A bullet fired into a fixed target loses half of its velocity after penetrating 3cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?

Options

A. (a) 2.0cm
B. (b) 3.0cm
C. (c) 1.0cm
D. (d) 1.5cm

Explanation

To solve the problem, we will apply the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- Let the initial velocity of the bullet be \( V \).
- After penetrating 3 cm into the target, the bullet's velocity reduces to \( \frac{V}{2} \).

2. **Applying the Work-Energy Theorem**:
- The work done by the resistive force \( F \) while the bullet penetrates the target can be expressed as:
\[ W = F \cdot S \]
- The change in kinetic energy as the bullet penetrates 3 cm (0.03 m) is given by:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} = \frac{1}{2} m V^2 - \frac{1}{2} m \left(\frac{V}{2}\right)^2 \]
- Simplifying the kinetic energy change:
\[ \Delta KE = \frac{1}{2} m V^2 - \frac{1}{2} m \frac{V^2}{4} = \frac{1}{2} m V^2 \left(1 - \frac{1}{4}\right) = \frac{1}{2} m V^2 \cdot \frac{3}{4} = \frac{3}{8} m V^2 \]

3. **Setting Up the Equation for Work Done**:
- The work done by the resistive force over the distance \( S_1 = 3 \, \text{cm} = 0.03 \, \text{m} \) is:
\[ W = F \cdot S_1 = F \cdot 0.03 \]
- According to the work-energy theorem:
\[ F \cdot 0.03 = \frac{3}{8} m V^2 \]
- Rearranging gives:
\[ F = \frac{3 m V^2}{8 \cdot 0.03} \]

4. **Finding Further Penetration**:
- Now, we need to find how much further the bullet penetrates before coming to rest. Let this additional distance be \( S_2 \).
- The initial velocity for this phase is \( \frac{V}{2} \) and the final velocity is 0.
- The change in kinetic energy for this phase is:
\[ \Delta KE = \frac{1}{2} m \left(\frac{V}{2}\right)^2 - 0 = \frac{1}{2} m \frac{V^2}{4} = \frac{1}{8} m V^2 \]
- The work done while penetrating \( S_2 \) is:
\[ W = F \cdot S_2 \]
- Setting the work done equal to the change in kinetic energy gives:
\[ F \cdot S_2 = \frac{1}{8} m V^2 \]

5. **Relating the Two Work Equations**:
- From the first part, we have \( F = \frac{3 m V^2}{8 \cdot 0.03} \).
- Substitute \( F \) into the second equation:
\[ \frac{3 m V^2}{8 \cdot 0.03} \cdot S_2 = \frac{1}{8} m V^2 \]
- Cancel \( m V^2 \) from both sides:
\[ \frac{3 S_2}{8 \cdot 0.03} = \frac{1}{8} \]
- Rearranging gives:
\[ 3 S_2 = 0.03 \quad \Rightarrow \quad S_2 = \frac{0.03}{3} = 0.01 \, \text{m} = 1 \, \text{cm} \]

6. **Final Answer**:
- The bullet will penetrate an additional distance of **1 cm** before coming to rest.