Q. 78

Expert-verifiedFound in: Page 334

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

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= mv_{b}r_{b ..................................................}.(1)_{}

where m = mass of bullet, v_{b} = velocity of bullet, r_{b} =1 m

Substitute the value we get

Li= (0.01 kg) (400 m/s) (1 m) = 4kgm^{2}/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 = 4kgm^{2}/s

ω = 1.197 rad/s = 1.2 rad/s

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