Answer
(a) $\frac{1}{16}$
(b) $\frac{5}{3}$
Work Step by Step
(a) We can find the ratio of final and kinetic energy of the system as follows:
$K_i=\frac{1}{2}m_1v^2+\frac{1}{2}m_2v^2$
$K_i=\frac{1}{2}(m_1+m_2)v^2$
and $K_f=\frac{1}{2}(m_1+m_2)v_f^2$
Given that $v_f=\frac{v}{4}$
$\implies K_f=\frac{1}{2}(m_1+m_2)(\frac{v}{4})^2$
$K_f=\frac{1}{32}(m_1+m_2)v^2$
Now $\frac{K_f}{K_i}=\frac{\frac{1}{32}(m_1+m_2)v^2}{\frac{1}{2}(m_1+m_2)v^2}=\frac{1}{16}$
(b) We can find the required ratio of mass as follows:
$P_i=P_f$
$\implies m_iv+(-m_2v)=(m_1+m_2)\frac{v}{4}$
This simplifies to:
$4m_1-4m_2=m_1+m_2$
$\implies 3m_1-5m_2=0$
$\implies \frac{m_1}{m_2}=\frac{5}{3}$
