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Step-by-Step Solution
Step 1: Understand the scenario
A bullet penetrates 3 cm into a fixed target and, in doing so, its velocity reduces to half of its initial velocity. We assume the bullet experiences a constant resistive force from the target. We want to determine how much farther the bullet will travel before coming to rest.
Step 2: Write the work–energy expressions
Let the initial velocity of the bullet be $v$, its mass be $m$, and the constant resistive force be $F$. The work done by the resistive force over a distance is equal to the loss in kinetic energy of the bullet.
When the bullet travels the first 3 cm (i.e., $3\,\text{cm}$) and slows down from $v$ to $\frac{v}{2}$, the loss in kinetic energy is:
$$
\Delta K_1 = \frac{1}{2} m v^2 - \frac{1}{2} m \left(\frac{v}{2}\right)^2.
$$
The work done by the resistive force over 3 cm is:
$$
W_1 = F \times 3.
$$
By the work–energy theorem, $W_1 = \Delta K_1$.
Step 3: Relate the loss in energy during additional penetration
Let the bullet penetrate an additional distance $x$ until it comes to rest (velocity becomes $0$). Then the loss in kinetic energy during this second phase is:
$$
\Delta K_2 = \frac{1}{2} m \left(\frac{v}{2}\right)^2 - \frac{1}{2} m \cdot 0^2.
$$
The work done by the resistive force over this additional distance $x$ is:
$$
W_2 = F \times x.
$$
Again, by the work–energy theorem, $W_2 = \Delta K_2$.
Step 4: Form the ratio and solve for x
From the first phase:
$$
F \times 3 = \frac{1}{2} m v^2 - \frac{1}{2} m \left(\frac{v}{2}\right)^2.
$$
From the second phase:
$$
F \times x = \frac{1}{2} m \left(\frac{v}{2}\right)^2.
$$
We take the ratio of the second equation to the first:
$$
\frac{x}{3}
= \frac{\frac{1}{2} m \left(\frac{v}{2}\right)^2}{\frac{1}{2} m v^2 - \frac{1}{2} m \left(\frac{v}{2}\right)^2}.
$$
Simplify the kinetic energy terms carefully:
$$
\frac{1}{2} m v^2 - \frac{1}{2} m \left(\frac{v}{2}\right)^2
= \frac{1}{2} m v^2 - \frac{1}{8} m v^2
= \frac{4}{8} m v^2 - \frac{1}{8} m v^2
= \frac{3}{8} m v^2.
$$
And
$$
\frac{1}{2} m \left(\frac{v}{2}\right)^2
= \frac{1}{2} m \cdot \frac{v^2}{4}
= \frac{1}{8} m v^2.
$$
Hence,
$$
\frac{x}{3} = \frac{\frac{1}{8} m v^2}{\frac{3}{8} m v^2} = \frac{1}{3}.
$$
Thus, $x = 1 \, \text{cm}.$
Step 5: Final Answer
The bullet will penetrate an additional 1 cm before coming to rest.
