A bullet of mass 4 ~g is fired horizontally with a speed of 300 ~m...

Given, $\mathrm{m}_1=4 \mathrm{~g}, \mathrm{u}_1=300 \mathrm{~m} / \mathrm{s}\mathrm{m}_2=0.8 \mathrm{~kg}=800 \mathrm{~g}, \mathrm{u}_2=0 \mathrm{~m} / \mathrm{s}$ From law of conservation of momentum, $\mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2$ Let the velocity of combined system $=\mathrm{v} \mathrm{m} / \mathrm{s}$ then, $4 \times 300+800 \times 0=(800+4) \times v\mathrm{v}=\frac{1200}{804}=1.49 \mathrm{~m} / \mathrm{s}$ Now, $\mu=0.3$ (given) $$\begin{aligned}&\mathrm{a}=\mu \mathrm{g}\\ &\mathrm{a}=0.3 \times10\left(\text{ take } \mathrm{g}=10 \mathrm{~m}/ \mathrm{s}^2\right) \\ &=3 \mathrm{~m}/ \mathrm{s}^2 \end{aligned}$$then,$from \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as}$ (1.49)^2=0+2 $\times$3 $\times$s $\mathrm{s}=\frac{\left(1.49^2\right)}{6} ; \mathrm{s}=\frac{2.22}{6}=0.379 \mathrm{~m}