The vector position of a 3.05 g particle moving in the xy plane varies in time a
ID: 1374099 • Letter: T
Question
The vector position of a 3.05 g particle moving in the xy plane varies in time according to
The vector position of a 3.05 g particle moving in the xy plane varies in time according to Vector F (e) Determine the net force exerted on the two-particle system at t = 2.60. Vector V = cm/s (d) Determine the acceleration of the center of mass at t = 2.60. Vector P (c) Determine the velocity of the center of mass at t = 2.60. Vector r = cm (b) Determine the linear momentum of the system at t = 2.60. t (a) Determine the vector position of the center of mass at t = 2.60.t^2 ? 6 ? 2 Vector r is in centimeters. At the same time, the vector position of a 5.45 g particle varies as t^2 where t is in seconds and t + 2 + 33 Vector r 1=Explanation / Answer
For particle 1 :
r = (3)i + (2it^2 + 6jt)
Differentiate the equation twice to get :
a1 = (4)j
Similarly differentiate the relation for the other particle twice :
a2 = (-4)j
Acceleration A of COM = (m1a1 + m2a2) / (m1 + m2) = -0.67j
Net force = (m1 + m2)A = -6j
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