We will use the principle of conservation of momentum. The total momentum before firing is zero, and the total momentum after firing must also be zero (since no external force is acting on the system).
Let:
- \( m_1 = 10 \, \text{g} = 0.01 \, \text{kg} \) (mass of the bullet)
- \( v_1 = 50 \, \text{m/s} \) (speed of the bullet)
- \( m_2 = 2 \, \text{kg} \) (mass of the gun)
- \( v_2 \) (recoil velocity of the gun)
Using conservation of momentum:
\[
m_1v_1 + m_2v_2 = 0
\]
\[
0.01 \times 50 + 2 \times v_2 = 0
\]
\[
0.5 + 2v_2 = 0
\]
\[
2v_2 = -0.5
\]
\[
v_2 = \frac{-0.5}{2} = -0.25 \, \text{m/s}
\]
Thus, the recoil speed of the gun is 0.25 m/s in the opposite direction.
Thus, the correct answer is (B) 0.25.