Consider a stick of mass, M = 14.0 kg and length, L = 1.60 m hanging vertically initially. A bullet of mass, m = 0.040 kg and speed, v = 70 m/s strikes and embeds itself at the bottom end of the stick. Magnitude of acceleration due to gravity is g = 10 m/s². Note: The moment of inertia of a rod about it's center of mass 1 = ML². You may need to use the parallel axis theorem. Closeup Part B: @max M What is the maximum angle, Omax achieved by the stick (with the bullet embedded in it)? Incorrect 2.17⁰ Correct: 6.97⁰ Incorrect 83° Incorrect 9.87⁰ Incorrect None of the above

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Consider a stick of mass, M = 14.0 kg and length, L = 1.60 m hanging vertically initially. A bullet of mass, m = 0.040 kg and speed, v = 70 m/s strikes and embeds
itself at the bottom end of the stick. Magnitude of acceleration due to gravity is g = 10 m/s². Note: The moment of inertia of a rod about it's center of mass
1 = ML². You may need to use the parallel axis theorem.
Closeup
Part B:
@max
M
What is the maximum angle, Omax achieved by the stick (with the bullet embedded in it)?
Incorrect 2.17⁰
Correct: 6.97⁰
Incorrect 83°
Incorrect 9.87⁰
Incorrect None of the above
Transcribed Image Text:Consider a stick of mass, M = 14.0 kg and length, L = 1.60 m hanging vertically initially. A bullet of mass, m = 0.040 kg and speed, v = 70 m/s strikes and embeds itself at the bottom end of the stick. Magnitude of acceleration due to gravity is g = 10 m/s². Note: The moment of inertia of a rod about it's center of mass 1 = ML². You may need to use the parallel axis theorem. Closeup Part B: @max M What is the maximum angle, Omax achieved by the stick (with the bullet embedded in it)? Incorrect 2.17⁰ Correct: 6.97⁰ Incorrect 83° Incorrect 9.87⁰ Incorrect None of the above
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