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 reduced to half its initial value .
B. The velocity of the bullet will be more than half of its earlier velocity .
C. The internal energy of the particles of th target will increase .
D. The bullet will move in a different parabolic path .

Answer

B, C, D

Explanation

Let v . be the velocity of the bullet with which it emerges after hitting thetarget .
K.E of the bullet emerging form soft target = `1/2 `
`1/2 mv^(2) = 1/2 (1/2 mv^(2))`
[ Air resistance is assumed to be equal to zero ]
or ` v. = v/(sqrt(2)) = 0.707 v `
i.e velocity of bullet is more than half of its earlier velocity
Since the bullet loses some of its initial vertical and horizontal velocities to the soft target , so the bullet after emerging out of the target moves in a different parabolic path .
A part of the K.E is transfered to the target , so internal energy of the particles of the target increases .