Question

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 :

• A. $\frac{r}{(3{)}^{1/3}}$
• B. $\frac{r}{3}$
• C. $\frac{r}{9}$
• D. $3^{1/2}r$

Answer 1

The correct option is A $\frac{r}{(3{)}^{1/3}}$

Option '$A$' is correct

Let us consider $m_1$ & $m_2$ are masses of the sphere and velocity is $v$ of the sphere before collision '$v$' is the velocity of other sphere and fist sphere at rest after collision.

From conservation of energy

$\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$

$\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2=\frac{1}{2}{m}_2{v}^{\prime 2}$ ....(1)

$\frac{1}{2}{m}_1{v}^2-\frac{1}{2}{m}_2{v}^2=\frac{1}{2}{m}_2{v}^{\prime 2}$

$v^{\prime }=\frac{n(m_1-m_2)}{m_2}$

Put the value of $v^{\prime }$ in eq (1)

$\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2=\frac{1}{2}{m}_2\times {\left(\frac{v(m_1-m_2)}{m_2}\right)}_2$

$m_1+m_2=\frac{(m_1-m_2{)}^2}{m_2}$

$m_1{m}_2+m_2^2=m_1^2+m_2^2-2m_1{m}_2$

$m_2=\frac{m_1}{3}$ ...(2)

The volume of the fiist sphere is $v$ and the volume of other is $v/3$

Now, the radius of the sphere is $r$ and other sphere is $r^{\prime }$

The volume of first sphere is

$\frac{4}{3}\pi r^{\prime }3=\frac{v}{3}$ ...(3)

the volume of radius other sphere is

$\frac{4}{3}\pi r^{\prime }3=\frac{v}{3}$ ....(4)

Divide eq. (4) by (3)

$r^{\prime }=\frac{r}{\left(3\right)\frac{1}{3}}$