A bullet is fired on a fixed target. It penetrates inside the target through distance `d = 3.75 cm` and then stops. mass of the bullet is `m = 1 kg` and of the target is `M = 4 kg`. Now an identical bullet moving with the same velocity is fired on the identical target which is placed at rest on a frictionless horizontal surface. Then find the distance (in cm) to which the bullet will penetrate inside the target?

Explanation

To solve the problem step by step, we will analyze the two scenarios given: the bullet penetrating a fixed target and the bullet penetrating a target that is free to move.

### Step 1: Understand the first scenario
In the first case, a bullet of mass \( m = 1 \, \text{kg} \) is fired at a fixed target of mass \( M = 4 \, \text{kg} \) and penetrates a distance \( d = 3.75 \, \text{cm} \). The work done by the bullet on the target is equal to the change in kinetic energy of the bullet.

**Equation:**
\[ W = -F \cdot d = -\frac{1}{2} m u^2 \]
Where \( F \) is the resistive force exerted by the target on the bullet.

### Step 2: Set up the equation for the first case
From the work-energy principle, we can write:
\[ -F \cdot d = -\frac{1}{2} m u^2 \]
This simplifies to:
\[ F \cdot d = \frac{1}{2} m u^2 \quad \text{(1)} \]

### Step 3: Understand the second scenario
In the second case, an identical bullet is fired at an identical target, but this time the target is free to move on a frictionless surface. When the bullet strikes the target, both the bullet and the target will move together after the collision.

### Step 4: Use conservation of momentum
Using the conservation of momentum, we have:
\[ m u = (M + m) V \]
Where \( V \) is the final velocity of both the bullet and the target after the collision. Rearranging gives:
\[ V = \frac{m u}{M + m} \quad \text{(2)} \]

### Step 5: Calculate the kinetic energy after the collision
The kinetic energy after the collision can be expressed as:
\[ KE = \frac{1}{2} (M + m) V^2 \]
Substituting \( V \) from equation (2):
\[ KE = \frac{1}{2} (M + m) \left(\frac{m u}{M + m}\right)^2 \]
This simplifies to:
\[ KE = \frac{1}{2} \frac{m^2 u^2}{M + m} \quad \text{(3)} \]

### Step 6: Set up the work-energy equation for the second case
The work done on the target in the second case can also be expressed as:
\[ -F \cdot d' = KE \]
Where \( d' \) is the distance penetrated in the second case. Thus:
\[ F \cdot d' = \frac{1}{2} \frac{m^2 u^2}{M + m} \quad \text{(4)} \]

### Step 7: Relate equations (1) and (4)
From equations (1) and (4), we can set up the ratio:
\[ \frac{F \cdot d}{F \cdot d'} = \frac{\frac{1}{2} m u^2}{\frac{1}{2} \frac{m^2 u^2}{M + m}} \]
This simplifies to:
\[ \frac{d}{d'} = \frac{M + m}{m} \]

### Step 8: Solve for \( d' \)
Rearranging gives:
\[ d' = d \cdot \frac{m}{M + m} \]
Substituting \( d = 3.75 \, \text{cm} \), \( m = 1 \, \text{kg} \), and \( M = 4 \, \text{kg} \):
\[ d' = 3.75 \cdot \frac{1}{4 + 1} = 3.75 \cdot \frac{1}{5} = 0.75 \, \text{cm} \]

### Final Answer
The distance to which the bullet will penetrate inside the target is \( d' = 0.75 \, \text{cm} \).

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