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 statments are correct in respect of bullet after it emerges out of the target ?

Options

A. The internal energy of the particles of the target will increase.
B. The velocity of the bullet will be more than half of its earlier velocity.
C. The bullet with continue to move along the same parablic path.
D. The bullet will move in a diffeent parabolic path.

Answer

A, B, D

Explanation

Assuming air resistance to be zero.
K.E. of bullet emerging from soft target `=(1)/(2)m upsilon'^(2)=(1)/(2)((1)/(2)m upsilon^(2))`
`upsilon^(')=(upsilon)/(sqrt(2))=0.707upsilon`
i.e. velocity is more than half of its earlier velocity.
As the bullet loses some of its vertical velocity component, therefore, velocity on emerging from the soft target changes. The bullet will move in a different parabolic path.
The energy lost in passing through the target is transferred to particles of the target. Therefore, their internal energy increases. Choice `(b), (d), (f),` are correct.