The vector position of a 3.15 g particle moving in the xy plane varies in time a

Question

The vector position of a 3.15 g particle moving in the xy plane varies in time according to r_1 = (3i + 3j)t + 2jt^2 where t is in seconds and r is in centimeters. At the same time, the vector position of a 5.20 g particle varies as r_2 = 3i - 2it^2 - 6jt. Determine the vector position of the center of mass at t = 3.00. Determine the linear momentum of the system at t = 3.00. Determine the velocity of the center of mass at t = 3.00. Determine the acceleration of the center of mass at t = 3.00. Determine the net force exerted on the two-particle system at t = 3.00.

Explanation / Answer

a) c.m. = 3.15 x r1 + 5.2 x r2 / (3.4 + 5.85)

put t = 2.60 , and u'll get ur answer.

b) p = mv

v = dr/dt

p = 3.15 dr1/dt + 5.20dr2/dt

= 3.15v1 + 5.20v2

here, v1 = dr1/dt = 3i + 3j + 4jt and   v2 = dr2/dt = - 4jt -6j

put the value of t= 3.00 s

c) v = p/(3.15 + 5.20)

p you wil get from part B.

d) a = 3.15dv1/dt + 5.20dv2/dt / (3.15 + 5.20)

= 3.15a1 + 5.20a2 / 9.3

here, a1 = dv1/dt = 4j and a2 = dv2/dt = - 4j

put the values, and solve a.

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