## Engineering Mechanics: Statics & Dynamics (14th Edition)

$v_{A_2}=0.353m/s$ $v_{B_2}=2.53m/s$ $T_1=T_2=0.5J$
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, $T_1=T_2$