A bullet on penetrating 30 cm into its target loses its velocity by 50%. What additional distanee will it penetrate into the target before it comes to rest?

Options

A. 30 cm
B. 20 cm
C. 10 cm
D. 5 cm

Explanation

To solve the problem step by step, we will use the equations of motion.

### Step 1: Understand the problem
A bullet penetrates 30 cm into a target and loses 50% of its velocity. We need to find the additional distance it will penetrate before coming to rest.

### Step 2: Determine initial and final velocities
Let the initial velocity of the bullet when it starts penetrating the target be \( v \). After penetrating 30 cm, its velocity reduces to \( \frac{v}{2} \) (50% reduction).

### Step 3: Use the equation of motion
We can use the third equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity after penetrating 30 cm = \( \frac{v}{2} \)
- \( u \) = initial velocity = \( v \)
- \( a \) = acceleration (deceleration in this case, hence negative)
- \( s \) = distance penetrated = 30 cm

Substituting the values into the equation:
\[ \left(\frac{v}{2}\right)^2 = v^2 + 2a(30) \]

### Step 4: Simplify the equation
Expanding the left side:
\[ \frac{v^2}{4} = v^2 + 60a \]
Rearranging gives:
\[ \frac{v^2}{4} - v^2 = 60a \]
\[ -\frac{3v^2}{4} = 60a \]
Thus, we can express acceleration \( a \):
\[ a = -\frac{3v^2}{240} = -\frac{v^2}{80} \]

### Step 5: Determine additional penetration distance
Now, we need to find the additional distance \( x \) the bullet penetrates before coming to rest. The initial velocity for this part is \( \frac{v}{2} \) and the final velocity is 0.

Using the same equation of motion:
\[ 0 = \left(\frac{v}{2}\right)^2 + 2(-\frac{v^2}{80})x \]

### Step 6: Substitute and simplify
Substituting the values:
\[ 0 = \frac{v^2}{4} - \frac{v^2}{40}x \]
Multiplying through by 40 to eliminate the fractions:
\[ 0 = 10v^2 - v^2x \]
Rearranging gives:
\[ v^2x = 10v^2 \]
Dividing both sides by \( v^2 \) (assuming \( v \neq 0 \)):
\[ x = 10 \text{ cm} \]

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