The vector position of a 3.65 g particle moving in the xy plane varies in time a

Question

The vector position of a 3.65 g particle moving in the xy plane varies in time according to (3i- 3i)*- 2j2 where t is in seconds and r is in centimeters. At the same time, the vector position of a 5.35 g particle varies as r2- 31 - 21t2 - 6jt. (a) Determine the vector position of the center of mass at t 2.80 cm (b) Determine the linear momentum of the system at t - 2.80. g cm/s (c) Determine the velocity of the center of mass at t- 2.80. cm cm/s (d) Determine the acceleration of the center of mass at t 2.80 acm cm/s2 (e) Determine the net force exerted on the two-particle system at t 2.80. net

Explanation / Answer

given

m = 3.65 g

vecotr position

r1 = (3i + 3j)t + 2jt^2

t is seconds, r in cm

for M = 5.35 g

r2 = 3i - 2it^2 - 6jt

a. location of com = (mr1 + Mr2)/(m + M)

COM = (10.95i*t + 10.95j*t + 7.3*j t^2 + 16.05i - 10.7 i t^2 - 32.1 j t)/(9)

for t = 2.8 s

COM = (-4.1308888i -0.2208 j) cm

b. linear momentum of the system at t = 2.8 s

mdr1/dt + Mdr2/dt = 3.65(3i + 3j + 4j*2.8) + 5.35(-4i*2.8 - 6j))

P = -48.97i + 19.73 j g cm /s

c. Vcom at t= 2.8 = (mdr1/dt + Mdr2/dt)/(m + M) = (-48.97i + 19.73 j)/9 = -5.4411111i + 2.192222 j cm/s

d. acceleration of come = (md^2r1/dt + Md^2r2/dt)/(M + m) = [3.65(4j) + 5.35*(-4i)]/9 = 1.62222 j - 2.377777 i cm/s/s

e. net force = acceleration *mass = 14.6j - 21.4i N

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