A 10 g bullet traveling at 400 m/s strikes a 10 kg, 1.0-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open. What is the angular velocity of the door just after impact?
Angular velocity of the door just after the impact is = 1.2 rad / sec
Mass of bullet = 10 gm = 0.01kg
speed of bullet = 400 m/sec
mass of door = 10 kg.
width of door = 1 m
bullet hits at edge of the door
Use conservation of momentum.
Initially momentum of door =0
momentum of bullet is total momentum
Li= mvbrb ...................................................(1)
where m = mass of bullet, vb = velocity of bullet, rb =1 m
Substitute the value we get
Li= (0.01 kg) (400 m/s) (1 m) = 4kgm2/s
Assume bullet get stuck in the door and traveling with door.
Momentum after strike is
Now find the inertia
moment of inertia of the door
and moment of inertia of the bullet
Substitute the values in equation(2), we get
Equate initial momentum and final momentum, we get
3.34 ω = 4kgm2/s
ω = 1.197 rad/s = 1.2 rad/s
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a. Straight out to his side, parallel to the floor?
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