A bullet fired into a trunk of a tree loses `1//4` of its kinetic energy in travelling a distance of 5 cm`. Before stopping it travels s further distance of

Options

A. 150 cm
B. 1.5 cm
C. 1.25 cm
D. 15 cm

Explanation

To solve the problem step by step, we need to analyze the situation where a bullet loses a portion of its kinetic energy while traveling through a tree trunk.

### Step 1: Understand the initial conditions
The bullet loses \( \frac{1}{4} \) of its kinetic energy after traveling a distance of \( 5 \, \text{cm} \). This means that after traveling \( 5 \, \text{cm} \), the bullet retains \( \frac{3}{4} \) of its initial kinetic energy.

### Step 2: Set up the kinetic energy equations
Let the initial kinetic energy of the bullet be:
\[ KE_i = \frac{1}{2} m u^2 \]
After traveling \( 5 \, \text{cm} \), the kinetic energy becomes:
\[ KE_f = \frac{3}{4} KE_i = \frac{3}{4} \left(\frac{1}{2} m u^2\right) = \frac{3}{8} m u^2 \]

### Step 3: Apply the work-energy theorem
According to the work-energy theorem, the work done by the force \( F \) while the bullet travels \( 5 \, \text{cm} \) is equal to the change in kinetic energy:
\[ W_1 = F \cdot S_1 = KE_i - KE_f \]
Substituting the values we have:
\[ F \cdot 5 \, \text{cm} = \frac{1}{2} m u^2 - \frac{3}{8} m u^2 \]
Calculating the right side:
\[ \frac{1}{2} m u^2 - \frac{3}{8} m u^2 = \left(\frac{4}{8} - \frac{3}{8}\right) m u^2 = \frac{1}{8} m u^2 \]
Thus, we have:
\[ F \cdot 5 \, \text{cm} = \frac{1}{8} m u^2 \]

### Step 4: Calculate the work done until the bullet stops
Now, we need to find the distance \( S_2 \) that the bullet travels until it comes to a stop. The final kinetic energy when the bullet stops is \( 0 \), so:
\[ W_2 = F \cdot S_2 = KE_f - 0 = KE_f = \frac{3}{8} m u^2 \]
Thus, we have:
\[ F \cdot S_2 = \frac{3}{8} m u^2 \]

### Step 5: Relate the two work equations
From the two equations we have:
1. \( F \cdot 5 \, \text{cm} = \frac{1}{8} m u^2 \)
2. \( F \cdot S_2 = \frac{3}{8} m u^2 \)

We can divide the second equation by the first:
\[ \frac{F \cdot S_2}{F \cdot 5 \, \text{cm}} = \frac{\frac{3}{8} m u^2}{\frac{1}{8} m u^2} \]
This simplifies to:
\[ \frac{S_2}{5 \, \text{cm}} = 3 \]
Thus, we find:
\[ S_2 = 3 \times 5 \, \text{cm} = 15 \, \text{cm} \]

### Final Answer
The bullet travels a further distance of \( 15 \, \text{cm} \) before stopping.

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