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 ?

Answer

(B,D,F)

Explanation

Consider the adjacent diagram for the given situation in the question.
(b) Conserving energy between "O" and "A"

`U_(i) + K_(i) = U_(f) + K_(f)`
`rArr`
`0 + (1)/(2) mv^(2) = mgh + (1)/(2) mv'`
`rArr ((v')^(2))/(2) =(v^(2))/(2) = -gh`
`rArr (v')^(2) =v^(2) -2 gh rArr v' = sqrt(v^(2) - 2gh)`
where `v'` is speed of the bullet just before hitting the target. Let speed after emerging from the target is `v"` then .
By question,
=`(1)/(2)(mv'')^(2) = (1)/(2)[(1)/(2)m(v')^(2)]`
`(1)/(2) m(v'') = (1)/(4) m(v')^(2) = (1)/(4) m'[v^(2) -2gh]`
`rArr (v'')^(2) = (v^(2) -2gh)/(2) = (v^(2))/(2) -gh`
`rArr v'' = sqrt((v^(2))/(2)-gh)`
From Eqs. (i) and (ii)
`(v')/(v'') = (sqrt(v^(2)-2gh))/(sqrt(v^(2)-2gh)/(sqrt(2))) = sqrt(2)`
`rArr v'' = (v')/(sqrt(2)) = v^(2) ((v')/(2))`
`rArr (v'')/((v')/(2)) = sqrt(2) = 1.414 gt1`
`rArr v'' gt (v')/(2)`
Hence, after emerging from target velocity of the bullet `(v'')` is more than half of its earlier velocity `v'` (velocity before emerging into the target)
(d) As the velocity of the bullet changes to `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 enery of the bullet is transferred to particles of the target. Therefore, their internal energy increases.