O -45 points SerPSE9 9 P054 My Notes O Ask Your The vector position of a 3.40 g

Question

O -45 points SerPSE9 9 P054 My Notes O Ask Your The vector position of a 3.40 g partide moving in the xy plane varies in time according to F, -(3i 3i)+2j2 where t is in seconds and r is in centimeters. At the same time, the vector position of a 6.00 g particle varies as (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 9 cmys (c) Determine the velocity of the center of mass at t-2.60. cm/s (d) Determine the acceleration of the center of mass at t 2.60. cm/s2 (e) Determine the net force exerted on the two-partidle system at t 2.60 Fnet- Need Help? LRead Subemt Assignment Save Assignment Progress Home My Assignments Extension Request

Explanation / Answer

a) vector position of particle of mass 3.4g at t=2.6 sec is r1 = (3i + 3j)2.6 +2(2.62)j = 7.8i + 7.8j + 13.52j = 7.8i+21.32j

  vector position of particle of mass 6g at t=2.6 sec is r2 = 3i - 2(2.62)i - 6(2.6)j = 3i - 13.52i - 15.6j = -10.52i - 15.6j

x coordinate of center of mass = (3.4(7.8) + 6(-10.52)) / (3.4 + 6) = -3.89cm

   y coordinate of center of mass = (3.4(21.32) + 6(-15.6)) / (3.4 + 6) = -2.25cm

So rcm = -3.89i - 2.25j

b) velocity of particle of mass 3.4g at t = 2.6 is dr1/dt = 3i + 3j + (4t)j = 3i + 13.4j

  velocity of particle of mass 6g at t = 2.6 is dr2/dt = (-4t)i - 6j = -10.4i - 6j

momentum of particle of mass 3.4g at t = 2.6 is (3.4*10-3)(3i + 13.4j) = (10.2i + 45.56j)10-3

  momentum of particle of mass 6g at t = 2.6 is (6*10-3)(-10.4i - 6j) = (-62.4i - 36j)10-3

Linear momentum of the system = (10.2i + 45.56j)10-3 + (-62.4i - 36j)10-3 = (-52.2i + 9.56j)10-3

c) velocity in the x direction of center of mass is (3.4(3) + 6(-10.4)) / (3.4 + 6) = -5.55

velociy direction of center of mass is (3.4(13.4) + 6(-6)) / (3.4 + 6) = 1.017

so vcm = -5.55i + 1.017j

d) Similarly we can find acceleration as dr/dt2

acceleration of particle of mass 3.4g is 4j

   acceleration of particle of mass 6g is -4i

acceleration in the x direction of center of mass is (3.4(0) + 6(-4)) / (3.4 + 6) = -2.55

acceleration in the y direction of center of mass is (3.4(4) + 6(0)) / (3.4 + 6) = 1.45

so acm = -2.55i + 1.45j

e) force exerted on the particle of mass 3.4 g is 'ma' = 3.4*10-3(4j) = 13.6*10-3j

  force exerted on the particle of mass 6 g is 'ma' = 6*10-3(-4i) = -24*10-3i

So net force exerted on the two particle system at t=2.6s is -24*10-3i + 13.6*10-3j

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