Question

In: Physics

The vector position of a 3.70 g particle moving in the xy plane varies in time...

The vector position of a 3.70 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 5.80 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 = 2.90 s.

rcm =

cm

(b)

Determine the linear momentum (in g · cm/s) of the system at t = 2.90 s.

p =

g · cm/s

(c)

Determine the velocity (in cm/s) of the center of mass at t = 2.90 s.

vcm =

cm/s

(d)

Determine the acceleration (in cm/s2) of the center of mass at t = 2.90 s.

acm =

cm/s2

(e)

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

Fnet =

µN

Solutions

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