d
By conservation of linear momentum :

\((0.1)(3 \hat{\mathrm{i}})+(0.1)(5 \hat{\mathrm{j}})=(0.1)(4)(\hat{\mathrm{i}}+\hat{\mathrm{j}})+(0.1) \overline{\mathrm{v}}\)

\(\Rightarrow \overrightarrow{\mathrm{v}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}\)

Speed of \(\mathrm{B}\) after collision \(|\overrightarrow{\mathrm{v}}|=\sqrt{2}\)

Now, kinetic energy \(=\frac{1}{2} \mathrm{mV}^{2}=\frac{1}{2}(0.1)(2)=\frac{1}{10}\)

\(\therefore \mathrm{x}=1\)