Question

A bullet of mass $m$ fired at $30^{°}$ 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 is correct in respect of bullet after it emerges out of the target?

• A. The velocity of the bullet remains the same
• B. The velocity of the bullet will be reduced to half its initial value
• C. The velocity of the bullet will be more than half of its earlier velocity
• D. The bullet will continue to move along the same parabolic path

Answer 1

Answer: (B)

Let $v^{\prime }$ be the velocity of the bullet after emerging from the target.
Kinetic energy of the bullet emerging from the target
$\frac{1}{2}m{v}^{\prime 2}=\frac{1}{2}\left(\frac{1}{2}m{v}^2\right)\text{ or }{v}^{\prime }=\frac{v}{\sqrt{2}}=0.707v$
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 target changes.
The bullet will move in a different parabolic path.