Ncert Exercises & Solutions

Question 7. 24. A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the center of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.(Hint: The moment of inertia of the door about the vertical axis at one end is ML²/3.)

Mass of the bullet, m = 10 g = 10 × 10^-3 kg
Velocity of the bullet, v = 500 m/s
The thickness of the door, L = 1 m
The radius of the door, r = 1/2m
Mass of the door, M = 12 kg
Angular momentum imparted by the bullet on the door:
α = mvr

= (10 \times 10^{- 3}) \times (500) \times \frac{1}{2} = 2.5 k g m^{2} s^{- 1}

...(i)
Moment of inertia of the door:

I = \frac{1}{3} M L^{2}

\begin{array}{l} = \frac{1}{3} \times 12 \times (1)^{2} = 4 {kgm}^{2} \\ But α = Iω \\ ∴ ω = \frac{α}{I} \\ = \frac{2 .5}{4} = 0 .625 {rads}^{- 1} \end{array}

Angular momentum imparted by the bullet, L = mv x r
=(10 x 10^-3) x 500 x 1/2 =2.5
Also,I=ML²/3=12 x (1.0)²/3=4 kg m²
Since L=Iw
w=L/I=2.5/4=0.625 rad / s

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