A bullet of mass 10 ; g and speed 500 ; m/s i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Mass of the bullet,$m = 10 \; g = 0.01 \; kg$. Velocity of the bullet,$v = 500 \; m/s$. Width of the door,$L = 1.0 \; m$. Mass of the door,$M = 12

Vedclass · Accepted Answer

$0.625 \; rad/s$

Vedclass · Answer

(A) Mass of the bullet,$m = 10 \; g = 0.01 \; kg$. Velocity of the bullet,$v = 500 \; m/s$. Width of the door,$L = 1.0 \; m$. Mass of the door,$M = 12 \; kg$. The bullet embeds at the centre,so the distance from the hinge is $r = L/2 = 0.5 \; m$. Angular momentum of the bullet about the hinge: $L_{bullet} = mvr = 0.01 	imes 500 	imes 0.5 = 2.5 \; kg \cdot m^2/s$. Moment of inertia of the door about the hinge: $I_{door} = \frac{1}{3}ML^2 = \frac{1}{3} 	imes 12 	imes (1.0)^2 = 4 \; kg \cdot m^2$. Moment of inertia of the bullet about the hinge: $I_{bullet} = mr^2 = 0.01 	imes (0.5)^2 = 0.0025 \; kg \cdot m^2$. Total moment of inertia $I = I_{door} + I_{bullet} = 4 + 0.0025 = 4.0025 \; kg \cdot m^2$. By conservation of angular momentum,$L_{bullet} = I \omega$. $\omega = \frac{2.5}{4.0025} \approx 0.6246 \; rad/s \approx 0.625 \; rad/s$.

Similar Questions

$A$ hot solid sphere is rotating about a diameter at an angular velocity $\omega_0$. If it cools so that its radius reduces to $\frac{1}{\eta}$ of its original value,its angular velocity becomes .............

Angular momentum of a particle rotating under a central force is constant due to

$A$ rod of length $L$ and mass $M$ is free to move on a frictionless surface. $A$ ball of mass $m$ moving with speed $v$ strikes the rod as shown in the figure. What should be the mass of the ball so that it remains at rest after the collision?

Vedclass Test Series

Exam Paper Generator

Online Exam Module