Question

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 $I = \frac{ML^2}{3}$)}

• A. $645\text{ m s}^{-1}$
• B. $342\text{ m s}^{-1}$
• C. $124\text{ m s}^{-1}$
• D. $500\text{ m s}^{-1}$

Hint:

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 $500\text{ m/s}$.

Answer 1

The Correct Option is D

Concept: Since no external turning torques act on the system about the vertical hinge reference line, the total angular momentum ($L$) of the system must be conserved during the collision: \[ L_{\text{initial}} = L_{\text{final}} \] Before impact, the angular momentum of the bullet moving at velocity $v$ striking at perpendicular distance $r$ is $L_i = m_b \cdot v \cdot r$. After the bullet embeds itself, the total combined moment of inertia is $I_{\text{total}} = I_{\text{door}} + I_{\text{bullet}}$.

Step 1: Calculate individual system moment of inertia parts.
Given data parameters:
• Door mass, $M = 12\text{ kg}$, door width, $L = 1.0\text{ m}$
• Bullet mass, $m_b = 10\text{ g} = 0.01\text{ kg}$
• Impact point distance, $r = \frac{L}{2} = 0.5\text{ m}$ Let's evaluate the door's moment of inertia: \[ I_{\text{door}} = \frac{M L^2}{3} = \frac{12 \times (1.0)^2}{3} = 4\text{ kg m}^2 \] The embedded bullet adds a point mass moment of inertia: \[ I_{\text{bullet}} = m_b \cdot r^2 = 0.01 \times (0.5)^2 = 0.01 \times 0.25 = 0.0025\text{ kg m}^2 \] Since $I_{\text{bullet}}$ is extremely small compared to $I_{\text{door}}$, we can approximate $I_{\text{total}} \approx I_{\text{door}} = 4\text{ kg m}^2$.

Step 2: Apply angular momentum conservation to solve for velocity ($v$).
Given final angular velocity, $\omega = 0.625\text{ rad s}^{-1}$: \[ m_b \cdot v \cdot r = I_{\text{total}} \cdot \omega \] \[ 0.01 \times v \times 0.5 = 4 \times 0.625 \] \[ 0.005 \cdot v = 2.5 \] \[ v = \frac{2.5}{0.005} = 500\text{ m s}^{-1} \]