A bullet fired into a wooden block loses half of its velocity after penetrating 40 cm. It comes to rest after penetrating a further distance of

Options

A. 22/3 cm
B. 40/3 cm
C. 20/3 cm
D. 22/5 cm

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Understand the Problem
A bullet is fired into a wooden block and loses half of its velocity after penetrating 40 cm. We need to find the additional distance it travels before coming to rest.

### Step 2: Define Initial Conditions
Let:
- Initial velocity of the bullet = \( u \)
- Velocity after penetrating 40 cm = \( v = \frac{u}{2} \)
- Distance penetrated = \( s_1 = 40 \) cm = 0.4 m

### Step 3: Use the Kinematic Equation
We can use the kinematic equation:
\[ v^2 - u^2 = 2as \]
Where:
- \( v \) = final velocity after 40 cm = \( \frac{u}{2} \)
- \( u \) = initial velocity
- \( a \) = acceleration (deceleration in this case, so it will be negative)
- \( s \) = distance = 0.4 m

Substituting the values:
\[ \left(\frac{u}{2}\right)^2 - u^2 = 2a(0.4) \]
This simplifies to:
\[ \frac{u^2}{4} - u^2 = 0.8a \]
\[ -\frac{3u^2}{4} = 0.8a \]
From this, we can solve for \( a \):
\[ a = -\frac{3u^2}{4 \times 0.8} = -\frac{3u^2}{3.2} = -\frac{3u^2}{8} \]

### Step 4: Calculate the Additional Distance
Now, we need to find the distance \( s_2 \) that the bullet travels after reaching \( v = 0 \).

Using the same kinematic equation:
\[ v^2 - u^2 = 2as \]
Here:
- Final velocity \( v = 0 \)
- Initial velocity \( u = \frac{u}{2} \)
- Acceleration \( a = -\frac{3u^2}{8} \)

Substituting these values:
\[ 0 - \left(\frac{u}{2}\right)^2 = 2\left(-\frac{3u^2}{8}\right)s_2 \]
This simplifies to:
\[ -\frac{u^2}{4} = -\frac{3u^2}{4}s_2 \]
Cancelling \( u^2 \) (assuming \( u \neq 0 \)):
\[ \frac{1}{4} = \frac{3}{4}s_2 \]
Solving for \( s_2 \):
\[ s_2 = \frac{1}{3} \]
Since we initially worked in meters, converting this to centimeters:
\[ s_2 = \frac{1}{3} \times 100 \text{ cm} = \frac{100}{3} \text{ cm} \approx 33.33 \text{ cm} \]

### Final Answer
The bullet comes to rest after penetrating an additional distance of \( \frac{100}{3} \) cm or approximately 33.33 cm.

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