A 2 ounce bullet is fired with a velocity of 1000 f/s at 0=10°into a 20 lb square panel with sides b = 6 in length. The bullet hits the panel at h = 4 in. Knowing that the panel is initially at rest, what is the velocity of the center of the panel immediately after the bullet becomes embedded?

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
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PROBLEM #3
NOTE: 16 ounces = 1 lb ; 12 inches = 1 foot
A 2 ounce bullet is fired with a velocity of 1000 f/s at 0=10°into a 20 lb square panel with sides b= 6 in
length. The bullet hits the panel at h = 4 in. Knowing that the panel is initially at rest, what is the velocity
of the center of the panel immediately after the bullet becomes embedded?
PROBLEM 17.101
A 45-g bullet is fired with a velocity of 400 m/s at e = 30° into a 9-kg
square panel of side b= 200 mm. Knowing that h=150 mm and that the
panel is initially at rest, determine (a) the velocity of the center of the panel
immediately after the bullet becomes embedded, (b) the impulsive reaction
at A, assuming that the bullet becomes embedded in 2 ms.
SOLUTION
mg =0.045 kg m, =9 kg lg =-mph² =-(9(0.200)² =
0.06 kg m?
Kinematics. After impact, the plate is rotating about the fixed Point A with angular velocity o = ()
w).
2
Principle of impulse and momentum. To simplify the analysis, neglect the mass of the bullet after impact.
Transcribed Image Text:PROBLEM #3 NOTE: 16 ounces = 1 lb ; 12 inches = 1 foot A 2 ounce bullet is fired with a velocity of 1000 f/s at 0=10°into a 20 lb square panel with sides b= 6 in length. The bullet hits the panel at h = 4 in. Knowing that the panel is initially at rest, what is the velocity of the center of the panel immediately after the bullet becomes embedded? PROBLEM 17.101 A 45-g bullet is fired with a velocity of 400 m/s at e = 30° into a 9-kg square panel of side b= 200 mm. Knowing that h=150 mm and that the panel is initially at rest, determine (a) the velocity of the center of the panel immediately after the bullet becomes embedded, (b) the impulsive reaction at A, assuming that the bullet becomes embedded in 2 ms. SOLUTION mg =0.045 kg m, =9 kg lg =-mph² =-(9(0.200)² = 0.06 kg m? Kinematics. After impact, the plate is rotating about the fixed Point A with angular velocity o = () w). 2 Principle of impulse and momentum. To simplify the analysis, neglect the mass of the bullet after impact.
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