b
(b)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
\({m_B}{v_B} + 0 = {m_{{\rm{sys}}{\rm{.}}}} \times {v_{{\rm{sys}}{\rm{.}}}}\)
==> \({v_{{\rm{sys}}{\rm{.}}}} = \frac{{{m_B}{v_B}}}{{{m_{{\rm{sys}}{\rm{.}}}}}} = \frac{{50 \times 10}}{{50 + 950}} = 0.5\;m/s\)
Fractional loss in K.E. = \(\frac{{\frac{1}{2}{m_B}v_B^2 - \frac{1}{2}{m_{{\rm{sys}}{\rm{.}}}}v_{{\rm{sys}}{\rm{.}}}^2}}{{\frac{1}{2}{m_B}v_B^2}}\)
By substituting \({m_B} = 50 \times {10^{ - 3}}kg,\;{v_B} = 10\;m/s\)
\({m_{{\rm{sys}}{\rm{.}}}} = 1kg,\;{v_s} = 0.5\;m/s\) we get
Fractional loss = \(\frac{{95}}{{100}}\)
Percentage loss \(= 95\%\)