Initial K.E. of system = K.E. of the bullet = $\frac{1}{2} m_{B} v_{B}^{2}$

By the law of conservation of linear momentum

$m_{B} v_{B} + 0 = m_{\text{sys.}} \times v_{\text{sys.}}$

⇒ ⇒ $v_{\text{sys.}} = \frac{m_B v_B}{m_{\text{sys.}}} = \frac{50 \times 10}{50 + 950} = 0.5 \, m/s$

Fractional loss in K.E. = $\frac{\frac{1}{2}m_{B}v_{B}^{2} - \frac{1}{2}m_{sys.}v_{sys.}^{2}}{\frac{1}{2}m_{B}v_{B}^{2}}$

By substituting $m_{B} = 50 \times 10^{-3} \, kg, \, v_{B} = 10 \, m/s$

$m_{\text{sys.}} = 1 \mathrm{kg}, v_{s} = 0.5 \mathrm{m/s}$ we get

Fractional loss = $\frac{95}{100}$ ∴ ∴ Percentage loss = 95%