The vector position of a 3.80 g particle moving in the XY plane varies in time a

Question

The vector position of a 3.80 g particle moving in the XY plane varies in time according to      (r_1= (3i + 3j)t + 2t^2)    where t is in seconds and r is in centimeters.  At the same time, the vector position of a 5.15 g particle varies at      (r_2 = 3i - 2it^2 -6jt)    

(a) Determine the vector position of the center of mass at t = 2.90.

    (b) Determine the linear momentum of the system at t = 2.90.

    (c) Determine the velocity of the center of mass at t = 2.90.

    (d) Determine the acceleration of the center of mass at t = 2.90.

    (e) Determine the net force exerted on the two-particle system at t = 2.90.

Explanation / Answer

m1 = 3.80 g, r1 = 3t i + (3t +2t2) j

r1' = 3 i + (3 + 4t) j

r1'' = 4 j

t = 2.90s,

r1 = 8.7 i + 25.52 j,

r1' = 3 i +14.6 j,

r1'' = 4 j

m2 = 5.50 g,

r2 = (3 - 2t2) i - 6tj,

r2' = -4t i - 6 j,

r2'' = -4 i,

t = 2.90s,

r2 = -13.82 i - 17.4 j,

r2' =-11.6 i - 6 j,

r2'' = -4 i,

a) rc = (m1r1 +m2r2)/(m1 + m2)

rc=(-4.258 i +0.823 j) cm


b) P = m1r1' + m2r2'

P=(-48.34 i + 24.58 j) g-cm/s

c) rc' = P/(m1 + m2) = (-5.401 i +2.746 j) cm/s

d) rc'' = (m1r1'' +m2r2'')/(m1 + m2) =(-2.301 i +1.68 j) cm/s2

e) (m1 + m2) rc'' = (-20.6+15.2 j)g-cm/s2 = (-20.6+15.2 j) * 10-5 N

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