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Asked by anushkagoyal656 | 28 Apr, 2022, 05:18: PM
Let m be mass of bullet moving with initial speed u.

Total initial momentum = m u

After penetrating wooden block , let the bullet move with speed v . Due to collision, wooden block moves with speed V just after collision.

Final momentum just after collision = ( M V ) + ( m v )
where M is mass of wooden block

Since momentum is conserved , we have

m u = ( M V ) + ( m v )

after simplification , we get ,   v = u - (M/m) V  ..................(1)

After collision , centre of mass of wooden block is raised 10 cm upwards

Hence by energy conservation , (1/2) M V2 = ( M g h )

Hence speed of wooden block just after collision , V =    ..................(2)

Using eqn.(2) , we rewrite eqn.(1) as

v = u - ( M / m )
v = 400 - ( 2 /.01 )
v = 120 m/s
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