Question

A rectangular door of mass \(18\,\mathrm{kg}\) and width \(90\,\mathrm{cm}\) is hinged at one end and can rotate about the vertical axis without friction. A bullet of mass \(15\,\mathrm{g}\) fired with a speed of \(450\,\mathrm{m\,s^{-1}}\) into the door gets embedded exactly at the center of the door. The angular speed of the door just after the bullet gets embedded is

• A. \(2.5~\mathrm{rad\,s^{-1}}\)
• B. \(0.625~\mathrm{rad\,s^{-1}}\)
• C. \(0.416~\mathrm{rad\,s^{-1}}\)
• D. \(1.25~\mathrm{rad\,s^{-1}}\)

Hint:

For perfectly inelastic rotational collisions, \[ \boxed{ mvr=I_{\text{final}}\omega } \] where \(I_{\text{final}}\) includes both the body and the embedded particle.

Answer 1

The Correct Option is B

Step 1: Apply conservation of angular momentum about the hinge. Initial angular momentum of the bullet is \[ L_i=mvr, \] where \[ m=0.015~\mathrm{kg},\qquad v=450~\mathrm{m\,s^{-1}},\qquad r=\frac{0.9}{2}=0.45~\mathrm{m}. \] Hence, \[ L_i=0.015\times450\times0.45 =3.0375~\mathrm{kg\,m^2\,s^{-1}}. \]

Step 2: Calculate the total moment of inertia. For the rectangular door about one edge, \[ I_{\text{door}} =\frac13 ML^2 =\frac13(18)(0.9)^2 =4.86~\mathrm{kg\,m^2}. \] Moment of inertia of the embedded bullet, \[ I_{\text{bullet}} =mr^2 =0.015(0.45)^2 =0.00304~\mathrm{kg\,m^2}. \] Thus, \[ I_{\text{total}} =4.86+0.00304 \approx4.863~\mathrm{kg\,m^2}. \]

Step 3: Determine the angular speed. Using \[ L_i=I_{\text{total}}\omega, \] \[ \omega =\frac{3.0375}{4.863} \approx0.625~\mathrm{rad\,s^{-1}}. \] Hence, \[ \boxed{\omega=0.625~\mathrm{rad\,s^{-1}}} \] Therefore, \[ \boxed{(B)} \] is the correct answer.