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Q. 78

Expert-verified
Found in: Page 334

### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

Book edition 4th
Author(s) Randall D. Knight
Pages 1240 pages
ISBN 9780133942651

# 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

See the step by step solution

## Step1: Given information

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

## Step2: Explanation

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