Animated Solution for Physics - Kinematics: If a body losses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest?

Imagine a high-speed bullet fired directly into a thick, dense wooden block. As soon as the bullet enters the wood at point A, it encounters a massive resistive force. This force acts like a brake, constantly sapping the bullet's speed.

The problem tells us that after penetrating just $3\text{ cm}$ into the block (let's call this point B), the bullet has lost exactly half of its initial velocity. If it started with a velocity $v$, it is now moving at $\frac{v}{2}$. But it hasn't stopped yet! It still has some kinetic energy left, and it will continue to push through the wood until it finally comes to a complete halt at point C. Our mission is to find out exactly how much further it travels—this unknown distance $x$.

Because the wooden block is uniform, it's safe to assume that the resistive force it applies is constant. A constant force means a constant deceleration, which we will call $a$.

Whenever we need to connect initial velocity, final velocity, acceleration, and distance without worrying about the time elapsed, the third equation of motion is our best friend:

We will apply this master equation twice: once for the first part of the journey (from A to B) to find the deceleration, and then again for the final stretch (from B to C) to find the remaining distance.

Let's focus on the journey from point A to point B. The bullet starts with an initial velocity $u=v$ and slows down to a final velocity $v_f=\frac{v}{2}$ over a distance $s=3\text{ cm}$.

Plugging these values into our equation:

Squaring the left side gives us $\frac{v^2}{4}$. Now, let's isolate the acceleration term:

Dividing both sides by 6, we find the deceleration: The negative sign is a beautiful confirmation of our physical intuition—the bullet is indeed slowing down!

Now, let's look at the final stretch from point B to point C. The bullet starts this phase with an initial velocity $u=\frac{v}{2}$ and finally comes to rest, meaning its final velocity $v_f=0$. The distance covered is our unknown, $x$, and the deceleration remains $a=-\frac{v^2}{8}$.

Substituting these into the third equation of motion again:

Let's simplify this expression:

Moving the $x$ term to the left side:

Notice how elegantly the $\frac{v^2}{4}$ terms cancel out on both sides! We are left with a beautifully simple result:

The bullet will penetrate exactly $1\text{ cm}$ more before coming to a complete stop.

Work-Energy Theorem

While the kinematic approach is rigorous and foolproof, there is a faster, more intuitive way to solve this using the Work-Energy Theorem.

The theorem states that the work done by the resistive force equals the change in kinetic energy. Since the force is constant, the work done is directly proportional to the distance traveled ($W=F\cdot s$). Therefore, the loss in kinetic energy is directly proportional to the penetration distance.

Let's look at the kinetic energy ($K=\frac{1}{2}m{v}^2$). When the velocity drops from $v$ to $\frac{v}{2}$, the kinetic energy drops from $K_i$ to $\frac{1}{4}{K}_i$. This means the bullet lost $\frac{3}{4}$ of its initial kinetic energy!

If it takes $3\text{ cm}$ to lose $\frac{3}{4}$ of its energy, how much distance will it take to lose the remaining $\frac{1}{4}$ of its energy? By simple proportions, if $\frac{3}{4}$ energy corresponds to $3\text{ cm}$, then $\frac{1}{4}$ energy corresponds to exactly $1\text{ cm}$.

Physics is truly beautiful when different concepts lead to the exact same elegant result!