The vector position of a 3.55 g particle moving in the xy plane varies in time a
ID: 1435516 • Letter: T
Question
The vector position of a 3.55 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 r is in centimeters. At the same time, the vector position of a 5.90 g particle varies as r_2 = 3i - 2it^2 - 6jt. (a) Determine the vector position of the center of mass at t = 3.00. (b) Determine the linear momentum of the system at t = 3.00. (c) Determine the velocity of the center of mass at t = 3.00. (d) Determine the acceleration of the center of mass at t = 3.00. (e) Determine the net force exerted on the two-particle system at t = 3.00.Explanation / Answer
rcm = (m1r1+m2r2)/(m1+m2) = 0.37566*(9i+27j)+ 0.62434 (-15i-18j)
=3.38 i +10.14 J - 9.365 i-11.238j
=-5.985i-1.098j
V1 = 3i+3j+4jt
V2 = -4it-6j
P= M1V1+m2V2 =3.55*(3i+15j)+5.90*(-12i-6j) = -60.15 i -17.85j
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