In his mission as a spy, Loid Forger fired a bullet weighing mB= 0.007 kg with a horizontal speed of VB= 550 m/s into thelower end of a gate which can be modeled into a slenderm=16-kg bar of length L=1.5m for simplified analysis only.Know that h=0.25 m and that the gate, which has a pinconnection at point C, is initially at rest. The bullet becomesembedded into the gate after impact.1.5 mсAD0.25 mVB= 550 m/s 2. Which of the following momentum equations about C is CORRECT and can be used to solve forthe angular velocity of the gate? ◆A. 1.25mßvß = 1.25mßvß - 0.5mvg — ÏÅ BwB. 1.25m BVB = 1.25mßvß +0.5mvg + IA BwC. 1.250mßvß = 1.250mßvß +0.500mvGD. 1.25mßvß = 1.25mBvB – 0.5mvG

Givenm=0.007kgVB=550 m/sm=16kgL=1.5mh=0.25m

In his mission as a spy, Loid Forger fired a bullet weigh = 0.007 kg with a horizontal speed of vB= 550 m/s into lower end of a gate which can be modeled into a slend m=16-kg bar of length L=1.5m for simplified analysis o Know that h=0.25 m and that the gate, which has a pin connection at point C, is initially at rest. The bullet bec

In his mission as a spy, Loid Forger fired a bullet weigh = 0.007 kg with a horizontal speed of vB= 550 m/s into lower end of a gate which can be modeled into a slend m=16-kg bar of length L=1.5m for simplified analysis o Know that h=0.25 m and that the gate, which has a pin connection at point C, is initially at rest. The bullet bec

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

In his mission as a spy, Loid Forger fired a bullet weighing mB
= 0.007 kg with a horizontal speed of VB= 550 m/s into the
lower end of a gate which can be modeled into a slender
m=16-kg bar of length L=1.5m for simplified analysis only.
Know that h=0.25 m and that the gate, which has a pin
connection at point C, is initially at rest. The bullet becomes
embedded into the gate after impact.
1.5 m
с
A
D
0.25 m
VB= 550 m/s

2. Which of the following momentum equations about C is CORRECT and can be used to solve for
the angular velocity of the gate? ◆
A. 1.25mßvß = 1.25mßvß - 0.5mvg — ÏÅ Bw
B. 1.25m BVB = 1.25mßvß +0.5mvg + IA Bw
C. 1.250mßvß = 1.250mßvß +0.500mvG
D. 1.25mßvß = 1.25mBvB – 0.5mvG

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