Subfield: Mechanics

A The Extremum Principle in Mechanics

Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts).

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Figure 1
Context question:
A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$.
Context answer:
\boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$}


Context question:
A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$.
Context answer:
\boxed{$v_2=\frac{v_1 \sin\theta_1}{\sin \theta_2}$}


Extra Supplementary Reading Materials:

We define a quantity called action $A=m \int v(s) d s$, where $d s$ is the infinitesimal length along the trajectory of a particle of mass $m$ moving with speed $v(s)$. The integral is taken over the path. As an example, for a particle moving with constant speed $v$ on a circular path of radius $R$, the action $A$ for one revolution will be $2 \pi m R v$. For a particle with constant energy $E$, it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which $A$ defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA).

Question: A.3 PLA implies that the trajectory of a particle moving between two fixed points in a region of constant <br> potential will be a straight line. Let the two fixed points $\mathrm{O}$ and $\mathrm{P}$ in Fig. 1 have coordinates $(0,0)$ and <br> $\left(x_{0}, y_{0}\right)$ respectively and the boundary point where the particle transits from region I to region II have <br> coordinates $\left(x_{1}, \alpha\right)$. Note that $x_{1}$ is fixed and the action depends on the coordinate $\alpha$ only. State the <br> expression for the action $A(\alpha)$. Use PLA to obtain the relationship between $v_{1} / v_{2}$ and these coordinates.

Solution:
By definition $A(\alpha)$ from $O$ to $P$ is

$$
A(\alpha)=m v_{1} \sqrt{x_{1}^{2}+\alpha^{2}}+m v_{2} \sqrt{\left(x_{0}-x_{1}\right)^{2}+\left(y_{0}-\alpha\right)^{2}}
$$

Differentiating w.r.t. $\alpha$ and setting the derivative of $A(\alpha)$ to zero

$$
\begin{gathered}
\frac{v_{1} \alpha}{\left(x_{1}^{2}+\alpha^{2}\right)^{1 / 2}}-\frac{v_{2}\left(y_{0}-\alpha\right)}{\left[\left(x_{0}-x_{1}\right)^{2}+\left(y_{0}-\alpha\right)^{2}\right]^{1 / 2}}=0 \\
\therefore \frac{v_{1}}{v_{2}}=\frac{\left(y_{0}-\alpha\right)\left(x_{1}^{2}+\alpha^{2}\right)^{1 / 2}}{\alpha\left[\left(x_{0}-x_{1}\right)^{2}+\left(y_{0}-\alpha\right)^{2}\right]^{1 / 2}}
\end{gathered}
$$

Note this is the same as A2, namely $v_{1} \sin \theta_{1}=v_{2} \sin \theta_{2}$.

Final answer: $\frac{v_{1}}{v_{2}}=\frac{(y_{0}-\alpha)(x_{1}^{2}+\alpha^{2})^{1 / 2}}{\alpha[(x_{0}-x_{1})^{2}+(y_{0}-\alpha)^{2}]^{1 / 2}}$