Here is my question:

A circular coin of radius a falls at speed u without rotating onto a smooth horizontal table. The perpendicular to its face makes an angle θ with the vertical. Determine the state of motion of the coin just after it strikes the table, assuming that the collision is elastic. Show that, when θ is small, the coin strikes the table a second time at an angle of 5 θ/11.

I think that after landing on the table the impulse $\delta p$ gives the centre of mass of the coin an upwards linear velocity $v$. The impulse also provides an angular impulse $\delta p* cos(theta)$ which sets the coin rotating about an axis in its plane through its centre of mass, so I imagine the other end of the coin will hit the table, it has downward velocity due to the rotation $ omega*r*cos(theta)$ and an upward velocity due to the upward motion of the centre of mass.
This end of the coin starts at a height of $2a* sin(theta)$ so takes a time T to reach the table, which is this distance over its velocity.
Setting angular momentum to  the angular impulse $\delta p* cos(theta)$ gives  an expression for omega, which can be integrated (note that the cosine term above is a constant!) to find the change in angle over a time T. Doing this and using small angles gives an expression for changes in theta In terms of u and $ v$. Using conservation of KE gives an expression $v$ in u, $v\approx3u/5$.
Is this approach ok?