Question

A bullet of mass m fired at ${30}^o$ to the horizontal leaves the barrel of the gun with a velocity $v$. 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?

• 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 bullet will move in a different parabolic path.
• D. The internal energy of the particles of the target will increase.

Answer 1

Answer: (B), (C), (D)

The velocity of the bullet will be more than half of its earlier velocity.
The bullet will move in a different parabolic path.
The internal energy of the particles of the target will increase.

As the K.E is reduced by one-half so the velocity will get reduced by $\sqrt{2}$ times as K.E. is directly proportional to the square of the velocity.

So, option A is incorrect and option B is correct

As the velocity gets reduced the path in which the bullet moves will also get changed as the equation of trajectory of the projectile motion depends on the velocity

So, option C is also correct

The KE lost by the bullet will be gained by the particles of the target.

So, option D is also correct

Hence correct answers are options $B,C,D$