Mock Midterm 2 Solution

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Complete two of three:

1. A proton is elastically scattered through an angle of 56° by a nucleus which recoils at an angle of α=60°. Find the atomic mass of the nucleus and the fraction of the kinetic energy transferred to it.


The first thing to do is draw the collision triangle

Collision.gif

We don't know p^{*} but q^{*}=p^{*}. Also \theta =56^{\circ } and \alpha =60^{\circ }. what we would like to know is the ratio of the two parts of the bottom of the triangle.

With geometry we can fill in the rest of the angles.

Collision2.gif

where \beta =\pi -\theta -2\alpha . Now we can use the law of sines to get the answer,

{\frac  {\sin \theta }{q^{*}}}={\frac  {\sin \beta }{m_{p}/m_{N}p^{*}}}

and

{\frac  {\sin \theta }{\sin \beta }}={\frac  {q^{*}}{m_{p}/m_{N}p^{*}}}={\frac  {m_{N}}{m_{p}}}.

We know that \theta =56^{\circ } and \beta =180^{\circ }-120^{\circ }-56^{\circ }=4^{\circ }. Taking the sines yields and atomic mass of 11.88 mp which is equal to 11.96 or essentially 12.

What about the sharing of the kinetic energy? To do this we need the value of \theta ^{*}=180^{\circ }-2\alpha =60^{\circ }. We also know that {\frac  {T_{2}}{T}}={\frac  {4m_{1}m_{2}}{M^{2}}}\sin ^{2}{\frac  {\theta ^{*}}{2}}={\frac  {4\left(11.88\right)}{12.88^{2}}}\left({\frac  {1}{2}}\right)^{2}={\frac  {11.88}{12.88^{2}}}=0.0716


2. Find the moment of inertia about an axis through its centre of a uniform hollow sphere of mass M and outer and inner radii a and b.


The easy way to look at this is to imagine the object to be a big sphere minus a little sphere. The moment of inertia of the inner sphere is

I_{b}={\frac  {2}{5}}M_{b}b^{2}

and its volume is

V_{b}={\frac  {4}{3}}\pi b^{3}

so we can write the moment of inertia in terms of the density, by writing M_{b}=\rho V_{b}

I_{b}={\frac  {2}{5}}\rho b^{2}V_{b}.

And similarly for the big sphere we get

I_{a}={\frac  {2}{5}}\rho a^{2}V_{a}.

so

I=I_{a}-I_{b}={\frac  {2}{5}}\rho \left(a^{2}V_{a}-b^{2}V_{b}\right)={\frac  {2}{5}}{\frac  {M}{V_{a}-V_{b}}}\left(a^{2}V_{a}-b^{2}V_{b}\right)={\frac  {2}{5}}M{\frac  {a^{5}-b^{5}}{a^{3}-b^{3}}}


3. Write down the kinetic energy of a particle in cylindrical coordinates in a frame rotating with angular velocity ω about the z-axis. Show that the terms proportional to ω and ω2 reproduce the Coriolis force and centrifugal force respectively.


Let's write out the kinetic energy of the particle in the inertial frame

T={\frac  {1}{2}}m\left({\dot  r}^{2}+{\dot  z}^{2}+r^{2}{\dot  \theta }_{i}^{2}\right)

and use \theta =\theta _{i}+\omega t to give

T={\frac  {1}{2}}m\left[{\dot  r}^{2}+{\dot  z}^{2}+r^{2}\left({\dot  \theta }+\omega \right)^{2}\right].

Now let's apply Lagrange's equations for the three coordinates

m{\ddot  r}=r\left({\dot  \theta }+\omega \right)^{2}=mr{\dot  \theta }^{2}+2mr{\dot  \theta }\omega +mr\omega ^{2}

m{\ddot  \theta }+2m{\dot  r}{\dot  \theta }+2m{\dot  r}\omega =0

m{\ddot  z}=0