d 1. More Practice with Collisions in 2D SPH4U 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 = 2.0[E] S V'=1.5 [30 N of E] S m =0.30 kg VB=0m VB' = ? S Determine the total East-West (x) and North-South (y) components of the momentum before: P TOTAL MAVAx+mBVE = AX = Bx P TOTALY MAVA+MBV By Ay = Break mv down into its East-West (x) and North-South (y) components: mv Ax= mv MV Ay Calculate the East-West (x) and North-South (y) components of the second ball's momentum: mv Bx=PTOTALX-mv Ax'= -mv Ay'= mv By A Find the momentum of the second ball (from the components) and the velocity (divide the momentum by the mass): Answer: 1.7 m/s [43 E of S]