Question

a bullet of mass 50g is fired below into the bob of mass 450g of a long simple pendulum. the bullet remains inside the bob and the bob rises through a height of 1.8m. find the speed of the bullet ?​

Answer 1

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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.

Answer 2

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