Course:PHYS350/Mock Midterm Solution

1. A particle of mass $m$ moves under a conservative force with a potential energy function given by

$V(x)=\left\{{\begin{matrix}{\frac {1}{2}}k\left(a^{2}-x^{2}\right)&{\rm {for}}|x|<a,\\0&{\rm {for}}|x|\geq a,\end{matrix}}\right.$

where $a$ and $k$ are constants, and $a>0$. What is the force on the particle? Sketch the function $V$, for both cases $k>0$ and $k<0$, and describe the possible types of motion.

Let's take $k=m\alpha ^{2}>0$ and we have a particle initial in the region $x<-a$, moving to the right with velocity $v$. When it emerges into the region $x>a$, will it do so earlier or later than if it were moving freely under no force? Find an expression for the time difference. (To do the required integral, try the substitution $x=A\sinh u$ where $A$ is some constant.

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Red is $k>0$ and blue is $k<0$

The different types of motion for $k>0$ are

1. $E<{\frac {1}{2}}ka^{2}$, motion is restricted to the left or right side. The particle will bounce off of the hill.
2. $E>{\frac {1}{2}}ka^{2}$, the particle can climb over the hill (slowing on the way up) and end up on the other side.

The different types of motion for $k<0$ are

1. $E<0$, motion is restricted to well. It oscillates harmonically back and forth.
2. $E>0$, the particle will escape to infinity.

The time to go pass the hill if it were not there is simply $2v/a$. The time to go over the hill is

$t_{\rm {hill}}=2\int _{0}^{a}{\frac {1}{\sqrt {v^{2}-\alpha ^{2}(a^{2}-x^{2})}}}dx={\frac {2}{\sqrt {v^{2}-\alpha ^{2}a^{2}}}}\int _{0}^{a}{\frac {1}{\sqrt {1+{\frac {\alpha ^{2}}{v^{2}-ka^{2}}}x^{2}}}}dx.$

Let's make the substitution that $x={\frac {\sqrt {v^{2}-\alpha ^{2}a^{2}}}{\alpha }}\sinh u$ to get

$t_{\rm {hill}}={\frac {2}{\alpha }}\int _{x=0}^{x=a}{\frac {\cosh u}{\sqrt {1+\sinh ^{2}u}}}du={\frac {2}{\alpha }}\int _{x=0}^{x=a}{\frac {\cosh u}{\cosh u}}du={\frac {2}{\alpha }}\sinh ^{-1}{\frac {\alpha a}{\sqrt {v^{2}-\alpha ^{2}a^{2}}}}={\frac {2}{\alpha }}\tanh ^{-1}{\frac {\alpha a}{v}}.$

Therefore the time difference is

$\Delta t={\frac {2a}{v}}-{\frac {2}{\alpha }}\tanh ^{-1}{\frac {\alpha a}{v}}.$

As $\alpha \rightarrow 0$ the time difference vanishes. As $v\rightarrow \alpha a$ from above, the time difference diverges.

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2. A particle of mass $m$ is attached to a string of length $l$. The particle moves in a horizontal plane without friction, and the string passes through a hole in the centre of the plane and is attached to another object of mass $M$ below the plane that only moves vertically under the Earth's gravitational field. Write out the Lagrangian for this system and derive the equations of motion. What quantities are conserved?

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The kinetic energy of two particles is

$T={\frac {1}{2}}\left(M+m\right){\dot {r}}^{2}+{\frac {1}{2}}mr^{2}{\dot {\theta }}^{2}$

and the potential energy is just

$V=Mgr$

where $r$ and $\theta$ are the polar coordinates of the particle that moves in the plane relative to the hole. The Lagrangian is

$L=T-V={\frac {1}{2}}\left(M+m\right){\dot {r}}^{2}+{\frac {1}{2}}mr^{2}{\dot {\theta }}^{2}-Mgr$

The total energy ($E=T+V$) is conserved and the angular momentum $mr^{2}{\dot {\theta }}$. The equations of motion are

$\left(M+m\right){\ddot {r}}=-Mg+mr{\dot {\theta }}^{2}$

and

${\ddot {\theta }}=-2{\frac {\dot {r}}{r}}{\dot {\theta }}$