A bullet of mass m=50 gm strikes `(Deltat~~0)`a sand bag of mass M =5 kg hanging from a fixed point, with a horizontal velocity `vec(v)_(p)`. If bullet sticks to the sand bag then the ratio of final & initial kinetic energy of the bullet is

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the given values
- Mass of the bullet, \( m = 50 \, \text{g} = 0.05 \, \text{kg} \)
- Mass of the sandbag, \( M = 5 \, \text{kg} \)
- Initial velocity of the bullet, \( v_p \) (unknown)
- Initial velocity of the sandbag, \( V_s = 0 \, \text{m/s} \) (since it is hanging and at rest)

### Step 2: Apply the conservation of momentum
Since the bullet sticks to the sandbag, we can use the conservation of momentum before and after the collision:

\[ \text{Initial momentum} = \text{Final momentum} \]

The initial momentum is only due to the bullet:

\[ m v_p + M \cdot 0 = (m + M) V_f \]

Substituting the known values:

\[ 0.05 v_p = (0.05 + 5) V_f \]

This simplifies to:

\[ 0.05 v_p = 5.05 V_f \]

### Step 3: Solve for the final velocity \( V_f \)
Rearranging the equation gives:

\[ V_f = \frac{0.05 v_p}{5.05} \]

### Step 4: Calculate the initial kinetic energy of the bullet
The initial kinetic energy \( KE_i \) of the bullet is given by:

\[ KE_i = \frac{1}{2} m v_p^2 = \frac{1}{2} (0.05) v_p^2 \]

### Step 5: Calculate the final kinetic energy of the combined system
After the collision, the kinetic energy \( KE_f \) of the bullet and sandbag together is:

\[ KE_f = \frac{1}{2} (m + M) V_f^2 = \frac{1}{2} (5.05) V_f^2 \]

Substituting \( V_f \):

\[ KE_f = \frac{1}{2} (5.05) \left( \frac{0.05 v_p}{5.05} \right)^2 \]

This simplifies to:

\[ KE_f = \frac{1}{2} (5.05) \cdot \frac{0.0025 v_p^2}{25.5025} = \frac{5.05 \cdot 0.0025 v_p^2}{2 \cdot 25.5025} \]

### Step 6: Find the ratio of final to initial kinetic energy
Now, we find the ratio \( \frac{KE_f}{KE_i} \):

\[ \frac{KE_f}{KE_i} = \frac{\frac{5.05 \cdot 0.0025 v_p^2}{2 \cdot 25.5025}}{\frac{1}{2} (0.05) v_p^2} \]

The \( \frac{1}{2} v_p^2 \) cancels out:

\[ \frac{KE_f}{KE_i} = \frac{5.05 \cdot 0.0025}{0.05 \cdot 25.5025} \]

Calculating this gives:

\[ \frac{KE_f}{KE_i} = \frac{5.05 \cdot 0.0025}{0.05 \cdot 25.5025} = \frac{0.012625}{1.275125} \approx 0.00000988 \approx 10^{-4} \]

### Final Answer
The ratio of final to initial kinetic energy of the bullet is approximately \( 10^{-4} \).

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