Question

two steel spheres approach each other with same speed and collide elastically. after the collision one ball of radius r comes to rest. the radius of other sphere is

Answer 1

Answer:The radius of the other sphere is  $r' = \dfrac{r}{(3)^{\dfrac{1}{3}}}$

Explanation:

Let us consider $m_{1}$ and $m_{2}$ are masses of the spheres and velocity is v of  the spheres before collision. v' is the velocity of  other sphere and first sphere at rest after collision.

From conservation of energy

$\dfrac{1}{2}m_{1}u_{1}^{2}+\dfrac{1}{2}m_{2}u_{2}^{2}=\dfrac{1}{2}m_{1}v_{1}^{2}+\dfrac{1}{2}m_{2}v_{2}^{2}$

$\dfrac{1}{2}m_{1}v^{2}+\dfrac{1}{2}m_{2}v^{2}=\dfrac{1}{2}m_{2}v'^{2}$.....(I)

$\dfrac{1}{2}m_{1}v^{2}-\dfrac{1}{2}m_{2}v^{2}=\dfrac{1}{2}m_{2}v'^{2}$

$m_{1}v-m_{2}v=m_{2}v'$

$v' = \dfrac{v(m_{1}-m_{2})}{m_{2}}$

Now, put the value of v' in equation (I)

$\dfrac{1}{2}m_{1}v^{2}+\dfrac{1}{2}m_{2}v^{2}=\dfrac{1}{2}m_{2}\times(\dfrac{v(m_{1}-m_{2})}{m_{2}})^{2}$

$m_{1}+m_{2}=\dfrac{(m_{1}-m_{2})^{2}}{m_{2}}$

$m_{1}m_{2}+m_{2}^{2}= m_{1}^{2}+m_{2}^{2}-2m_{1}m_{2}$

$m_{2} = \dfrac{m_{1}}{3}$......(II)

The mass of the other sphere is one by third of the first sphere.

So, the volume of the other sphere will be one by third of the first sphere.

Therefore, the volume of the first sphere is V and the volume of other sphere is V/3

Now, the radius of the first sphere is r and other sphere is r'

The volume of first sphere is

$\dfrac{4}{3}\pi r^{3} = V$.....(III)

The volume of other sphere is

$\dfrac{4}{3}\pi r'^{3} = \dfrac{V}{3}$.....(IV)

Divided equation (IV) by (III)

$r' = \dfrac{r}{(3)^{\dfrac{1}{3}}}$

Hence, the radius of the other sphere is $r' = \dfrac{r}{(3)^{\dfrac{1}{3}}}$

Answer 2

According to the given data, conservation of momentum and conservation of energy is to be applied on the given situation to form the equation and then be compared to rule out the relation.

Thereby,

Initial momentum = Final Momentum

$\Rightarrow m_{ 1 }v-m_{ 2 }v=m_{ 1 }v" -m_{ 2 }v'$

$\Rightarrow v"=0$

$\Rightarrow v'=\frac { v(m_{ 1 }-m_{ 2 }) }{ m_{ 2 } }$

$\Rightarrow \frac { 1 }{ 2 } m_{ 1 }v_{ 2 }+\frac { 1 }{ 2 } m_{ 2 }v_{ 2 }=\frac { 1 }{ 2 } m_{ 2 }{ v }_{ 2 }'$

Thereby $\frac { 1 }{ 2 } m_{ 1 }v_{ 2 }+\frac { 1 }{ 2 } m_{ 2 }v_{ 2 }=\frac { 1 }{ 2 } m_{ 2 }(\frac { m_{ 1 }-m_{ 2 } }{ m_{ 2 } } )v_{ 2 }$

$\Rightarrow \frac { 1 }{ 2 } v_{ 2 }\left( m_{ 1 }+m_{ 2 } \right) =\frac { 1 }{ 2 } m_{ 2 }(\frac { m_{ 1 }-m_{ 2 } }{ m_{ 2 } } )v_{ 2 }$

$\Rightarrow m_{ 1 }+m_{ 2 }=\frac { (m_{ 1 }-m_{ 2 })^{ 2 } }{ m_{ 2 } }$

$\Rightarrow m_{ 1 }m_{ 2 }+m_{ 2 }^{ 2 }=m_{ 1 }^{ 2 }+m_{ 2 }^{ 2 }-2m_{ 1 }m_{ 2 }$

$\Rightarrow 3m_{ 1 }m_{ 2 }=m_{ 1 }^{ 2 }$

$\Rightarrow m_{ 2 }=\frac { m_{ 1 } }{ 3 }$

Then the Volume of sphere $=\frac { 4 }{ 3 } \pi r^{ 3 }$ and so now the V of the second sphere becomes $\frac { V }{ 3 }$

$\Rightarrow \frac { V }{ 3 } =\frac { 4 }{ 3 } \pi { r' }^{ 3 }$

$\Rightarrow \frac { V }{ \frac { V }{ 3 }  } =\frac { \frac { 4 }{ 3 } \pi r^{ 3 } }{ \frac { 4 }{ 3 } \pi { r' }^{ 3 } }$

$\Rightarrow \frac { 1 }{ 3 } =\frac { r'^{ 3 } }{ r^{ 3 } }$

$\Rightarrow r'=\frac { r }{ \sqrt [ 3 ]{ 3 } }$