A bullet fired into a target loses half its velocity after penetrating 3 cm. Then it comes to rest after moving further a distance of

Options

A. 6 cm
B. 3cm
C. 2 cm
D. 1cm

Answer

To solve the problem, we need to determine the distance the bullet travels after losing half its velocity and coming to rest. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the initial conditions
The bullet loses half of its initial velocity after penetrating 3 cm. Let's denote:
- Initial velocity \( U \)
- Final velocity after 3 cm \( V = \frac{U}{2} \)

### Step 2: Apply the kinematic equation
We can use the kinematic equation:
\[ V^2 = U^2 - 2AS \]
where:
- \( V \) is the final velocity,
- \( U \) is the initial velocity,
- \( A \) is the acceleration (deceleration in this case),
- \( S \) is the distance traveled (3 cm).

Substituting the known values:
\[ \left(\frac{U}{2}\right)^2 = U^2 - 2A(3) \]

### Step 3: Simplify the equation
Expanding the left side:
\[ \frac{U^2}{4} = U^2 - 6A \]
Rearranging gives:
\[ 6A = U^2 - \frac{U^2}{4} \]
This simplifies to:
\[ 6A = \frac{4U^2 - U^2}{4} = \frac{3U^2}{4} \]

### Step 4: Solve for acceleration \( A \)
Now, we can solve for \( A \):
\[ A = \frac{3U^2}{24} = \frac{U^2}{8} \text{ cm/s}^2 \]

### Step 5: Calculate the distance to come to rest
Now we need to find the distance \( S' \) the bullet travels after the first 3 cm until it comes to rest. The final velocity \( V' = 0 \) and the initial velocity for this segment is \( V = \frac{U}{2} \).

Using the same kinematic equation:
\[ V'^2 = V^2 - 2AS' \]
Substituting the known values:
\[ 0 = \left(\frac{U}{2}\right)^2 - 2\left(\frac{U^2}{8}\right)S' \]
This simplifies to:
\[ 0 = \frac{U^2}{4} - \frac{U^2}{4}S' \]

### Step 6: Rearranging to find \( S' \)
Rearranging gives:
\[ \frac{U^2}{4}S' = \frac{U^2}{4} \]
Dividing both sides by \( \frac{U^2}{4} \) (assuming \( U \neq 0 \)):
\[ S' = 1 \text{ cm} \]

### Final Answer
The bullet comes to rest after moving a further distance of **1 cm**.

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