A bullet of mass mB = 0.01 kg is moving with a speed of 106 m/s when it collides
ID: 2094242 • Letter: A
Question
A bullet of mass mB = 0.01 kg is moving with a speed of 106 m/s when it collides with a rod of mass mR = 7 kg and length L = 2 m (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 imbeds itself in the rod at a distance L/4 from the pivot point. As a result, the bullet-rod system starts rotating.
(a) Find 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.
(b) How much kinetic energy is lost in the collision?
Explanation / Answer
A)
initial angular momentum = mb vb L/4 = 0.01 x 106 x 0.5 = 0.53
final angular momentum = I = [Mr L^2 /12 + Mb (L/4)^2]
[7 x 2^2 /12 + 0.01 x (2/4)^2] = 2.336
2.336 = 0.53
= 0.227
B)
initial KE = 1/2 mb vb^2 = 1/2 x 0.01 x 106^2 = 56.18 J
final KE = 1/2 I ^2 = 1/2 x 2.336 x 0.227^2 = 0.06J
energy lost = 56.12 J
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.