[asy]size(8cm); real w = 2.718; // width of block real W = 13.37; // width of the floor real h = 1.414; // height of block real H = 7; // height of block + string real t = 60; // measure of theta pair apex = (w/2, H); // point where the strings meet path block = (0,0)--(w,0)--(w,h)--(0,h)--cycle; // construct the block draw(shift(-W/2,0)*block); // draws white block path arrow = (w,h/2)--(w+W/8,h/2); // path of the arrow draw(shift(-W/2,0)*arrow, EndArrow); // draw the arrow picture pendulum; // making a pendulum... draw(pendulum, block); // block fill(pendulum, block, grey); // shades block draw(pendulum, (w/2,h)--apex); // adds in string add(pendulum); // adds in block + string add(rotate(t, apex) * pendulum); // adds in rotated block + string dot("$\theta$", apex, dir(-90+t/2)*3.14); // marks the apex and labels it with theta draw((apex-(w,0))--(apex+(w,0))); // ceiling draw((-W/2-w/2,0)--(w+W/2,0)); // floor[/asy] A block of mass $m=\text{4.2 kg}$ slides through a frictionless table with speed $v$ and collides with a block of identical mass $m$, initially at rest, that hangs on a pendulum as shown above. The collision is perfectly elastic and the pendulum block swings up to an angle $\theta=12^\circ$, as labeled in the diagram. It takes a time $ t = \text {1.0 s} $ for the block to swing up to this peak. Find $10v$, in $\text{m/s}$ and round to the nearest integer. Do not approximate $ \theta \approx 0 $; however, assume $\theta$ is small enough as to use the small-angle approximation for the period of the pendulum. [i](Ahaan Rungta, 6 points)[/i]

One hundred billion light years from Earth is planet Glorp. The inhabitants of Glorp are intelligent, uniform, amorphous beings with constant density which can modify their shape in any way, and reproduce by splitting. Suppose a Glorpian has somehow formed itself into a spinning cylinder in a frictionless environment. It then splits itself into two Glorpians of equal mass, which proceed to mold themselves into cylinders of the same height, but not the same radius, as the original Glorpian. If the new Glorpians' angular velocities after this are equal and the angular velocity of the original Glorpian is $\omega$, let the angular velocity of the each of the new Glorpians be $\omega'$. Then, find $ \left( \frac {\omega'}{\omega} \right)^{10} $. [i](B. Dejean, 3 points)[/i]