Let us suppose that the mass of the bullet is, $m=10 \mathrm{g}={10}^{-2 }\mathrm{kg}$ and $M_A=\frac{1}{2} \mathrm{kg}, M_B=1.49 \mathrm{kg}$,

now the loss of kinetic energy of a bullet is,

$$\begin{gathered} ∆K.E.\times 100=-\left(\frac{\frac{1}{2}m{v_1}^2-{\displaystyle \frac{1}{2}}m{u}^2}{\frac{1}{2}m{u}^2}\times 100\right)=-\left(\frac{{v_1}^2-u^2}{u^2}\times 100\right) \\ ∆K.E.\times 100=-\left(\frac{{v_1}^2}{u^2}-1\right)\times 100 \end{gathered}$$

for plate $A$ conservation of momentum,

$$\begin{gathered} mu=m{v}_1+M_{A}v \\ \Rightarrow m\left(u-v_1\right)=M_{A}v \end{gathered}$$

for the plate, $B$ the bullet is combined with plate,

$m{v}_1=\left(m+M_B\right)v$

now from above two-equation of plates,

$\frac{u-v_1}{v_1}=\frac{M_A}{m+M_B}=\frac{0.5}{0.01+1.49}=\frac{0.5}{1.50}=\frac{1}{3}$

$$\begin{gathered} 3u-3v_1=v_1 \\ \Rightarrow v_1=\frac{3}{4}u \end{gathered}$$

$∆K.E.\times 100=-\left(\frac{{\left({\displaystyle \frac{3}{4}}u\right)}^2}{u^2}-1\right)\times 100=43.75 \%$