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2. In your own words, explain the solution to the “Clacking Block” problem.The solution started off by taking two physics equations: conservation of energy and total momentum and both those equations are set to a constant. With these equations, they created a phase space and represented v1 as the x-coordinate plane and v2 as the y-coordinate plane. The conservation of energy equation results in the shape of an ellipse, and every set of velocity is a point on that ellipse. Since we are dealing with pi they changed the ellipse to a circle by multiplying the x-coordinate by root m1 and the y-coordinate by root m2. Now we can say that the conservation of energy equation can be represented by the equation of a circle which equals a constant(1/2(x2+y2) = const). When the block starts moving it is the right block moving left the velocity is negative and it means that we are on the left side of the plane (1st point). To find the next point on the circle, they used the equation for the conservation of momentum (√m1x+√m2y=const) which represents the equation for the slope (−√𝑚1/𝑚2). If we draw that slope line it will have a point on the circle which represents the velocities at the collision. The next point will be when the second block bounces off the wall, the speed of the block remains the same, but the velocity goes from negative to positive. This point will be the exact same point as the second point only reflected about the x-axis. Using the slope, you can repeat this process and find the rest of the points of collision. ^Using the inscribed angle theorem we are shown the angles of the archs. We can see that that every arch angle is 2θ except for the one in the end zone which is less than 2θ. This means that the two blocks are both moving to the right without colliding again. If we add another 2θ, it will overlap with the last one. This means that the blocks would collide as many times as possible before it covers more than 2π which is a full circle. The equation looks like:

2θ+2θ+2θ+...+2θ <2π which can be simplified to N× θ < π

Now to find θ they used tanθ=√m2/√m1 and θ =tan-1√m2√m1, m2 being the smaller block and m1 being the bigger block.

Plugging in different mass ratios we see that the arctan of the ratio is very close to the ratio value itself.
Clacking Block Problem:
The "Clacking Block" problem involves two sliding blocks in an idealized world with no friction and perfectly elastic collisions. The surprising fact is that when the mass ratio between the blocks is a power of 100, such as 1,000,000, the total number of collisions (including those with the wall) has the same starting digits as pi. The explanation lies in the concept of a phase space (configuration space), which is a way to represent the state of a system in a geometric manner.
To understand the solution, we start by using the conservation of energy and momentum. The total kinetic energy of the system and the total momentum remain constant during collisions. By representing the velocities of the blocks as coordinates on a phase diagram, with the x-coordinate representing √(m1v1 )  and the y-coordinate representing √(m2v2 ), we obtain a circle equation for the conservation of energy.
Each collision corresponds to a jump in the phase diagram, either along a vertical line when the smaller block hits the wall or along a line with a slope of -√(m1/m2  ) when the blocks collide. These jumps alternate between vertical and diagonal lines until the smaller block's velocity becomes positive and smaller than the velocity of the larger block.
The number of collisions is related to the angle between the diagonal line and the vertical line in the phase diagram. By using the inscribed angle theorem, the arc length between two points on the circle is twice this angle. Thus, the number of collisions is determined by how many times this angle can be added to itself before surpassing 2π radians. 
The angle itself is approximated by the inverse tangent of the square root of the small mass divided by the square root of the large mass. For small values, the tangent is approximately equal to the angle. Therefore, when the mass ratio is a power of 100, such as 1,000,000, the number of collisions has the same starting digits as pi. The appearance of pi in the number of collisions is a result of the evenly spaced points on the circle in the phase diagram and the approximation of the tangent function for small angles. 




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