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

Answer

To solve the problem step by step, we will analyze the motion of the bullet as it penetrates the target and loses velocity due to constant resistance.

### Step 1: Understand the initial conditions
The bullet is fired with an initial velocity \( V \) and after penetrating \( 3 \, \text{cm} \) (or \( 0.03 \, \text{m} \)), it loses half of its velocity. Thus, after penetrating \( 3 \, \text{cm} \), its velocity becomes:
\[ V_f = \frac{V}{2} \]

### Step 2: Use the kinematic equation to find acceleration
We can use the kinematic equation:
\[ V_f^2 = V_i^2 + 2a s \]
where:
- \( V_f = \frac{V}{2} \) (final velocity after 3 cm)
- \( V_i = V \) (initial velocity)
- \( s = 0.03 \, \text{m} \) (distance penetrated)
- \( a \) is the acceleration (which will be negative since it is deceleration).

Substituting the values:
\[ \left(\frac{V}{2}\right)^2 = V^2 + 2a(0.03) \]
This simplifies to:
\[ \frac{V^2}{4} = V^2 + 0.06a \]

### Step 3: Rearranging the equation
Rearranging the equation gives:
\[ \frac{V^2}{4} - V^2 = 0.06a \]
\[ -\frac{3V^2}{4} = 0.06a \]
Thus, we can express acceleration \( a \) as:
\[ a = -\frac{3V^2}{4 \times 0.06} = -\frac{3V^2}{0.24} = -12.5V^2 \]

### Step 4: Determine the further penetration distance
Now, we need to find out how much further the bullet penetrates before coming to rest. The initial velocity for this part of the motion is \( V_f = \frac{V}{2} \) and the final velocity \( V_f = 0 \).

Using the same kinematic equation:
\[ 0 = \left(\frac{V}{2}\right)^2 + 2(-12.5V^2)s \]
This simplifies to:
\[ 0 = \frac{V^2}{4} - 25V^2s \]

### Step 5: Solve for \( s \)
Rearranging gives:
\[ 25V^2s = \frac{V^2}{4} \]
\[ s = \frac{1}{100} \, \text{m} = 0.01 \, \text{m} = 1 \, \text{cm} \]

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