# EXTREME(um) Problem from the International Physics Olympiad in India

TLDRThis video delves into a problem from the International Physics Olympiad held in India. It begins with a particle on a frictionless plane, split into two regions with different potentials. The video explains how to use conservation of energy to express the velocity in the second region. It then introduces the principle of least action, deriving a relationship between the initial and final velocities using advanced calculus. The discussion touches on the connection between this problem and Snell's law in optics, showcasing the beauty of physics through intricate mathematical reasoning.

###### Takeaways

- π The problem discussed is from the International Physics Olympiad in India, involving a particle moving on a frictionless plane with different potential energy regions.
- π The first part of the problem uses the conservation of energy principle to express the potential energy in region two (V2) in terms of the initial kinetic energy (mv1^2/2) and potential (V0).
- π In the second part, the problem introduces the concept of forces and potential gradients, emphasizing that only the x-component of velocity changes when transitioning between regions due to potential differences.
- π The y-component of the velocity (V1y = V2y) remains unchanged, highlighting the independence of motion in the y-direction from potential changes.
- π― The principle of least action is introduced as a method to determine the path a particle takes, where the action (integral of speed along the path) is minimized.
- βοΈ The action integral is split into two parts, one for each region, and differentiated with respect to a variable alpha to find the extremum, which corresponds to the actual path taken by the particle.
- π The differentiation process involves setting the derivative of the action with respect to alpha to zero, leading to a relationship between the velocities in the two regions and the coordinates of the transition point.
- π The final relationship derived is a ratio of V1 to V2, expressed in terms of the coordinates of the transition point and the endpoints, which simplifies to a trigonometric relationship involving the angles of incidence and refraction.
- π¬ The script concludes with a connection to Snell's law in optics, suggesting that the principles discussed can be applied to derive this law in a future video.
- π¨βπ« The video script serves as an educational tool, teaching unique concepts of physics through the analysis of an Olympiad problem.
- π The detailed mathematical derivations and principles discussed provide a comprehensive understanding of the problem and the underlying physics.

###### Q & A

### What is the initial condition of the particle in the problem?

-The particle starts at the origin on a frictionless plane.

### How is the plane divided, and what are the potential energies in each region?

-The plane is divided into two regions. In Region 1, the potential energy is V = 0, and in Region 2, the potential energy is V = Vβ.

### How do you express the velocity V2 in terms of mass, initial velocity V1, and potential energy Vβ?

-Using the conservation of energy, V2 can be expressed as V2 = sqrt(V1^2 - 2Vβ/m).

### How does the y-component of velocity change as the particle moves between the two regions?

-The y-component of velocity remains unchanged as the particle moves between the two regions.

### How can V2 be expressed in terms of V1, Theta 1, and Theta 2?

-Using trigonometric relationships, V2 can be expressed as V2 = V1 * (sin(Theta 1) / sin(Theta 2)).

### What is the principle of least action, and how is it defined in this problem?

-The principle of least action states that the path taken by a particle is the one where the action, defined as A = M * integral(v * ds), is an extremum (minimum or maximum).

### What coordinates are used to describe the transition point and the final position of the particle?

-The transition point where the particle crosses from Region 1 to Region 2 is described by coordinates (X1, Alpha), and the final position is described by coordinates (Xβ, Yβ).

### How is the action integral split for the two regions, and what does each part represent?

-The action integral is split into two parts: one for Region 1 (A1 = M * V1 * sqrt(X1Β² + AlphaΒ²)) and one for Region 2 (A2 = M * V2 * sqrt((Xβ - X1)Β² + (Yβ - Alpha)Β²)). Each part represents the contribution to the action from the respective regions.

### What is the result of differentiating the action with respect to Alpha and setting it to zero?

-Differentiating the action with respect to Alpha and setting it to zero results in an equation that relates V1, V2, Alpha, X1, Yβ, and Xβ, which can be used to find the ratio V1/V2.

### How does the final derived expression for V1/V2 relate to Snell's law in optics?

-The final expression V1/V2 = (sin(Theta 2) / sin(Theta 1)) is analogous to Snell's law in optics, where the ratio of the sines of the angles is equal to the ratio of the velocities (or indices of refraction).

###### Outlines

##### π― Introduction to Conservation of Energy and Kinetic-Potential Relationship

The problem begins with a particle starting at the origin on a frictionless plane, transitioning between two regions with different potentials. The task is to express the velocity V_2 in terms of the initial velocity V_1 and potential V_0 using conservation of energy. The first region has only kinetic energy, while the second region has both kinetic and potential energy. By applying conservation principles, the relationship between V_2, V_1, and V_0 is derived, concluding the first part of the question.

##### π Velocity Components and Force Consideration

The script then discusses the velocity components and the forces involved as the particle transitions between the regions. It emphasizes that the change in potential only affects the x-axis component, meaning the y-component of velocity remains unchanged. This leads to the derivation that the sine components of angles Ξ_1 and Ξ_2 are related through the velocities V_1 and V_2, adding another layer to the solution.

##### π Principle of Least Action and Path Calculation

Next, the concept of 'action,' a quantity defined as A = m β« v ds, is introduced. The principle of least action is applied, which states that the path taken by the particle is the one that minimizes this action. Coordinates are defined for the transition point and final position, and the integral is split into two parts for each region. This sets up the problem to find the action as a function of the coordinate alpha, which represents the y-coordinate at the boundary.

##### π Differentiation to Minimize Action

To find the extremum of the action function, the derivative with respect to alpha is calculated. The script walks through the process of differentiating the action expression, using the chain rule and simplifying terms. By setting the derivative equal to zero, the relationship between V_1 and V_2 in terms of the coordinates and velocities is established, which forms the basis for minimizing the action.

##### π Final Derivation and Connection to Snell's Law

The final derivation connects the principles of least action to trigonometric relationships, showing that the velocity ratios relate to the sine of angles, mirroring Snell's Law in optics. The result reinforces the earlier findings, providing a deeper understanding of the problem. The video ends by hinting at the application of these concepts in deriving Snell's Law in a future video, encouraging viewers to explore more Olympiad problems for unique insights into physics.

###### Mindmap

###### Keywords

##### π‘Conservation of Energy

##### π‘Potential Energy

##### π‘Kinetic Energy

##### π‘Principle of Least Action

##### π‘Snell's Law

##### π‘Line Integral

##### π‘Gradient of Potential

##### π‘Trigonometric Functions

##### π‘Chain Rule

##### π‘Pythagorean Theorem

###### Highlights

A particle starts at the origin on a frictionless plane, which is divided into two regions with different potentials.

In region one, the potential is zero, and in region two, the potential is V naught.

The problem begins by expressing V2 in terms of mv1 and the potential using conservation of energy.

In region one, the particle's total energy is purely kinetic, described by 1/2 mv1 squared.

In region two, the total energy includes both kinetic energy and potential energy, leading to the equation: 1/2 mv2 squared + V naught.

Using conservation of energy, V2 is solved in terms of V1 and the potential V naught.

The problem progresses by expressing V2 in terms of V1, Theta1, and Theta2, focusing on changes along the x-axis only.

The y-component of velocity remains unchanged, leading to the relationship V1y = V2y or V1 sin Theta1 = V2 sin Theta2.

The action principle is introduced, defined as the integral of velocity over a path, with the true path being the one that minimizes action.

Coordinates are set for when the particle crosses from region one to two, aiming to find action as a function of alpha.

The integral for action is split into two parts: one for region one and another for region two, due to constant speeds in each region.

The expression for action involves Pythagorean distances, with Mv1 and Mv2 as constants, leading to a differentiation with respect to alpha.

By setting the derivative of action with respect to alpha to zero, the principle of least action is applied.

Terms are simplified and rearranged, showing V1/V2 as a function of geometric terms involving angles and distances.

This mathematical derivation relates to Snell's Law in optics, illustrating how principles in physics can intersect and derive familiar laws.

###### Transcripts

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