A door 1 m side, having a mass of 15 kg is hinged at one side so that it can rotatate without friction about the vertical axis. A bullet having mass 10 gm and speed 400 m/s is fired into the door in a direction perpendicular to the plane of the door and embeds itself at the exact centre of the door. Find the angular velocity of the door just after the bullet embeds itself. Is K.E conserved?
Asked by ayushikas | 14th Jan, 2012, 07:50: PM
Here you can use conservation of angular momentum.
Initially the door was at rest so the angular momentum of door was zero.
The angular momentum of bullet plus door system = 2Kgm2/sec.------(1)
After collision the bullet stick in to the door and they both rotate about the axis of the door.
So the angular momentum of the system= I(M.I. of the door bullet system)* w -----(2)
now we compare (1) and (2)
Here the length of door is not given so it is not possible for us to fiond the M.I. of the door.
Please check that and try it further.
Answered by | 17th Jan, 2012, 02:49: PM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number