Answer
(a) $v_{Cx} = 1.75~m/s$
$v_{Cy} = 0.26~m/s$
(b) The change in kinetic energy is -0.0925 J.
Work Step by Step
(a) We can use conservation of momentum to solve this question.
$p_x = (m_A+m_B+m_C)~v_x$
$m_Av_{Ax}+m_Bv_{Bx}+m_Cv_{Cx} = (m_A+m_B+m_C)~v_x$
$v_{Cx} = \frac{(m_A+m_B+m_C)~v_x - m_Av_{Ax}-m_Bv_{Bx}}{m_C}$
$v_{Cx} = \frac{(0.020~kg+0.030~kg+0.050~kg)(0.50~m/s) - (0.020~kg)(-1.50~m/s)-(0.030~kg)(-0.50~m/s)~cos(60^{\circ})}{0.050~kg}$
$v_{Cx} = 1.75~m/s$
$p_y = (m_A+m_B+m_C)~v_y = 0$
$m_Av_{Ay}+m_Bv_{By}+m_Cv_{Cy} = 0$
$0+m_Bv_{By}+m_Cv_{Cy} = 0$
$v_{Cy} = \frac{-m_Bv_{By}}{m_C}$
$v_{Cy} = \frac{-(0.030~kg)(-0.50~m/s)~sin(60^{\circ})}{0.050~kg}$
$v_{Cy} = 0.26~m/s$
(b) We can find the initial speed of sphere C.
$v_C = \sqrt{(1.75~m/s)^2+(0.26~m/s)^2}$
$v_C = 1.77~m/s$
$K_1 = \frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2+\frac{1}{2}m_Cv_C^2$
$K_1 = \frac{1}{2}(0.020~kg)(1.50~m/s)^2+\frac{1}{2}(0.030~kg)(0.50~m/s)^2+\frac{1}{2}(0.050~kg)(1.77~m/s)^2$
$K_1 = 0.105~J$
$K_2 = \frac{1}{2}(m_A+m_B+m_C)v_f^2$
$K_2 = \frac{1}{2}(0.10~kg)(0.50~m/s)^2$
$K_2 = 0.0125~J$
$\Delta K = K_2-K_1$
$\Delta K = 0.0125~J - 0.105~J$
$\Delta K = -0.0925~J$
The change in kinetic energy is -0.0925 J.
