Answer
(a) The final speed of block A is 3.60 m/s.
(b) The initial potential energy stored in the compressed spring is 8.64 J.
Work Step by Step
(a) $p = 0$
$m_Av_A+m_Bv_B = 0$
$m_Av_A=-m_Bv_B$
$v_A=\frac{-m_Bv_B}{m_A}$
$v_A=\frac{-(3.00~kg)(1.20~m/s)}{1.00~kg}$
$v_A = -3.60~m/s$
The final speed of block A is 3.60 m/s.
(b) The initial potential energy stored in the compressed spring is equal to the total kinetic energy of the two blocks.
$U_s = \frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2$
$U_s = \frac{1}{2}(1.00~kg)(3.60~m/s)^2+\frac{1}{2}(3.00~kg)(1.20~m/s)^2$
$U_s = 8.64~J$
The initial potential energy stored in the compressed spring is 8.64 J.
