Two steel spheres approach each other head-on with the same speed and collide elastically. After the collision one of the sphere's of radius r comes to rest. The radius of the other sphere is

${\text{m}}_2\to {\text{u}}_{\text{2}}=-\text{u}$

${\text{m}}_1\to {\text{v}}_{\text{1}}=0$

${\text{v}}_{\text{1}}=\frac{\left({\text{m}}_{\text{1}}-{\text{m}}_{\text{2}}\right){\text{u}}_{\text{1}}}{\left({\text{m}}_{\text{1}}+{\text{m}}_{\text{2}}\right)}+\frac{{\text{2m}}_{\text{2}}{\text{u}}_{\text{2}}}{\left({\text{m}}_{\text{1}}+{\text{m}}_{\text{2}}\right)}$

$0=\frac{\left({\text{m}}_{\text{1}}-{\text{m}}_{\text{2}}\right)\text{u}+2{\text{m}}_{\text{2}}\left(-\text{u}\right)}{\left({\text{m}}_{\text{1}}+{\text{m}}_{\text{2}}\right)}$

$0={\text{m}}_{\text{1}}\text{u}-{\text{m}}_{\text{2}}\text{u}-2{\text{m}}_{\text{2}}\text{u}$

$\text{0}={\text{m}}_1\text{u}-3{\text{m}}_{\text{2}}\text{u}$

${\text{m}}_1\text{u}={\text{3m}}_{\text{2}}$

${\text{d}}_{\text{1}}\times \frac{4}{3}\pi {\text{r}}_{\text{1}}^{\text{3}}=3\times {\text{d}}_{\text{2}}\times \frac{4}{3}\pi {\text{r}}_{\text{2}}^{\text{3}}$

$\text{d}=\frac{\text{m}}{\text{v}}$

$\text{m}=\text{d}\times \text{v}$

$=\text{d}\times \frac{4}{3}\pi {\text{r}}^{\text{3}}\left(\because \text{Sphere}\right)$

(${\text{d}}_{\text{1}}={\text{d}}_{\text{2}},\because$  both are steel spheres)

${\text{r}}_{\text{1}}^{\text{3}}={\text{r}}_{\text{2}}^{\text{3}}$

$\frac{{\text{r}}_1^3}{3}={\text{r}}_{\text{2}}^{\text{3}}$

$\frac{\text{r}}{{\left(\text{3}\right)}^{\text{1/3}}}={\text{r}}_{\text{2}}$