Answer
(a) The maximum energy stored in the springs is 4.00 J
The velocity of each block at this time is zero because all the mechanical energy of the system is stored in the springs.
(b) The final velocity of block A is -1.00 m/s so it is moving to the left.
The final velocity of block B is 1.00 m/s so it is moving to the right.
Work Step by Step
(a) The maximum energy stored in the springs is equal to the initial kinetic energy of block A.
$E = \frac{1}{2}m_A~v_A^2$
$E = \frac{1}{2}(2.00~kg)(2.00~m/s)^2$
$E = 4.00~J$
The maximum energy stored in the springs is 4.00 J
The velocity of each block at this time is zero because all the mechanical 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 $m_B = 6.00~kg$.
We can use conservation of momentum to set up an equation.
$m_A~v_A + 0 = m_A~v_A' + m_B~v_B'$
Since the collision is elastic, we can use equation (8-27) to set up another equation.
$v_A - 0 = v_B' - v_A'$
$v_A' = v_B' - v_A$
We can use this expression for $v_A'$ in the first equation.
$m_A~v_A + 0 = m_A~v_B' - m_Av_A + m_B~v_B'$
$v_B' = \frac{2m_A~v_A}{m_A+m_B}$
$v_B' = \frac{(2)(2.00~kg)(2.00~m/s)}{(2.00~kg)+(6.00~kg)}$
$v_B' = 1.00~m/s$
We can use $v_B'$ to find $v_A'$.
$v_A' = v_B' - v_A$
$v_A' = 1.00~m/s - 2.00~m/s$
$v_A' = -1.00~m/s$
The final velocity of block A is -1.00 m/s so it is moving to the left.
The final velocity of block B is 1.00 m/s so it is moving to the right.
