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 A B 1.5 m
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 A B 1.5 m
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Question
Which of the following gives the correct momentum equation about point A for the slender rod and the block?
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
A
B
1.5 m
Transcribed Image Text:A.
+0: mcvc (1.500 m) + MABVG (0.750 m) + TABW = mcv'c (1.500 m) + MABV'G (0.750 m) + TABW'
B.
+V: mcvc (1.500 m) = mcv'c (1.500 m) + MABV'G (0.750 m) + ĪABW'
Mabv'g
C.
+U: mcvc (1.500 m) = mcv'c (1.500 m) + MABV'G (0.750 m) + TABW'(0.750 m)
C
D.
+5: mcvc (1.500 m) = mcv'c (1.500 m) + MABV'G (0.750 m)
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