The vector position of a 3.40 g particle moving in the xy plane varies in time a

Question

The vector position of a 3.40 g particle moving in the xy plane varies in time according to the following equation. At the same time, the vector position of a 5.05 g particle varies according to the following equation. For each equation, t is in s and r is in cm. Solve the following when t = 2.80 (a) Find the vector position of the center of mass. cm cm (b) Find the linear momentum of the system. g-cm/s g-cm/s (c) Find the velocity of the center of mass. cm/s cm/s (d) Find the acceleration of the center of mass. cm/s2 cm/s2 (e) Find the net force exerted on the two-particle system. ?N ?N

Explanation / Answer

you have m1 & m2 r1 & r2 center of mass=R R=(r1m1+r2m2)/(m1+m2) t=2.60 r1= (3i + 3j)(2.60) + 2j(2.60)^2 = r2=-10.52i + 15.6 [(7.8i + 21.32j)(3.40)+(-10.52i + 15.6)(5.65)]/(3.40+5.65)= __i + __j

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