Answer
(a) The speed of puck A before the collision was 0.790 m/s.
(b) The change in kinetic energy during the collision is -0.0023 J.
Work Step by Step
(a) $m_Av_{A1} = m_Av_{A2}+m_Bv_{B2}$
$v_{A1} = \frac{m_Av_{A2}+m_Bv_{B2}}{m_A}$
$v_{A1} = \frac{(0.250~kg)(-0.120~m/s)+(0.350~kg)(0.650~m/s)}{0.250~kg}$
$v_{A1} = 0.790~m/s$
The speed of puck A before the collision was 0.790 m/s.
(b) $K_1 = \frac{1}{2}m_Av_{A1}^2$
$K_1 = \frac{1}{2}(0.250~kg)(0.790~m/s)^2$
$K_1 = 0.0780~J$
$K_2 = \frac{1}{2}m_Av_{A2}^2+ \frac{1}{2}m_Bv_{B2}^2$
$K_2 = \frac{1}{2}(0.250~kg)(0.120~m/s)^2+ \frac{1}{2}(0.350~kg)(0.650~m/s)^2$
$K_2 = 0.0757~J$
We can find the change in kinetic energy during the collision.
$\Delta K = K_2-K_1$
$\Delta K = (0.0757~J)-(0.0780~J)$
$\Delta K = -0.0023~J$
The change in kinetic energy during the collision is -0.0023 J.
