The vector position of a 4.00 g particle moving in the xy plane varies in time a
ID: 1435450 • Letter: T
Question
The vector position of a 4.00 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 5.80 g particle varies as r_2 = 3i - 2it^2 - 6jt. (a) Determine the vector position of the center of mass at t = 2.60. (b) Determine the linear momentum of the system at t = 2.60. (c) Determine the velocity of the center of mass at t = 2.60. (d) Determine the acceleration of the center of mass at t = 2.60. (e) Determine the net force exerted on the two-particle system at t = 2.60.Explanation / Answer
a) We have m1 & m2 and r1 & r2
center of mass=R
R=(r1m1+r2m2)/(m1+m2)
t=2.60
r1= (3i + 3j)(2.60) + 2j(2.60)^2 = 7.8i + 21.32j
r2=-10.52i + 15.6j
[(7.8i + 21.32j)(4.0)+(-10.52i + 15.6j)(5.8)]/(4.0+5.8)= -3.04i + 17.93j
b) p = m1v1 + m2v2 = [(3i + 13.4j)4 + ( -10.4i -6j)*5.8] = -48.3i + 18.82j
c) v1 = (3i + 3j) + 4tj = 3i + 13.4j
v2 = -4ti - 6j = -10.4i -6j
V=(v1m1+v2m2)/(m1+m2) = [(3i + 13.4j)4 + ( -10.4i -6j)*5.8]/9.8 = -4.93i + 1.92j
d) a1 = 4j
a2 = -4i
A = (a1m1+a2m2)/(m1+m2) = [(4j*4) -(4i*5.8)]/9.8 = -2.37i + 1.63j
e) F = m1a1 + m2a2 = [(4j*4) -(4i*5.8)] = -23.2i + 16j
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