The vector position of a 3.25 gram particle moving in the xy plane varies in tim
ID: 2181420 • Letter: T
Question
The vector position of a 3.25 gram particle moving in the xy plane varies in time according tor1=(3i+3j)t+2jt2
at the same time the vector position of a 5.65 gram particle varies according to
r2=3i?2it2?6jt
a. determine the vector position at t=2.6
(in cm)
b. determine the linear momentum of the system at t=2.6
(in g*cm/s)
c. determine the velocity of the center of mass at t=2.6
(in cm/s)
d. determine the acceleration of the center of mass at t=2.6
(in cm/s^2)
e. determine the net force exerted on the two-particle system at t=2.6
Explanation / Answer
Put t = 2.6 in both the Particales For 3.25 g r1 = 7.8i + 16.52j Magnitude = sqrt(7.8^2 + 16.52^2) = 18.26 cm Momentum = mass*dr1/dt = 3.25*(3i + 3j +4*2.6j) = 9.75i + 43.55j Magnitude = sqrt(9.75^2 + 43.55^2) = 44.63 gcm/sec
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