Answer :

Given:

Mass of Bullet, m = 10g = 10-3 kg


Speed of bullet, v = 500m/s


Width of the door, w = 1m


Distance from hinge where bullet hits, r = 0.5 m


Mass of door, M = 12 Kg


We know that angular moment is given by,


α = mvr


where,


α = angular momentum


m = mass of the body


v = velocity of the body


r = radius of the body


by putting the value, we get,


α = 10 × 10-3 kg × 500 ms-1 × 0.5 m


α = 2.5 Kgm2s-1 …(i)


Moment of inertia is given by,


I = ML2/3 (For rectangle)


Where,


M = mass of door


L = width of door


I = 1/3 × 12 Kg × (1m)2


I = 4 Kgm2


We also have,


α = Iω


where is α is the angular momentum


l is the linear momentum


ω is the angular velocity


ω = α /I


By putting the values, we get,


ω = 2.5/ 4


ω = 0.625 rad s-1


Note: It is a good practice to memorise the moment of inertia for common shapes like rectangle, circle, triangle, cylinder and sphere.


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