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

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.

r_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.

v_cm =

cm/s

(d)

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

a_cm =

cm/s²

(e)

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

F_net =

µN

Solutions

Expert Solution

genius_generous answered 1 year ago

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