Answer
$v_{A_2}=0.353m/s$
$v_{B_2}=2.53m/s$
$T_1=T_2=0.5J$
Work Step by Step
According to the law of conservation of linear momentum
$m_Av_{A_1}+m_Bv_{B_1}=m_Av_{A_2}+m_Bv_{B_2}$
We plug in the known values to obtain:
$(0.25)(2)+0=(0.25)v_{A_2}+(0.175)v_{B_2}$
$\implies 0.5=0.25v_{A_2}+0.175v_{B_2}~~~~$[eq(1)]
Since the collision is perfectly elastic, therefore $e=1$
$\implies e=\frac{v_{B_2}-v_{A_2}}{v_{A_1}-v_{B_1}}$
$\implies 1=\frac{v_{B_2}-v_{A_2}}{2-0}$
$\implies 2=v_{B_2}-v_{A_2}~~~~$[eq(2)]
After solving eq(1)and eq(2), we obtain:
$v_{A_2}=0.353m/s$
and $v_{B_2}=2.53m/s$
Now the kinetic energy of the disks before the collision is given as
$T_1=\frac{1}{2}m_Av_{A_1}+\frac{1}{2}m_Bv_{B_1}$
We plug in the known values to obtain:
$T_1=\frac{1}{2}(0.25)(2)^2=0.5J$
The kinetic energy of the disks after the collision is given as
$T_2=\frac{1}{2}(0.25)(0.353)^2+\frac{1}{2}(0.175)(2.35)^2=0.5J$
Thus, $T_1=T_2$
