A bullet fired into a fixed target loses half of its velocity after penetrating `1 cm`. How much further it will penetrate before coming to rest, assuming that it faces constant resistance to motion

Options

A. `1.5 cm`
B. `1.0 cm`
C. `3.0 cm`
D. `2.0 cm`

Explanation

To solve the problem step by step, we will use the equations of motion and the concept of constant resistance.

### Step 1: Understand the initial conditions
The bullet loses half of its velocity after penetrating 1 cm. Let the initial velocity of the bullet be \( V \). After penetrating 1 cm, the velocity becomes \( \frac{V}{2} \).

### Step 2: Use the equation of motion
We can use the equation of motion which relates initial velocity, final velocity, acceleration, and displacement:
\[ v^2 = u^2 + 2as \]
Here,
- \( v \) is the final velocity (\( \frac{V}{2} \)),
- \( u \) is the initial velocity (\( V \)),
- \( a \) is the acceleration (which will be negative since it is retardation),
- \( s \) is the displacement (1 cm or 0.01 m).

Substituting the known values:
\[ \left(\frac{V}{2}\right)^2 = V^2 + 2(-a)(0.01) \]

### Step 3: Simplify the equation
Expanding the equation gives:
\[ \frac{V^2}{4} = V^2 - 0.02a \]
Rearranging leads to:
\[ 0.02a = V^2 - \frac{V^2}{4} \]
\[ 0.02a = \frac{3V^2}{4} \]
Thus, we find:
\[ a = \frac{3V^2}{0.08} = \frac{3V^2}{8} \]

### Step 4: Determine further penetration before coming to rest
Now, we need to find how much further the bullet penetrates before coming to rest. The bullet's velocity at this point is \( \frac{V}{2} \), and it will come to rest (final velocity \( v = 0 \)).

Using the same equation of motion:
\[ 0 = \left(\frac{V}{2}\right)^2 + 2(-a)(x) \]
Substituting \( a = \frac{3V^2}{8} \):
\[ 0 = \frac{V^2}{4} - 2\left(\frac{3V^2}{8}\right)x \]
Rearranging gives:
\[ 2\left(\frac{3V^2}{8}\right)x = \frac{V^2}{4} \]
\[ \frac{3V^2}{4}x = \frac{V^2}{4} \]
Dividing both sides by \( \frac{V^2}{4} \) (assuming \( V \neq 0 \)):
\[ 3x = 1 \]
Thus:
\[ x = \frac{1}{3} \text{ cm} \]

### Conclusion
The bullet will penetrate an additional \( \frac{1}{3} \) cm before coming to rest.