A bullet fired into a fixed target loses half of its velocity in penetrating 15 cm. How much further it will penetrate before coming to rest?

Options

A. 5 cm
B. 15 cm
C. 7.5 cm
D. 10 cm

Answer

To solve the problem step by step, we will use the equations of motion to find how much further the bullet penetrates before coming to rest.

### Step 1: Understand the given data
- The bullet loses half of its velocity when it penetrates 15 cm.
- Let the initial velocity of the bullet be \( u \).
- After penetrating 15 cm, the final velocity \( v \) becomes \( \frac{u}{2} \).

### Step 2: Use the third equation of motion
We will use the third equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( v = \frac{u}{2} \)
- \( u = u \)
- \( s = 15 \, \text{cm} \)
- \( a \) is the acceleration (deceleration in this case).

Substituting the values into the equation:
\[ \left(\frac{u}{2}\right)^2 = u^2 + 2a(15) \]
This simplifies to:
\[ \frac{u^2}{4} = u^2 + 30a \]

### Step 3: Rearranging the equation
Rearranging the equation gives:
\[ \frac{u^2}{4} - u^2 = 30a \]
\[ -\frac{3u^2}{4} = 30a \]
Thus, we can solve for \( a \):
\[ a = -\frac{3u^2}{120} = -\frac{u^2}{40} \]

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

Using the same equation of motion:
\[ v^2 = u^2 + 2as \]
Substituting the known values:
\[ 0 = \left(\frac{u}{2}\right)^2 + 2\left(-\frac{u^2}{40}\right)s_0 \]
This simplifies to:
\[ 0 = \frac{u^2}{4} - \frac{u^2}{20}s_0 \]

### Step 5: Solve for \( s_0 \)
Rearranging gives:
\[ \frac{u^2}{20}s_0 = \frac{u^2}{4} \]
Dividing both sides by \( u^2 \) (assuming \( u \neq 0 \)):
\[ \frac{s_0}{20} = \frac{1}{4} \]
Thus:
\[ s_0 = 20 \times \frac{1}{4} = 5 \, \text{cm} \]

### Conclusion
The bullet will penetrate a further distance of **5 cm** before coming to rest.

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