More Practice with Collisions in 2D SPH4U 1 . A steel ball of mass 0.50 kg, moving with a velocity of 2.0 m/s [E], strikes a second ball of mass 0.30 kg, initially at rest. The collision is a glancing one, causing the steel ball to have a velocity of 1.5 m/s [30 N of E] after the collision. Determine the velocity of the second ball after the collision. m=0.50 kg V 1=2.0 " [E] V1'=1.5 [30' N of E] m :=0.30 kg V B= Om VB ' = ? Determine the total East-West (x) and North-South (y) components of the momentum before: PTOTALX=MAVAX+ MBV BX= PTOTALymAVASTmBV By Break mvA' down into its East-West (x) and North-South (y) components: "' = my Ax my AV Calculate the East-West (x) and North-South (y) components of the second ball's momentum: mv BX '= PTOTALX - MV Ax my By '= PTOTALy - my Ay Find the momentum of the second ball (from the components) and the velocity (divide the momentum by the mass): Answer: 1.7 m/s [430 E of S]