ring, the bullet embedded block swings to position 4 which is making an angle al. Due to gravitational force, the bullet embedded block then swings backwards irst about O and then about the peg P after the cord comes in contact with the ne energy lost during the impact and aerodynamic drag, determine ken for the bullet embedded block measured from position A to swing one cycle, um angle by using vibration method that the bullet embedded block will swing r side after the cord hits the peg P, and in the cord 0.25 second after the bullet embedded block swings backwards from
ring, the bullet embedded block swings to position 4 which is making an angle al. Due to gravitational force, the bullet embedded block then swings backwards irst about O and then about the peg P after the cord comes in contact with the ne energy lost during the impact and aerodynamic drag, determine ken for the bullet embedded block measured from position A to swing one cycle, um angle by using vibration method that the bullet embedded block will swing r side after the cord hits the peg P, and in the cord 0.25 second after the bullet embedded block swings backwards from
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Subject: Engg. Mechanics.

Transcribed Image Text:Figure Ql shows a bullet of mass m= 6g is fired into a wood block of mass M =1 kg which is
at equilibrium at the instance shown. The block is suspended by an inextensible cord with a length
1=1.2 m. After firing, the bullet embedded block swings to position 4 which is making an angle
0= 5' with vertical. Due to gravitational force, the bullet embedded block then swings backwards
from position A first about O and then about the peg P after the cord comes in contact with the
peg. Neglecting the energy lost during the impact and aerodynamic drag, determine
(a) the time taken for the bullet embedded block measured from position A to swing one cycle,
the maximum angle by using vibration method that the bullet embedded block will swing
(b)
to the other side after the cord hits the peg P, and
(c)
the tension in the cord 0.25 second after the bullet embedded block swings backwards from
its position A.
d = 1/2
A
Figure Q1
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