In: Physics
The vector position of a 3.95 g particle moving in the xy plane varies in time according to
r1 = (3î + 3ĵ)t + 2ĵt2
where t is in seconds and r is in centimeters. At the same time, the vector position of a 5.70 g particle varies as
r2 = 3î − 2ît2 − 6ĵt.
(a)Determine the vector position (in cm) of the center of mass of the system at t = 2.90 s.
rcm = cm
(b)Determine the linear momentum (in g · cm/s) of the system at t = 2.90 s.
p = g · cm/s
(c)Determine the velocity (in cm/s) of the center of mass at t = 2.90 s.
vcm = cm/s
(d)
Determine the acceleration (in cm/s2) of the center of mass at t = 2.90 s.
acm = cm/s2
(e)Determine the net force (in µN) exerted on the two-particle system at t = 2.90 s.
Fnet = µN