A bullet of mass `m` fired at `30^(@)` to the horizontal leaves the barrel of the gun with a velocity `upsilon`. The bullet hits a soft target at a height `h` above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target.
Which of the following statements are correct in respect of bullet after it emerges out of the target ?

Options

A. The velocity of the bullet will be more than half of its earlier velocity
B. The bullet will move in a different parabolic path
C. The internal energy of the particles of the target will increase
D. The bullet will fall vertically downward after hiting the target

Explanation

(b,d,f) Consider th adjacent diagram for the given situation in the question

(b) Conserving energy between "O" and "A"
`=(1)/(2)(mv")^(2)=(1)/(2)[(1)/(2)m(v')^(2)]`
`=(1)/(2)(mv")^(2)=(1)/(4)m(v')^(2)=(1)/(4)m[v^(2)-2gh]`
`(v")^(2)=(v^(2)-2gh)/(2)=(v^(2))/(2)-gh`
`Rightarrow v"=sqrt((v^(2))/(2)-gh)`
From eqs. (i) and (ii)
`v"=sqrt((v^(2)-2gh))/((sqrt(v^(2)-2gh))/(sqrt2))=sqrt2`
`Rightarrow v"=(v')/(sqrt2)=v^(2)((v')/(2))`
`Rightarrow (v")/((v')/(2))=sqrt2=1.414 gt 1`
`Rightarrow v" gt (v')/(2)`
Hence after emerging from the target of the bullet (v") is more than half of the earlier velocity v' (velocity before emerging into the target)
(d) As the velocity of the bullet changes v' which is less than `v^(1)` hence path followed will change and the bullet reaches at point B instead of A', as shown in the figure
(f) As the bullet is passing through the target the loss in energy of the bullet is transferred to particles of the target. Therefore their internal energy increases.