\text { Given: } \mathrm{u}_1=3 \mathrm{u}, \mathrm{u}_2=0 \\ \therefore \quad & \mathrm{e}=\frac{\mathrm{v}_2-\mathrm{v}_1}{3 \mathrm{u}}=\frac{\mathrm{v}_2}{3 \mathrm{u}}-\frac{\mathrm{v}_1}{3 \mathrm{u}} \end{array} \end{aligned} $$

$$ v_2-v_1=3 u e \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$$

By law of conservation of momentum,

$$ \begin{aligned} & \mathrm{mu}_1+\mathrm{mu}_2=\mathrm{mv}_1+\mathrm{mv}_2 \\ & 3 \mathrm{mu}=\mathrm{mv}_1+\mathrm{mv}_2 \\ & \mathrm{v}_1+\mathrm{v}_2=3 \mathrm{u} \\ & \mathrm{v}_1=3 \mathrm{u}-\mathrm{v}_2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $$

From equation (i) and (ii),

$$ \begin{aligned} & v_2-\left(3 u-v_2\right)=3 u e \Rightarrow 2 v_2-3 u=3 u e \\ & v_2=\frac{3 u(1+e)}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $$

For $\mathrm{v}_1$

$$ \begin{aligned} & v_1=3 u-v_2=3 u-\frac{3 u(1+e)}{2} \\ & v_1=\frac{3 u(1-e)}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $$

From equation (iii) and (iv),

$$ \text { Ratio }=\frac{v_2}{v_1}=\frac{\frac{3 u(1+e)}{2}}{\frac{3 u(1-e)}{2}}=\frac{1+e}{1-e} $$

Center Of Mass

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