Question

Two bodies of masses 50g and 30g moving in the same direction, along the same straight line with velocities 50cm/s and 30cm/s respectively. Suffer one dimensional elastic collision . calculate their velocities after the collision.

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Let}{\mathrm{v}}_{1\mathrm{f}}\&{\mathrm{v}}_{2\mathrm{f}}\mathrm{be}\mathrm{the}\mathrm{final}\mathrm{velocities}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{and}\mathrm{the}\mathrm{second}\mathrm{body}\mathrm{after}\mathrm{collision}. \\ {\mathrm{m}}_1=50\mathrm{g} \\ {\mathrm{m}}_2=30\mathrm{g} \\ {\mathrm{v}}_{1\mathrm{i}}=50\mathrm{cm}/\mathrm{s} \\ {\mathrm{v}}_{2\mathrm{i}}=30\mathrm{cm}/\mathrm{s} \\ \mathrm{By}\mathrm{usual}\mathrm{notation}, \\ {\mathrm{v}}_{1\mathrm{f}}=\left(\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\times {\mathrm{v}}_{1\mathrm{i}}\right)+\left(\frac{2{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\times {\mathrm{v}}_{2\mathrm{i}}\right) \\ {\mathrm{v}}_{1\mathrm{f}}=\left(\frac{50-30}{50+30}\times 50\right)+\left(\frac{2\times 30}{50+30}\times 30\right) \\ {\mathrm{v}}_{1\mathrm{f}}=\left(\frac{20}{80}\times 50\right)+\left(\frac{60}{80}\times 30\right) \\ {\mathrm{v}}_{1\mathrm{f}}=\frac{50}{4}+\frac{90}{4} \\ {\mathrm{v}}_{1\mathrm{f}}=\frac{140}{4}=35\mathrm{cm}/\mathrm{s} \\ {\mathrm{v}}_{2\mathrm{f}}=\left(\frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_2}\times {\mathrm{v}}_{2\mathrm{i}}\right)+\left(\frac{2{\mathrm{m}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_2}\times {\mathrm{v}}_{1\mathrm{i}}\right) \\ {\mathrm{v}}_{2\mathrm{f}}=\left(\frac{30-50}{50+30}\times 30\right)+\left(\frac{2\times 50}{50+30}\times 50\right) \\ {\mathrm{v}}_{2\mathrm{f}}=-\left(\frac{20}{80}\times 30\right)+\left(\frac{100}{80}\times 50\right) \\ {\mathrm{v}}_{2\mathrm{f}}=-\frac{30}{4}+\frac{250}{4} \\ {\mathrm{v}}_{2\mathrm{f}}=\frac{220}{4}=55\mathrm{cm}/\mathrm{s} \end{gathered}$$