The 2.00-kg slender rod shown is hanging in a vertical position and is pin-supported at point A. The slender rod is initially at rest until a 1.000-kg block C, strikes it at its end at point B. The block slides on a frictionless surface with a velocity of 3.50 m/s to the right. After the impact, it slides with a velocity of 1.250 m/s to the right, and the bar rotates with an angular velocity, w'.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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The 2.00-kg slender rod shown is hanging in a vertical
position and is pin-supported at point A. The slender rod is
initially at rest until a 1.000-kg block C, strikes it at its end
at point B.
The block slides on a frictionless surface with a velocity of
3.50 m/s to the right. After the impact, it slides with a
velocity of 1.250 m/s to the right, and the bar rotates with
an angular velocity, w'.
3.5 m/s
с
B
1.5 m
Transcribed Image Text:The 2.00-kg slender rod shown is hanging in a vertical position and is pin-supported at point A. The slender rod is initially at rest until a 1.000-kg block C, strikes it at its end at point B. The block slides on a frictionless surface with a velocity of 3.50 m/s to the right. After the impact, it slides with a velocity of 1.250 m/s to the right, and the bar rotates with an angular velocity, w'. 3.5 m/s с B 1.5 m
2. Which of the following gives the correct momentum equation about point A for the slender rod and the block?
A. +5: mcvc (1.500 m) + mabVg (0.750 m) + ĪABW = mcv'c (1.500 m) + MABV'G (0.750 m) + ĪABW'
B. +U: mcvc (1.500 m) = mcv'c (1.500 m) + MABV'g (0.750 m) + ĪABW'
G
c. +5: mcvc (1.500 m) = mcv'c (1.500 m) + Mabvʼg (0.750 m) + Īabw'(0.750 m)
G
D. +5: mcvc (1.500 m) = mcv'c (1.500 m) + mabv'g (0.750 m)
G
Transcribed Image Text:2. Which of the following gives the correct momentum equation about point A for the slender rod and the block? A. +5: mcvc (1.500 m) + mabVg (0.750 m) + ĪABW = mcv'c (1.500 m) + MABV'G (0.750 m) + ĪABW' B. +U: mcvc (1.500 m) = mcv'c (1.500 m) + MABV'g (0.750 m) + ĪABW' G c. +5: mcvc (1.500 m) = mcv'c (1.500 m) + Mabvʼg (0.750 m) + Īabw'(0.750 m) G D. +5: mcvc (1.500 m) = mcv'c (1.500 m) + mabv'g (0.750 m) G
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