A bullet of mass m B = 1.69 ? 10 ? 2 kg is moving with a speed of 102 . m / s wh
ID: 2262054 • Letter: A
Question
A bullet of mass mB=1.69?10?2kg is moving with a speed of 102.m/s when it collides with a rod of mass mR=7.69kg and length L=1.05m (shown in the figure). The rod is initially at rest, in a vertical position, and pivots about an axis going through its center of mass. The bullet embeds itself in the rod at a distance L4 from the pivot point. As a result, the bullet-rod system starts rotating.
a) Find the magnitude of the angular velocity, ? , of the bullet-rod system after the collision. You can neglect the width of the rod and can treat the bullet as a point mass.
Please include a multiplication sign (*) to indicate that two units are being multiplied.
b) What is the change in kinetic energy in the collision?
Please include a multiplication sign (*) to indicate that two units are being multiplied.
Explanation / Answer
apllying conservation of angular momentum
before collision
A.M of bullet=l=mB*v*L/4--------1
after collision AM of bullet=l'=mb*(L/4)^2*w-------2
where w if angular velocity of rod fater collision
AM of rod=L=Iw
I=1/3m(L/2)^2
I=MI of rod=1/3 ML^2--------3
from 1,2,3,
l=L+l'
from above
w=mB*v*L/4/(I+m(L/2)^2)
w=mB*v/(4L(M/12+mB/16))=.638 rad/sec
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.