Answer
(a) The magnitude of the velocity is 0.226 m/s, and they are moving to the left after the collision.
(b) The amount of mechanical energy that dissipates is 197 J.
Work Step by Step
(a) We can use conservation of momentum to solve this question. Let the right be the positive direction.
$(m_A+m_B)~v_2 = m_A~v_{A1}+m_B~v_{B1}$
$v_2 = \frac{m_A~v_{A1}+m_B~v_{B1}}{m_A+m_B}$
$v_2 = \frac{(7.50~kg)(-5.00~m/s)+(5.75~kg)(6.00~m/s)}{7.50~kg+5.75~kg}$
$v_2 = -0.226~m/s$
The magnitude of the velocity is 0.226 m/s, and they are moving to the left after the collision.
(b) $K_1 = \frac{1}{2}m_A~v_A^2+\frac{1}{2}m_B~v_B^2$
$K_1 = \frac{1}{2}(7.50~kg)(5.00~m/s)^2+\frac{1}{2}(5.75~kg)(6.00~m/s)^2$
$K_1 = 197.3~J$
$K_2 = \frac{1}{2}m_2~v_2^2$
$K_2 = \frac{1}{2}(13.25~kg)(0.226~m/s)^2$
$K_2 = 0.3384~J$
$\Delta K = K_2 - K_1$
$\Delta K = 0.3384~J - 197.3~J$
$\Delta K = -197~J$
The amount of mechanical energy that dissipates is 197 J.
