A bullet loses 25% of its initial kinetic energy after penetrating through a distance of 2cm in a target. The further distance it travels before coming to rest is

Options

A. 1cm
B. 2cm
C. 8/3cm
D. 6cm

Answer

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

### Step 1: Understand the Initial Conditions
The bullet loses 25% of its initial kinetic energy after penetrating 2 cm into the target. We denote the initial kinetic energy as \( KE_i = \frac{1}{2} mv^2 \), where \( m \) is the mass of the bullet and \( v \) is its initial velocity.

### Step 2: Calculate the Kinetic Energy after Penetration
After losing 25% of its kinetic energy, the remaining kinetic energy is:
\[ KE_f = KE_i - 0.25 KE_i = 0.75 KE_i = 0.75 \left(\frac{1}{2} mv^2\right) = \frac{3}{8} mv^2 \]

### Step 3: Relate Initial and Final Velocities
The final kinetic energy can also be expressed in terms of the final velocity \( u \):
\[ KE_f = \frac{1}{2} mu^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ \frac{3}{8} mv^2 = \frac{1}{2} mu^2 \]
Cancelling \( m \) from both sides:
\[ \frac{3}{8} v^2 = \frac{1}{2} u^2 \]
Multiplying both sides by 2:
\[ \frac{3}{4} v^2 = u^2 \]

### Step 4: Use the Third Equation of Motion
Now, we will use the third equation of motion to find the acceleration \( a \):
\[ v^2 = u^2 + 2as \]
Substituting \( s = 0.02 \, \text{m} \) (2 cm) and \( u^2 = \frac{3}{4} v^2 \):
\[ v^2 = \frac{3}{4} v^2 + 2a(0.02) \]
Rearranging gives:
\[ v^2 - \frac{3}{4} v^2 = 0.04a \]
\[ \frac{1}{4} v^2 = 0.04a \]
\[ a = \frac{v^2}{4 \times 0.04} = \frac{v^2}{0.16} = 25v^2 \]

### Step 5: Calculate the Further Distance Traveled
Now, we will find the distance \( s' \) the bullet travels after it has lost its initial kinetic energy. The initial velocity for this part is \( u = \sqrt{\frac{3}{4}} v = \frac{\sqrt{3}}{2} v \) and the final velocity \( v = 0 \).

Using the third equation of motion again:
\[ 0 = u^2 + 2(-a)s' \]
Substituting \( u^2 = \frac{3}{4} v^2 \) and \( a = 25v^2 \):
\[ 0 = \frac{3}{4} v^2 - 2(25v^2)s' \]
Rearranging gives:
\[ 2(25v^2)s' = \frac{3}{4} v^2 \]
Cancelling \( v^2 \) from both sides:
\[ 50s' = \frac{3}{4} \]
\[ s' = \frac{3}{200} = 0.015 \, \text{m} = 1.5 \, \text{cm} \]

### Step 6: Total Distance Traveled
The total distance the bullet travels before coming to rest is:
\[ \text{Total Distance} = 2 \, \text{cm} + 1.5 \, \text{cm} = 3.5 \, \text{cm} \]

### Final Answer
The further distance it travels before coming to rest is **3.5 cm**.