come about?
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. Te
hinged at one end and rotates about a vertical axis practically without friction. Find
the angular speed of the door just after the bullet embeds into it.
Answers
Given,
Given, Mass of bullet (m) = 10g = 0.01 Kg
Given, Mass of bullet (m) = 10g = 0.01 Kg Speed of bullet (v) = 500m/s
Given, Mass of bullet (m) = 10g = 0.01 Kg Speed of bullet (v) = 500m/s Width of wire(L) = 1m
Given, Mass of bullet (m) = 10g = 0.01 Kg Speed of bullet (v) = 500m/s Width of wire(L) = 1m Mass of door ( M) = 12 Kg
Given, Mass of bullet (m) = 10g = 0.01 Kg Speed of bullet (v) = 500m/s Width of wire(L) = 1m Mass of door ( M) = 12 Kg A/c to question,
Given, Mass of bullet (m) = 10g = 0.01 Kg Speed of bullet (v) = 500m/s Width of wire(L) = 1m Mass of door ( M) = 12 Kg A/c to question, Gets embedded exactly at the centre of door , so , distance from the hinged end of the door,
Given, Mass of bullet (m) = 10g = 0.01 Kg Speed of bullet (v) = 500m/s Width of wire(L) = 1m Mass of door ( M) = 12 Kg A/c to question, Gets embedded exactly at the centre of door , so , distance from the hinged end of the door, (r) = L/2 = 1/2 = 0.5 m
Given, Mass of bullet (m) = 10g = 0.01 Kg Speed of bullet (v) = 500m/s Width of wire(L) = 1m Mass of door ( M) = 12 Kg A/c to question, Gets embedded exactly at the centre of door , so , distance from the hinged end of the door, (r) = L/2 = 1/2 = 0.5 m Angular momentum transferred by the bullet to the door = mvr= 0.01 × 500 × 0.5 = 2.5 js______(1)
Given, Mass of bullet (m) = 10g = 0.01 Kg Speed of bullet (v) = 500m/s Width of wire(L) = 1m Mass of door ( M) = 12 Kg A/c to question, Gets embedded exactly at the centre of door , so , distance from the hinged end of the door, (r) = L/2 = 1/2 = 0.5 m Angular momentum transferred by the bullet to the door = mvr= 0.01 × 500 × 0.5 = 2.5 js______(1)Moment of inertia of the door about the vertical axis at one end of its end ,
Given, Mass of bullet (m) = 10g = 0.01 Kg Speed of bullet (v) = 500m/s Width of wire(L) = 1m Mass of door ( M) = 12 Kg A/c to question, Gets embedded exactly at the centre of door , so , distance from the hinged end of the door, (r) = L/2 = 1/2 = 0.5 m Angular momentum transferred by the bullet to the door = mvr= 0.01 × 500 × 0.5 = 2.5 js______(1)Moment of inertia of the door about the vertical axis at one end of its end , I = ML²/3 = 12 × 1/3 = 4kgm²
Given, Mass of bullet (m) = 10g = 0.01 Kg Speed of bullet (v) = 500m/s Width of wire(L) = 1m Mass of door ( M) = 12 Kg A/c to question, Gets embedded exactly at the centre of door , so , distance from the hinged end of the door, (r) = L/2 = 1/2 = 0.5 m Angular momentum transferred by the bullet to the door = mvr= 0.01 × 500 × 0.5 = 2.5 js______(1)Moment of inertia of the door about the vertical axis at one end of its end , I = ML²/3 = 12 × 1/3 = 4kgm² Now,
Given, Mass of bullet (m) = 10g = 0.01 Kg Speed of bullet (v) = 500m/s Width of wire(L) = 1m Mass of door ( M) = 12 Kg A/c to question, Gets embedded exactly at the centre of door , so , distance from the hinged end of the door, (r) = L/2 = 1/2 = 0.5 m Angular momentum transferred by the bullet to the door = mvr= 0.01 × 500 × 0.5 = 2.5 js______(1)Moment of inertia of the door about the vertical axis at one end of its end , I = ML²/3 = 12 × 1/3 = 4kgm² Now, angular momentum = Iw
Given, Mass of bullet (m) = 10g = 0.01 Kg Speed of bullet (v) = 500m/s Width of wire(L) = 1m Mass of door ( M) = 12 Kg A/c to question, Gets embedded exactly at the centre of door , so , distance from the hinged end of the door, (r) = L/2 = 1/2 = 0.5 m Angular momentum transferred by the bullet to the door = mvr= 0.01 × 500 × 0.5 = 2.5 js______(1)Moment of inertia of the door about the vertical axis at one end of its end , I = ML²/3 = 12 × 1/3 = 4kgm² Now, angular momentum = Iw 2.5 = 4×w { from equation (1)
Given, Mass of bullet (m) = 10g = 0.01 Kg Speed of bullet (v) = 500m/s Width of wire(L) = 1m Mass of door ( M) = 12 Kg A/c to question, Gets embedded exactly at the centre of door , so , distance from the hinged end of the door, (r) = L/2 = 1/2 = 0.5 m Angular momentum transferred by the bullet to the door = mvr= 0.01 × 500 × 0.5 = 2.5 js______(1)Moment of inertia of the door about the vertical axis at one end of its end , I = ML²/3 = 12 × 1/3 = 4kgm² Now, angular momentum = Iw 2.5 = 4×w { from equation (1) w = 0.625 rad/s
Answer:
ANSWER
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=
3
1
×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