3. A bullet of mass M - 2.59 x 10 kg is moving with a speed of 374.5 m/s when it

Question

3. A bullet of mass M - 2.59 x 10 kg is moving with a speed of 374.5 m/s when it collides with a pole of mass M1-3 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. (a) What is the angular velocity of the rod and the bullet immediately after the collision? (b) What is the initial kinetic energy of the system before collision? (c) What is the rotational kinetic energy immediately after the impact? (d) Is the kinetic energy conserved? If not, what is the energy loss? L/3 ation of ans

Explanation / Answer

a

. initial angular momentum of the bullet= mass x speed of bullet x distance from axis of rotation

= 2.59 x 10^-3 x 374.5 x 2/3

= 0.646 kg. m^2/s

Final angular momentum of pole and bullet system

= ( moment of inertia of system) x angular velocity

= ( mass x length of pole^2/3 + mass of bullet x distance from axis^2) x w

= ( 3 x 2^2/3 + 2.59 x 10^-3 x (2/3)^2) x w

= 4.001 w kg.m^2/s

As angular momentum is conserved, initial angular momentum=final angular momentum

so, 4.001 w= 0.646

so, w= 0.161 rad/s

so angular velocity of pole and bullet system is 0.161 rad/s

b.

initial kinetic energy of the system = 1/2 x mass of bullet x bullet speed^2

=1/2 x 2.59 x 10^-3 x 374.5^2

= 181.62 j

c.

rotational kinetic energy immediately after impact = 1/2 x moment of inertia of system x w^2

=1/2 x 4.001 x 0.161^2

= 0.052 j

d.

kinetic energy is not conserved as kinetic energy before collision is not equal to kinetic energy after collision.

Energy loss= 181.62 - 0.052 = 181.568J

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