A bullet of mass 20 g moving horizontally at speed of 300 m/s is fired into wooden block of mas 500 suspended by a long string. The bullet crosses the block and emerges on the other side. If the centre of mass of the block rises through a height of 20.0 cm, find the speed of the bullet as it emerges from the block.

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`