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=10g=0.01kg, v=500m//s`
`M=2kg, l=5m, h=0.1m`
It V is velocity gainied by the block, then from KE of block `=` PE of block.
`(1)/(2)MV^(2)=mgh`
`V=sqrt(2gh)=sqrt(2xx9.8xx0.1)`
`V=1.4m//s`
If `v^(')` is speed of the bullet when it emerges from the block, then applying the principle of conservation of linear momentum,
`mxxv+Mxx0=MxxV+mv^(')`
`0.01xx500+0=2xx1.4+0.01v^(')`
`5=2.8++(v^('))/(100)`
or `v^(')=100(5-2.8)=220m//s`