When a bullet is fired a target, its velocity dwecreases by half after penetrating 30 cm into it. The additional thickness it will penetrate before coming to rest is

Options

A. 3m cm
B. 40 cm
C. 10 cm
D. 50 cm

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Understand the Problem
We know that when a bullet is fired at a target, its velocity decreases by half after penetrating 30 cm into it. We need to find out how much additional thickness the bullet will penetrate before coming to rest.

### Step 2: Initial Conditions
Let the initial velocity of the bullet be \( V \). After penetrating 30 cm, the velocity becomes \( \frac{V}{2} \).

### Step 3: Apply the Equation of Motion
Since the bullet is experiencing uniform retardation, we can use the third equation of motion:
\[ V^2 = U^2 - 2AS \]
where:
- \( V \) is the final velocity,
- \( U \) is the initial velocity,
- \( A \) is the acceleration (retardation in this case), and
- \( S \) is the distance traveled.

For the first part of the motion (30 cm penetration):
- \( V = \frac{V}{2} \)
- \( U = V \)
- \( S = 30 \, \text{cm} \)

Substituting these values into the equation:
\[ \left(\frac{V}{2}\right)^2 = V^2 - 2A(30) \]
This simplifies to:
\[ \frac{V^2}{4} = V^2 - 60A \]

### Step 4: Solve for Acceleration
Rearranging the equation gives:
\[ 60A = V^2 - \frac{V^2}{4} \]
\[ 60A = \frac{4V^2 - V^2}{4} = \frac{3V^2}{4} \]
Thus, we find:
\[ A = \frac{3V^2}{240} = \frac{V^2}{80} \, \text{cm/s}^2 \]

### Step 5: Additional Penetration Calculation
Now, we need to find the additional thickness \( x \) the bullet will penetrate before coming to rest. The initial velocity for this part is \( \frac{V}{2} \) and the final velocity \( V_f = 0 \).

Using the same equation of motion:
\[ 0 = \left(\frac{V}{2}\right)^2 - 2Ax \]
Substituting \( A = \frac{V^2}{80} \):
\[ 0 = \frac{V^2}{4} - 2\left(\frac{V^2}{80}\right)x \]
This simplifies to:
\[ \frac{V^2}{4} = \frac{V^2}{40}x \]

### Step 6: Solve for \( x \)
Dividing both sides by \( V^2 \) (assuming \( V \neq 0 \)):
\[ \frac{1}{4} = \frac{x}{40} \]
Cross-multiplying gives:
\[ x = 10 \, \text{cm} \]

### Conclusion
The additional thickness the bullet will penetrate before coming to rest is **10 cm**.

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