob 2's Journey

Minimum Speed for Bob 2

Equating the Velocities

Solving for the Ratio

The Sigma Insight: Vertical Circular Motion

Analyzing the Setup

The Master Equation

The Perfect Swap

Bob 2's Awakening

Final Calculation

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