**Given:** m_{1} = 0.02 kg, m_{2} = 100 kg, v_{1} = 80 m s^{-1}

**To find: **Recoil speed (v_{2})

**Formula: **m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}

**Calculation:** Initially gun and shell are at rest.

∴ m_{1}u_{1 }+ m_{2}u_{2} = 0

Final momentum = m_{1}v_{1} – m_{2}v_{2}

Using formula,

0 = 0.02 (80) – 100(v_{2})

∴ v_{2} = \(\frac{0.02\times80}{100}\) = 0.016 ms^{-1}

The recoil speed of the gun is 0.016 m s^{-1}.