Chapter 8, Problem 17P

A ball with a mass of 0.600 kg is initially at rest. It is struck by a second ball having a mass of 0.400 kg, initially moving with a velocity of 0.250 m/s toward the right along the x axis. After the collision, the 0.400 kg ball has a velocity of 0.200 m/s at an angle of 36.9° above the x axis in the first quadrant. Both balls move on a frictionless, horizontal surface. (a) What are the magnitude and direction of the velocity of the 0.600 kg ball after the collision? (b) What is the change in the total kinetic energy of the two balls as a result of the collision?

8.17. Set Up: Before and after diagrams are sketched in the figure below. m_A = 0.600 kg, m_B = 0.400 kg. v_A,i,x = 0, v_B,i,x = 0.250m/s. v_B,f = 0.200 m/s, θ = 36.9°.

Solve: (a) Conservation of momentum gives P_i,x = P_f,x and P_i,y = P_f,y. P_i,x = P_f,x says m_Bv_B,i,x = m_Bv_B,f,x + m_Av_A,f,x.