In a ballistics demonstration, a police officer fires a bullet mass `50.0g` with speed `200ms^(-1)` on soft plywood of thickness 2.00cm. The bullet emerges only with `10%` of its initial kinetic energy. What is the emergent speed of the bullet ?

Explanation

Here, `m=50.0g`
`=(50)/(1000)kg=(1)/(20)kg`
`v_(i)=200ms^(-1)`
`:.` Initial K.E., `K_(i)=(1)/(2)mv_(i)^(2)`
`=(1)/(2)xx(1)/(20)(200)^(2)=1000J`
Final K.E., `K_(f)=10% (K_(i))`
`=(10)/(100)xx1000J=100J`
If `v_(f)` is emergent speed of the bullet, then
`(1)/(2)mv_(f)^(2)=K_(f)`
`v_(f)=sqrt((2K_(f))/(m))=sqrt((2xx100)/(1//20))=63.2m//s`
Note that K.E. is reduced by `90%` but speed is reduced by nearly `68%`.