A bullet of mass M = 2.59 Times 10^3 kg is moving with a speed of 374.5 m/s when
ID: 1418951 • Letter: A
Question
A bullet of mass M = 2.59 Times 10^3 kg is moving with a speed of 374.5 m/s when it collides with a pole of mass M_1 = 3 kg and length L = 2.0 m. The pole 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 pole at a point that is 1/3 the length of the pole above the pivot point. As a result the bullet-pole system starts to rotate. What is the angular velocity of the rod and the bullet immediately after the collision? What is the initial kinetic energy of the system before collision? What is the rotational kinetic energy immediately after the impact? Is the kinetic energy conserved? If not, what is the energy loss?Explanation / Answer
V = initial speed of bullet
w = angular velocity of the combination
Using conservation of angular momentum
MV(L/3) + 0 = (mL2/12 + M (L/3)2) w
(0.00259) (374.5) (2/3) = (3 (2)2/12 + (0.00259) (2/3)2) w
w = 0.65 rad/s
b)
initial KE = (0.5) M V2 = (0.5) (0.00259) (374.5)2 = 181.62 J
c)
Rotational KE = (0.5) (mL2/12 + M (L/3)2) w2 = (0.5) (3 (2)2/12 + (0.00259) (2/3)2) (0.65)2 = 0.21
d)
no KE is not conserved
loss = 181.62 - 0.21 = 181.41 J
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