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

Options

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

Explanation

To solve the problem, we will use 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. **Identify Initial Conditions:**
- Let the initial velocity of the bullet be \( v \).
- After penetrating 3 cm (which is \( s_1 = 3 \) cm), the bullet loses half of its velocity, so its new velocity is \( \frac{v}{2} \).

2. **Calculate Initial and Final Kinetic Energy:**
- The initial kinetic energy (\( KE_i \)) when the bullet is fired:
\[ KE_i = \frac{1}{2} m v^2 \]
- The kinetic energy (\( KE_f \)) after penetrating 3 cm:
\[ KE_f = \frac{1}{2} m \left(\frac{v}{2}\right)^2 = \frac{1}{2} m \cdot \frac{v^2}{4} = \frac{1}{8} mv^2 \]

3. **Calculate Work Done During First Penetration:**
- The work done (\( W_1 \)) during the first 3 cm penetration is equal to the change in kinetic energy:
\[ W_1 = KE_i - KE_f = \frac{1}{2} mv^2 - \frac{1}{8} mv^2 = \frac{4}{8} mv^2 - \frac{1}{8} mv^2 = \frac{3}{8} mv^2 \]
- The work done can also be expressed as:
\[ W_1 = F \cdot s_1 \]
- Therefore:
\[ F \cdot 3 = \frac{3}{8} mv^2 \quad \text{(1)} \]

4. **Calculate Work Done During Further Penetration:**
- Now, consider the second part where the bullet penetrates further until it comes to rest. Let the additional distance be \( s_2 \).
- The initial kinetic energy for this part is:
\[ KE_i = \frac{1}{2} m \left(\frac{v}{2}\right)^2 = \frac{1}{8} mv^2 \]
- The final kinetic energy when the bullet comes to rest is 0, so:
\[ W_2 = KE_i - 0 = \frac{1}{8} mv^2 \]
- The work done can also be expressed as:
\[ W_2 = F \cdot s_2 \]
- Therefore:
\[ F \cdot s_2 = \frac{1}{8} mv^2 \quad \text{(2)} \]

5. **Divide the Two Equations:**
- From equations (1) and (2):
\[ \frac{F \cdot s_2}{F \cdot s_1} = \frac{\frac{1}{8} mv^2}{\frac{3}{8} mv^2} \]
- This simplifies to:
\[ \frac{s_2}{3} = \frac{1}{3} \]
- Hence:
\[ s_2 = 1 \text{ cm} \]

### Final Answer:
The bullet will penetrate **1 cm** further before coming to rest.