A bullet of mass 10g moving horizontally with a velocity of` 400 ms^-1`strikes a wooden block of mass 2kg which is suspended by a light inextensible string of length 5m .As a result , the centre of gravity of the block is found to rise a vertical distance of 10 cm . The speed of the bullet after it emerges out horizontally from thr block will be

Explanation

Mass of bullet `m=20 m =0.02kg `
Mass of wooden block
`M=500 gm=0.5kg`
Velocity of the bullet with which it strikes
`u=300m/sec`
Let tehh bullet emerges out with velociyt V and the velocity of block `v^1`
As per law of conservation of momentum
`mu=Mv^1+m_1v`.......i
Again applying work energy principle for the block after the collision
`0-(1/2)Mx(v^10^2=-M.g.h`
(where h=0.2m)
`rarr (v^2)^2=2gh`
`=sqrt(20xx2.0)=2m/sec`
Substituting the value `v^1` in equation i we get
`0.02xx300=2+0.2xxv`
`rarr v=(6-1)/0.02`
`=5/0.02=250m/sec`