Answer:
[tex]\omega_f = 0.4\ rad/s[/tex]
Explanation:
given,
door width = 1 m
mass of = 15 Kg
Mass of bullet = 0.01 Kg
speed of the bullet = 400 m/s
[tex]I_{total} =I_{door} + I_{bullet}[/tex]
[tex]I_{total} =\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2[/tex]
a) from conservation of angular momentum
[tex]L_i = L_f[/tex]
[tex] mv\dfrac{W}{2} = I_{total}\omega_f[/tex]
[tex] mv\dfrac{W}{2}= \dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2)\omega_f[/tex]
[tex]\omega_f= \dfrac{mv\dfrac{W}{2}}{\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2}[/tex]
[tex]\omega_f= \dfrac{\dfrac{mv}{2}}{\dfrac{MW }{3}+(\dfrac{mW}{4})}[/tex]
[tex]\omega_f= \dfrac{\dfrac{0.01\times 400}{2}}{\dfrac{15\times 1 }{3}+(\dfrac{0.01\times 1}{4})}[/tex]
[tex]\omega_f = 0.4\ rad/s[/tex]
