Please explain 5 th question.

5. A spherical ball A of mass m is released from rest on a smooth bowl 0.2 m high, The sphere slides down and colliede  elastically with another sphere  B of mass m/4 placed on the bottom of the bowl. If the ball B has to just reach the top and escape the bowl, calculate from where A should be released ?

Dear student

Let the ball A is released at a height h from ground. The potential energy at this point given by mgh.
when ball A comes to the bottom of the bowl it has only kinetic energy which is given by
$$\begin{gathered} \frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2 \\ but for sphere I=\frac{2}{5}m{r}^2 where r is the radius of sphere and v=r\omega so \\ \frac{1}{2}m{v}^2+\frac{1}{2}\frac{2}{5}m{r}^2{\left(\frac{v}{r}\right)}^2 \\ \frac{7}{10}m{v}^2 \end{gathered}$$
And by conservation of energy the potential energy is completely convert to its kinetic energy so 
$$\begin{gathered} mgh=\frac{7}{10}m{v}^2 \\ v=\sqrt{\frac{10gh}{7}} \end{gathered}$$
When two balls collide elastically their final velocity is given by
 
$$\begin{gathered} for ball 1 \\ {v}_{1f}=\left(\frac{m_1-m_2}{m_1+m_2}\right)v_{1i}+\left(\frac{2m_2}{m_1+m_2}\right)v_{2i} \\ and for ball 2 \\ {v}_{2f}=\left(\frac{2m_1}{m_1+m_2}\right)v_{1i}+\left(\frac{m_2-m_1}{m_1+m_2}\right)v_{2i} \\ So velocity of sphere B is given by where v_{1i}=v v_{2i}=0, m_1=m and m_2=\frac{m}{4} \\ {v}_{2f}=\left(\frac{2m}{m+{\displaystyle \frac{m}{4}}}\right)v \\ {v}_{2f}=\frac{8}{5}v \\ so the kinetic energy of sphere 2 after collision is \\ \frac{7}{10}\frac{m}{4}\frac{64}{25}{v}^2 \\ this energy will converted into potential energy of the sphere so \\ \frac{7}{10}\frac{m}{4}\frac{64}{25}{v}^2=\frac{m}{4}g\times 0.2 \\ u\sin g value of v we get \\ \frac{7}{10}\frac{64}{25}\frac{10gh}{7}=0.2\times g \\ \Rightarrow h=\frac{5}{64}m \end{gathered}$$