Given, the mass of the balls, m1 = 0.44 kg and m2 = 0.22 kg initial velocities, u1 = 3.30 m/s towards east and u2 = 0 m/s Let the velocities of m1 and m2 after the collision are v1 and v2 respectively. By the conservation of momentum, we get m1v1 + m2v2 = m1u1 + m2u2 or, 0.44v1+0.22v2= 0.44*3.3 + 0.22*0 or, v2 = 6.6-2v1 (1) And by the conservation of kinetic energy, we get 1/2∗m1v12+1/2∗m2v22=1/2∗m1u12 +0 or, m1v12+m2v22=m1u12 or, 0.44* v12 + 0.22* v22 = 4.7916 (2) Substituting the equation (1) to (2), we get 0.44* v12 + 0.22 (6.6−2v1)2 =...