A bullet of mass 5 g is fired at a velocity of `900 ms^(-1)` from a rifle of mass 2.5 kg. What is the recoil velocity of the rifle?

Options

A. `0.9 ms^(-1)`
B. `180 ms^(-1)`
C. `900 ms^(-1)`
D. `1.8 ms^(-1)`

Explanation

To find the recoil velocity of the rifle 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
Initially, both the rifle and the bullet are at rest, which means their initial velocities are zero. Therefore, the initial momentum of the system (rifle + bullet) is zero.

### Step 2: Define the Variables
- Mass of the bullet, \( m_B = 5 \, \text{g} = 5/1000 \, \text{kg} = 0.005 \, \text{kg} \)
- Mass of the rifle, \( m_R = 2.5 \, \text{kg} \)
- Velocity of the bullet after firing, \( v_B = 900 \, \text{m/s} \)
- Recoil velocity of the rifle, \( v_R \) (this is what we need to find)

### Step 3: Apply Conservation of Momentum
According to the conservation of momentum:
\[ \text{Initial Momentum} = \text{Final Momentum} \]
Since the initial momentum is zero:
\[ 0 = m_R v_R + m_B v_B \]

### Step 4: Substitute the Known Values
Substituting the known values into the equation:
\[ 0 = (2.5 \, \text{kg}) v_R + (0.005 \, \text{kg})(900 \, \text{m/s}) \]

### Step 5: Solve for \( v_R \)
Rearranging the equation to solve for \( v_R \):
\[ (2.5 \, \text{kg}) v_R = - (0.005 \, \text{kg})(900 \, \text{m/s}) \]
Calculating the right side:
\[ (2.5 \, \text{kg}) v_R = -4.5 \, \text{kg m/s} \]
Now, divide both sides by \( 2.5 \, \text{kg} \):
\[ v_R = \frac{-4.5 \, \text{kg m/s}}{2.5 \, \text{kg}} = -1.8 \, \text{m/s} \]

### Step 6: Interpret the Result
The negative sign indicates that the rifle recoils in the opposite direction to the bullet. Thus, the magnitude of the recoil velocity of the rifle is:
\[ |v_R| = 1.8 \, \text{m/s} \]

### Final Answer
The recoil velocity of the rifle is \( 1.8 \, \text{m/s} \).

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