Q.A bullet of mass $m$ fired at $30^\circ$ 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? (Note: more than one of the given options may be correct.)

Concept: Work–Energy Theorem — the loss in kinetic energy of the bullet equals the work done on the target, and the bullet’s path after the target depends on its velocity components.

Step 1: Kinetic energy after impact

Initial KE = $\frac{1}{2} m v^2$. After emerging, KE = $\frac{1}{2} \times \frac{1}{2} m v^2 = \frac{1}{4} m v^2$.

Step 2: Speed after impact

Let $v'$ be the speed after emerging. Then $\frac{1}{2} m v'^2 = \frac{1}{4} m v^2 \implies v'^2 = \frac{v^2}{2} \implies v' = \frac{v}{\sqrt{2}}$.

Since $\frac{v}{\sqrt{2}} > \frac{v}{2}$, the speed is more than half the initial speed. So (A) is false, (B) is true.

Step 3: Path after impact …

1. The core idea: Work–Energy Theorem

The bullet loses kinetic energy because the target does negative work on it — that work is the force of the target multiplied by the distance over which it acts. The Work–Energy Theorem says:

The net work done on an object equals its change in kinetic energy.

So if the bullet emerges with half its original kinetic energy, the target has removed exactly half of that energy. No other forces (like gravity) do work during the extremely brief collision, because the displacement during the hit is horizontal and gravity acts vertically — so the work done by gravity during the collision is negligible.

2. Step-by-step reasoning

Step 1: Find the speed after the hit

Let the speed just before impact be $u$. Then:

After emerging, $K_{\text{after}} = \frac12 K_{\text{before}} = \frac14 m u^2$.

But $K_{\text{after}} = \frac12 m v_{\text{after}}^2$, so:

This is not half the speed — it’s about $0.707\,u$. So option (A) is false, and option (B) is true: the velocity is more than half of its earlier value.

Step 2: Does the path remain the same?

The bullet is a projectile. Its path is determined by the velocity vector (magnitude and direction) at the moment it leaves the target. The target changes both:

• The magnitude drops from $u$ to $u/\sqrt{2}$. …

Concept: Work-Energy Theorem and Projectile Trajectories

The target does negative work on the bullet, which is what removes half

its kinetic energy; since $KE\propto v^2$, halving the energy changes the

speed by a factor of $1/\sqrt2$, not $1/2$ — and that changed velocity

vector determines a different parabola afterward.

Step 1: Find the speed just after emerging from the target

Let $u$ be the speed just before impact, so $K_{\text{before}}=\tfrac12 mu^2$.

Given $K_{\text{after}} = \tfrac12 K_{\text{before}}$:

This is more than half of $u$ (since $0.707 > 0.5$), so (a) is false, (b) is true.

Step 2: Decide whether the path stays the same

A projectile's parabola is fixed entirely by its velocity vector at that

point. The target has reduced the bullet's speed to $u/\sqrt2$ (a smaller

magnitude, and generally not purely in the original direction either, since

the target resists along the bullet's motion). A projectile launched with a

different velocity vector traces a different parabola, so (c) is false, (d) is true. …