Answer
$11.2\space kg\rightarrow1.25\space m/s$
$21.7\space kg\rightarrow 0.645\space m/s$
Work Step by Step
Let's apply the principle of conservation of momentum for the system.
$m_{1}u_{1}+m_{2}u_{2}=m_{1}V_{1}+m_{2}V_{2}$
$m_{1}u_{1}+m_{2}u_{2}=0$
$u_{2}=-\frac{m_{1}u_{1}}{m_{2}}-(1)$
Let's apply the principle of conservation of mechanical energy for the system.
Kinetic energy + Potential energy = Constant
$\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}+0=0+0+\frac{1}{2}kx^{2}$
$m_{1}u_{1}^{2}+m_{2}u_{2}^{2}=kx^{2}-(2)$
(1)=>(2),
$m_{1}u_{1}^{2}+m_{2}(-\frac{m_{1}u_{1}}{m_{2}})^{2}=kx^{2}$
$u_{1}^{2}[\frac{m_{1}(m_{2}+m_{1})}{m_{2}}]=kx^{2}$
$u_{1}=x\sqrt {\frac{m_{2}k}{m_{1}(m_{1}+m_{2})}}$ ; Let's plug known values into this equation.
$u_{1}=0.141\space m\sqrt {\frac{21.7\space kg(1330\space N/m)}{11.2\space kg(21.7\space kg+11.2\space kg)}}=1.25\space m/s$
Speed of the 11.2 kg block = 1.25 m/s
From (1)
$u_{2}=-\frac{(11.2\space kg)(1.25\space m/s)}{21.7\space kg}\approx-0.645\space m/s$
Speed of the 21.7 kg block = 0.645 m/s
