A bullet of mass m B = 0.0149 kg is moving with a speed of 102 m/s when it colli
ID: 2243292 • Letter: A
Question
A bullet of mass mB = 0.0149 kg is moving with a speed of 102 m/s when it collides with a rod of mass mR = 7.85 kg and length L= 1.08 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?
(Hint: Please enter a positive number for your answer! If ?K < 0, then the amount of kinetic energy lost is a positive number.)
Explanation / Answer
Here
Angular momentum remmians COnstant
Therefore
mvr = Iw
(0.0149*102*1.08/4) = ((1/12)*7.85*1.08^2 + 0.0149*(1.08/4)^2)*w
w = 0.537 rad/sec
Loss in Kientic energy = 0.5*m*v^2 - 0.5*I*w^2
= 0.5*0.0149*102^2 - 0.5*((1/12)*7.85*1.08^2 + 0.0149*(1.08/4)^2)*0.537^2
= 77.4 J
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