Physics - NLM Question with Solution | TestHub

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 $=\mathrm{vm}/\mathrm{s}$ 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) $$\begin{array}{rl}& \mathrm{a}=\mu \mathrm{g}\\ & \mathrm{a}=0.3\times 10\left(\text{ take }\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2\right)\\ & =3\mathrm{m}/{\mathrm{s}}^2\end{array}$$ then, from ${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{as}$ $$(1.49{)}^2=0+2\times 3\times s$$ $\mathrm{s}=\frac{\left(1.49^2\right)}{6};\mathrm{s}=\frac{2.22}{6}=0.379\mathrm{m}$