Answer
(a) $U_s = 5.25~J$
The velocity of each block at this time will be zero because all the energy of the system is stored in the springs.
(b) Block A is moving 2.17 m/s to the left.
Block B is moving 0.333 m/s to the right.
Work Step by Step
(a) The maximum energy stored in the springs will be equal to the kinetic energy of the two blocks.
$U_s = K$
$U_s = \frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2$
$U_s = \frac{1}{2}(2.00~kg)(2.00~m/s)^2+\frac{1}{2}(10.00~kg)(0.500~m/s)^2$
$U_s = 5.25~J$
The velocity of each block at this time will be zero because all the energy of the system is stored in the springs.
(b) Let $v_A'$ be the final velocity of block A.
Let $v_B'$ be the final velocity of block B.
Let $m_A = 2.00~kg$ and let $m_B = 10.00~kg$
We can use conservation of momentum to set up an equation.
$m_A~v_A + m_B~v_B = m_A~v_A' + m_B~v_B'$
Since the collision is elastic, we can use equation (8.27) to set up another equation.
$v_A - v_B = v_B' - v_A'$
$v_A' = v_B' - v_A + v_B$
We can use this expression for $v_A'$ in the first equation.
$m_A~v_A + m_B~v_B = m_A~v_B' - m_Av_A + m_Av_B+ m_B~v_B'$
$v_B' = \frac{2m_A~v_A+m_B~v_B -m_Av_B}{m_A+m_B}$
$v_B’ = \frac{(2)(2.00~kg)(2.00~m/s)+(10.00~kg)(-0.500~m/s)- (2.00~kg)(-0.500~m/s)}{(2.00~kg)+(10.00~kg)}$
$v_B' = 0.333~m/s$
We can use $v_B'$ to find $v_A'$.
$v_A' = v_B' - v_A + v_B$
$v_A' = 0.333~m/s - 2.00~m/s + (-0.500)~m/s$
$v_A' = -2.17~m/s$
Block A is moving 2.17 m/s to the left.
Block B is moving 0.333 m/s to the right.
