The vector position of a 3.20 g particle moving in the xy plane varies in time a
ID: 1466660 • Letter: T
Question
The vector position of a 3.20 g particle moving in the xy plane varies in time according to r_1 = (3i + 3j)t + 2jt^2 where t is in seconds and the same time, the vector position of a 5.80 g particle varies as r_2 = 3i - 2it^2 - 6jt. Determine the vector position of the center of mass at t = 2.40. Determine the linear momentum of the system at t = 2.40. Determine the velocity of the center of mass at t = 2.40. Determine the acceleration of the center of mass at t = 2.40. Determine the net force exerted on the two-particle system at t = 2.40.Explanation / Answer
a) at t = 2.40 s
r1 = (3x2.40)i + ( 3x2.40 + 2x2.40^2)j
r1 = 7.2i + 18.72j cm
r2 = (3 - 2x2.40^2)i + ( -6 x 2.40)j = - 8.52 i - 14.4 j cm
rcm = m1r1 + m2r2 / (m1 + m2)
rcm = - 2.93 i - 2.62 j cm
b) v1 = dr1/dt = 3i + ( 3 + 4t)j
v1 = 3i + 12.6j cm/s
v2 = dr2/dt = -4ti - 6j
v2 = - 9.6i - 6j cm/s
p = (m1v1 + m2v2)
vcm = -46.08i + 5.52j g cm/s
c) vcm = p / (m1 + m2) = - 5.12i + 0.613 j cm/s
d) a1 =dv1/ dt = 0 + 4j
a2 = dv2 /dt = - 4i + 0 j
acm = (m1a1 + m2a2) / (m1 +m2)
= - 2.58i + 1.42 j
e) F = (m1 +m2)a
= - 23.22i + 12.78j g cm / s^2
= - 0.232 i + 0.1278 j uN
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