The vector position of a 3.85 g particle moving in the xy plane varies in time according to r1 = (3î + 3ĵ)t + 2ĵt2 where t is in seconds and r is in centimeters. At the same time, the vector position of a 5.40 g particle varies as r2 = 3î − 2ît2 − 6ĵt.(a)Determine the vector position (in cm) of the center of mass of the system at t = 2.80 s.rcm = cm(b)Determine the linear momentum (in g · cm/s) of the system at t = 2.80 s.p = g · cm/s(c)Determine the velocity (in cm/s) of the center of mass at t = 2.80 s.vcm = cm/s(d)Determine the acceleration (in cm/s2) of the center of mass at t = 2.80 s.acm = cm/s2(e)Determine the net force (in µN) exerted on the two-particle system at t = 2.80 s.Fnet =  µN

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Asked Dec 6, 2019
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The vector position of a 3.85 g particle moving in the xy plane varies in time according to r1 = (3î + 3ĵ)t + 2ĵt2 where t is in seconds and r is in centimeters. At the same time, the vector position of a 5.40 g particle varies as r2 = 3î − 2ît2 − 6ĵt.
(a)
Determine the vector position (in cm) of the center of mass of the system at t = 2.80 s.
rcm = cm
(b)
Determine the linear momentum (in g · cm/s) of the system at t = 2.80 s.
p = g · cm/s
(c)
Determine the velocity (in cm/s) of the center of mass at t = 2.80 s.
vcm = cm/s
(d)
Determine the acceleration (in cm/s2) of the center of mass at t = 2.80 s.
acm = cm/s2
(e)
Determine the net force (in µN) exerted on the two-particle system at t = 2.80 s.
Fnet =  µN
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Expert Answer

Step 1

Mass of the first particle (m1),

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т, 3 3.85 g

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Step 2

Position, velocity and acceleration vector of the first particles,

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;=(3ì + 3)t + 2jr = 3i + 3j+4jt ā, = 4j

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Step 3

Mass of the second particl...

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= 5.40 g т,

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