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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 $\theta$ 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 $5 \hspace{.5mm} \theta/11$.

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 $=\delta p* a*cos(\theta)$

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 $2\hspace{.5mm} a* sin(\theta)$ 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?

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 $\theta$ 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 $5 \hspace{.5mm} \theta/11$.

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 $=\delta p* a*cos(\theta)$

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 $2\hspace{.5mm} a* sin(\theta)$ so takes a time T to reach the table, which is this distance over its velocity.

Setting angular momentum to the angular impulse $=\delta p* a cos(\theta)$ 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?

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 $\theta$ 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 $5 \hspace{.5mm} \theta/11$.

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 $2\hspace{.5mm} a* sin(\theta)$ 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?

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 $\theta$ 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 $5 \hspace{.5mm} \theta/11$.

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 $=\delta p* a*cos(\theta)$

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 $2\hspace{.5mm} a* sin(\theta)$ 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?