Rotational Motion Question 31
Question: A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 Kg. It is hinged at one end and rotates about a vertical axis practically without friction. The angular speed of the door just ager the bullet embeds into it will be:
[JEE ONLINE 09-04-2013]
Options:
A) 6.25 rad/sec
B) 0.625 rad/sec
C) 3.35 rad/sec
D) 0.335 rad/sec
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] From conservation of angular momentum (from parallel axes theorem) Angular momentum imparted by bullet on the door=mvr
=$(10×10^{−3})×500×0.5kgm^2/s$
Moment of inertia of the door, $I=ML^2/3=\frac{1}{3}×12×1^2=4kgm^2$
Angular momentum of the system after the bullet gets embedded≈Iω
From conservation of angular momentum about the rotation axis,
$mvr=Iω$
⟹$ω=0.625rad/s$
BETA
