A bullet of mass m = 50 gm strikes a sand bag of mass M = 5 kg hanging from a fixed point, with a horizontal velocity `vecv_(p)` . If bullet sticks to the sand bag then the ratio of final & initial kinetic energy of the bullet is (approximately) :

Options

A. `10^(-2)`
B. `10^(-3)`
C. `10^(-6)`
D. `10^(-4)`

Answer

To solve the problem, we need to find the ratio of the final kinetic energy to the initial kinetic energy of the bullet after it strikes the sandbag and sticks to it. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Problem
We have a bullet of mass \( m = 50 \, \text{g} = 0.05 \, \text{kg} \) and a sandbag of mass \( M = 5 \, \text{kg} \). The bullet strikes the sandbag with a horizontal velocity \( \vec{v}_p \). After the collision, the bullet sticks to the sandbag.

### Step 2: Apply Conservation of Momentum
Since the bullet sticks to the sandbag, we can use the conservation of momentum in the horizontal direction. The initial momentum of the system (bullet + sandbag) before the collision is equal to the final momentum after the collision.

**Initial Momentum:**
\[ p_{\text{initial}} = m \cdot v_p \]

**Final Momentum:**
After the collision, the bullet and the sandbag move together with a common velocity \( v \):
\[ p_{\text{final}} = (m + M) \cdot v \]

Setting the initial momentum equal to the final momentum:
\[ m \cdot v_p = (m + M) \cdot v \]

### Step 3: Substitute Known Values
Substituting the values:
\[ 0.05 \cdot v_p = (0.05 + 5) \cdot v \]
\[ 0.05 \cdot v_p = 5.05 \cdot v \]

### Step 4: Solve for Final Velocity \( v \)
Rearranging the equation to solve for \( v \):
\[ v = \frac{0.05 \cdot v_p}{5.05} \]

### Step 5: Calculate the Ratio of Kinetic Energies
The initial kinetic energy of the bullet is given by:
\[ KE_{\text{initial}} = \frac{1}{2} m v_p^2 \]

The final kinetic energy of the bullet and sandbag system is:
\[ KE_{\text{final}} = \frac{1}{2} (m + M) v^2 \]

### Step 6: Substitute \( v \) into Final Kinetic Energy
Substituting \( v \) into the final kinetic energy expression:
\[ KE_{\text{final}} = \frac{1}{2} (5.05) \left(\frac{0.05 \cdot v_p}{5.05}\right)^2 \]
\[ = \frac{1}{2} (5.05) \cdot \frac{0.0025 \cdot v_p^2}{(5.05)^2} \]
\[ = \frac{0.0025}{2 \cdot 5.05} v_p^2 \]

### Step 7: Find the Ratio of Final to Initial Kinetic Energy
Now, we can find the ratio:
\[ \frac{KE_{\text{final}}}{KE_{\text{initial}}} = \frac{\frac{0.0025}{2 \cdot 5.05} v_p^2}{\frac{1}{2} \cdot 0.05 v_p^2} \]
The \( v_p^2 \) and \( \frac{1}{2} \) cancel out:
\[ = \frac{0.0025}{0.05 \cdot 5.05} \]
\[ = \frac{0.0025}{0.2525} \approx 0.00988 \]

### Step 8: Final Calculation
To find the approximate ratio:
\[ \frac{KE_{\text{final}}}{KE_{\text{initial}}} \approx 10^{-4} \]

### Conclusion
Thus, the ratio of the final kinetic energy to the initial kinetic energy of the bullet is approximately \( 10^{-4} \).