A long simple pendulum has a bob of mass 450 gm. A bullet of mass 50 gm is fired from below into the bob. As a result of this Bob along with the bullet rises through a distance of 4.9 metre. Calculate the velocity of the bullet.
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Let the velocity of bullet be v m/s.
Then before collision:
the momentum of bullet =0.05×v Kg-m/s.
the momentum of Bob= 0.45× 0 = 0 .
=> initial momentum of bullet+Bob system= 0.05v+0=0.05v.
further after collision bullet stuck to the Bob hence the momentum of the system is given by:(0.05+0.45)×V kg-m/s.
(where V is the velocity after the impact of both bullet and Bob)
APPLYING LAW OF CONSERVATION OF LINEAR MOMENTUM:
0.05v=0.5V
=> v=10V.............(1)
as the system (bullet+bob ) rises to height 4.9m.
then ,
APPLYING LAW OF CONSERVATION OF MECHANICAL ENERGY:
KE (just after impact)= PE of system.
(1/2)×0.5×V×V=0.5×g×4.9
(g=acceleration due to gravity)
solving above equation we get :
V =9.8m/s.
now place this value of V in (1)
we get velocity of bullet =10×9.8=98m/s.
Then before collision:
the momentum of bullet =0.05×v Kg-m/s.
the momentum of Bob= 0.45× 0 = 0 .
=> initial momentum of bullet+Bob system= 0.05v+0=0.05v.
further after collision bullet stuck to the Bob hence the momentum of the system is given by:(0.05+0.45)×V kg-m/s.
(where V is the velocity after the impact of both bullet and Bob)
APPLYING LAW OF CONSERVATION OF LINEAR MOMENTUM:
0.05v=0.5V
=> v=10V.............(1)
as the system (bullet+bob ) rises to height 4.9m.
then ,
APPLYING LAW OF CONSERVATION OF MECHANICAL ENERGY:
KE (just after impact)= PE of system.
(1/2)×0.5×V×V=0.5×g×4.9
(g=acceleration due to gravity)
solving above equation we get :
V =9.8m/s.
now place this value of V in (1)
we get velocity of bullet =10×9.8=98m/s.
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