The bullet of mass 4×10-^2 kg moving with a velocity of 20m/s pierces through a sand bag and comes to rest in 0.25 sec . What is the initial momentum of the body in (Kg-m/s)
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Answer:
Mass of the bullet is m
b
=50gm=0.05Kg
Initial velocity of bullet u
b
=100m/s
Final velocity is v
b
=0m/s
(1) Initial momentum of bullet p
i
=m
b
u
b
=5Kgm/s
(2) Final momentum of bullet p
f
=m
b
v
b
=0.05×0=0
(3) Retardation occurred by the wooden block is
v
2
−u
2
=2as
−(100)
2
=2a(0.02)
a=−250000m/s
2
(4) Resistive force
F=ma
=0.05×−250000
=12500N
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