A bullet when fixed at a target with a velocity of `100 ms^(-1)`, penetrates one metre into it. If the bullet is fired with the same velocity as a similar target with a thickness `0.5` metre, then it will emerge from it with a velocity of

Options

A. `50 sqrt(2) m//s`
B. `(50)/(sqrt(2)) m//s`
C. `50 m//s`
D. `10 m//s`

Explanation

Let `v` be the velocity with which the bullet will energy . Now, change in kinetic energy = work done
For first case : `(1)/(2) m (100)^(2) - (1)/(2) m (0)^(2) = F xx 1` (i)
For second case : `(1)/(2) m (100)^(2) - (1)/(2) mv^(2) = F xx 0.5` (ii)
Dividing equation (ii) by (i), we get
`((100)^(2) - v^(2))/((100)^(2)) = (0.5)/(1) = (1)/(2) or v = (100)/(sqrt(2)) = 50 sqrt(2) m//s`