The vector position of a 3.30 g particle moving in the xy plane varies in time according to r, = (31 + 3j)t + 2jt2 where t is in seconds and r is in centimeters. At the same time, the vector position of a 5.65 g particle varies as r2 = 31 - 2it2 - 6jt. (a) Determine the vector position (in cm) of the center of mass of the system at t = 2.90 s. rom = -7.621 - 1.57j * cm (b) Determine the linear momentum (in g . cm/s) of the system at t = 2.90 s. P = -65.54i + 14.28j x g . cm/s (c) Determine the velocity (in cm/s) of the center of mass at t = 2.90 s. Vom -7.581 + 1.59j x cm/s (d) Determine the acceleration (in cm/s2) of the center of mass at t = 2.90 s. acm = -2.53i + 1.47j cm/s2 (e) Determine the net force (in UN) exerted on the two-particle system at t = 2.90 s. Fnet = -2261 + 132; (10-6) X HN