A bullet of mass `0.1` kg is fired with a speed of 100 `m//sec` , the mass of gun is 50 kg . The velocity of recoil is

Options

A. `0.05ms^(-1)`
B. `0.5 ms^(-1)`
C. `0.1 ms^(-1)`
D. `0.2 ms^(-1)`

Explanation

To find the velocity of recoil of the gun when a bullet is fired, we can use the principle of conservation of momentum. Here’s a step-by-step solution:

### Step 1: Understand the system
- We have a bullet of mass \( m = 0.1 \, \text{kg} \) that is fired with a speed \( u = 100 \, \text{m/s} \).
- The mass of the gun is \( M = 50 \, \text{kg} \).
- Initially, both the gun and bullet are at rest, so the initial momentum of the system is zero.

### Step 2: Apply the conservation of momentum
- According to the law of conservation of momentum, the total momentum before firing must equal the total momentum after firing.
- Before firing, the total momentum \( P_i = 0 \) (since both the gun and bullet are at rest).
- After firing, the momentum of the bullet is \( p_b = m \cdot u \) and the momentum of the gun is \( p_g = -M \cdot V \) (the negative sign indicates that the gun moves in the opposite direction to the bullet).

### Step 3: Set up the momentum equation
- The equation based on conservation of momentum can be written as:
\[ P_i = P_f \implies 0 = m \cdot u - M \cdot V \]
Rearranging gives us:
\[ M \cdot V = m \cdot u \]

### Step 4: Solve for the recoil velocity \( V \)
- We can express the recoil velocity \( V \) as:
\[ V = \frac{m \cdot u}{M} \]
- Substituting the values:
\[ V = \frac{0.1 \, \text{kg} \cdot 100 \, \text{m/s}}{50 \, \text{kg}} \]
\[ V = \frac{10 \, \text{kg m/s}}{50 \, \text{kg}} = 0.2 \, \text{m/s} \]

### Final Answer
- The velocity of recoil of the gun is \( V = 0.2 \, \text{m/s} \).

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