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creativ13 [48]
1 year ago
15

a system of four particles moves along one dimension. the center of mass of the system is at rest, and the particles do not inte

ract with any objects outside of the system. find the velocity ????4 of particle 4 at ????1
Physics
1 answer:
Zinaida [17]1 year ago
5 0

The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

The given parameters;

m1 = 1.45 kg, v1(t) = (6.09m/s) + (0.299m/s^2) × t

m2 = 2.81 kg, v2(t) = (7.83m/s) + (0.357m/s^2) × t

m3 = 3.89 kg, v3(t) = (8.09m/s) + (0.405m/s^2) × t

m4 = 5.03kg

The velocity of the center mass of the particles is calculated as;

McmVcm = m1v1 + m2v2 +m3v3+m4v4

Vcm= m1v1 + m2v2 +m3v3 +m4v4/ Mcm

0 = m1v1 + m2v2 +m3v3 +m4v4/ Mcm

m1v1 + m2v2 +m3v3+m4v4 = 0

m4v4 = -(m1v1 + m2v2 +m3v3)

v4 =-(m1v1 + m2v2 +m3v3)/ m4

The velocity of particle 1 at time, t = 2.83 s;

vi = 6.09 + 0.299× 2.83

v1 = 6.94 m/s

The velocity of particle 3 at time, t = 2.83 s;

v2 = 7.83 + 0.357 × 2.83

v2 = 8.84 m/s

The velocity of particle 3 at time, t = 2.83 s;

v3 = 8.09 + 0.405 × 2.83

v3 = 9.24 m/s

The velocity of particle 3 at time, t = 2.83 s;

v4 = - (m1v1 + m2v2 + m3v3)/m4

v4 = -(1.45×6.94 + 2.81×8.84 + 3.89×9.24)/5.03

v4 = -14.4 m/s

Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

Learn more about Velocity here:

brainly.com/question/18084516

#SPJ4

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