A bullet fired into a door gets embedded exactly at its center, causing the door to rotate about its vertical hinge axis practically without friction with an angular velocity of $625\text{ rad s}^{-1}$. The door is $0\text{ m}$ wide and weighs $12\text{ kg}$. If the mass of the bullet is $10\text{ g}$, find the speed with which it was fired. (Hint: The moment of inertia of the door about the vertical axis at one end is $I = \frac{ML^2}{3}$)}
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Because the bullet's mass ($10\text{ g}$) is tiny compared to the door ($12\text{ kg}$), ignoring its contribution to the final moment of inertia simplifies the math immensely. The right side evaluation yields exactly $4 \times 0.625 = 2.5$. Dividing this by the bullet's factor ($0.005$) instantly gives $500\text{ m/s}$.