A bullet of mass 0.01J kg and travelling at a speed of `500m//s` strikes a block of mass 2 kg which is suspended by a string of length 5m. The cnetre of gravity of the block is found to rise a vertical distance of 0.1m, figure. What is the speed of the bullet after it emerges from the block ? `(g=9.8m//s^(2)).`

Explanation

Here, `m=0.01kg, v=500m//s`
`M=2kg, l=5m, h=0.1m, g=9.8m//s^(2)`
Let V be the velocity acquired by the block.
`:. (1)/(2)MV^(2)=Mgh`
`V=sqrt(2gh)=sqrt(2xx9.8xx0.1)=1.4m//s`
If `v^(')` is speed of bullet on emerging out of block, Figure, then by law of conservation of momentum, we get
`mv+Mxx0=MV+mv^(')`
or `v^(')=(mv-MV)/(m)=(0.01xx500-2xx1.4)/(0.01)`
`:. v^(')=200m//s`