Two particles are moving in the xz-plane. Particle #1 has a mass of m1 = 3.35 kg

Question

Two particles are moving in the xz-plane. Particle #1 has a mass of m1 = 3.35 kg and a position which is a function of time given by r1 = (4 m/s)ti + [(3 m/s)t + (2 m/s2)t2]k. Particle #2 has a mass of m2 = 5.90 kg and a position which is a function of time given by r2 = [(2 m) (3 m/s2)t2]i (4 m/s)tk. For this system of particles, determine the following at a time t = 2.70 s.

a.) position of the center of mass of the two particles

b.) velocity of the center of mass of the two particles

c.)acceleration of the center of mass of the two particles

d.) total momentum of the two particles

e.) net force exerted on the two particles

Explanation / Answer

a) r1 = 4t i + (3t + 2t^2)k

r2 = (2 - 3t^2)i - 4t k

at t = 2.7 s

r1 = (4 x 2.7)i + (3x2.7 + 2(2.7^2))k = 10.8i + 22.68k m

r2 = (2 - 3(2.7^2))i - 4(2.7)k = -19.87i - 10.8k m

Rcm = (m1r1 + m2r2 ) / (m1 + m2)

= (3.35 (10.8i + 22.68k)) + (5.90 (-19.87i - 10.8k)) / (3.35 + 5.90)

= - 8.76i + 1.33k m

b) v1 = dr1/dt = 4i + (3 + 4t) k

v2 = dr2/dt = -6t i - 4k

at t = 2.70 s

v1 = 4i + (3 + 4(2.7))k = 4i + 13.8k m/s

v2 = (-6 x 2.7)i - 4k = -16.2i - 4k m/s

Vcm = (m1v1 + m2v2) / (m1 + m2)

= - 8.88i + 2.45 k m/s

c) a1 = dv1 /dt = 0i + 4k

a2 = dv2/dt = -6i + 0k

a_cm = (3.35 x 4k + 5.90 x-6i) / (3.35+ 5.9)

= - 3.83i + 1.45k m/s^2

d) momentum = total mass x Vcm

= (3.35+ 5.90) ( - 8.88i + 2.45 k )

= -82.14i + 22.66k kg m/s

e) F = m a_cm

= (3.35 + 5.90) ( - 3.83i + 1.45k)

= - 35.43i + 13.42 k N

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