Answer
(a) The magnitude of block A's acceleration is $162~m/s^2$ and the magnitude of block B's acceleration is $54 ~m/s^2$.
(b) The final speed of block A is 5.22 m/s.
The final speed of block B is 1.74 m/s.
Work Step by Step
(a) $F = kx = (720~N/m)(0.225~m)$
$F = 162~N$
$F = m_A~a$
$a = \frac{F}{m_A} = \frac{162~N}{1.00~kg}$
$a = 162~m/s^2$
$F = m_B~a$
$a = \frac{F}{m_B} = \frac{162~N}{3.00~kg}$
$a = 54~m/s^2$
The magnitude of block A's acceleration is $162~m/s^2$ and the magnitude of block B's acceleration is $54 ~m/s^2$.
(b) Since the initial momentum is zero, we can use conservation of momentum to note that the magnitude of momentum will be the same for each block.
$m_Av_A = m_B v_B$
$v_A = \frac{m_B v_B}{m_A}$
$v_A = \frac{3.00~kg~ v_B}{1.00~kg}$
$v_A = 3.00~v_B$
After leaving the spring, the sum of the kinetic energy of the blocks will be equal to the initial potential energy in the spring.
$K = U_s$
$\frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2 = \frac{1}{2}kx^2$
$m_A~(3.00~v_B)^2+m_Bv_B^2 = kx^2$
$v_B^2 = \frac{kx^2}{9.00~m_A+m_B}$
$v_B = \sqrt{\frac{kx^2}{9.00~m_A+m_B}}$
$v_B = \sqrt{\frac{(720~N/m)(0.225~m)^2}{(9.00)(1.00~kg)+(3.00~kg)}}$
$v_B = 1.74~m/s$
$v_A = 3.00~v_B = (3.00)(1.74~m/s)$
$v_A = 5.22~m/s$
The final speed of block A is 5.22 m/s.
The final speed of block B is 1.74 m/s.
