Two wooden plank of mass `M_(1)= 1 kg, M_(2) = 2.98 kg` smooth surface. A bullet of mass `m = 20 gm` strikes the block `M_(1)` and pierces through it, then strikes the plank `B` and sticks to its. Consequently both the planks move with equal velocities. Find the percentage change in speed of the bullet when it escapes from the first block.

Answer

`33.3%`

Explanation

Conservation of linear momentum of the system yield
`Mv_(0)=M_(1)v+mv'=(M_(2)+m)v`
`impliesv=(mv_(0))/(M_(2)+m)`
and `mv'=mv_(0)-M_(1)v`

`=mv_(0)-(M_(1)mv_(0))/(M_(2)+m)`
`=mv_(0)[1-M_(1)/(M_(2)+m)]`
`impliesv'=v_(0)-(v_(0)M_(1))/(M_(2)+m)`
now find `(|v'-v_(0)|)/v_(0)xx100=M_(1)/((M+m))xx100=33.3%`