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The magnitude and direction of net angular momentum of particles is [tex]52.7kg.m^2/s[/tex] out of the page.
Given,
particle 1:-
Mass ,m1 = 6.5 kg
speed , v1 = 2.2 m/s
distance from point O, d1 = 1.5 m
Particle 2:-
Mass ,m2 = 3.1 kg
speed, v2 = 3.6 m/s
distance from point O, d2 = 2.8 m
Angular momentum can be determined by formula,
p=mvd
angular momentum of particle-1, [tex]\rho_1=6.5*2.2*1.5=21.45 kg.m^2/s[/tex]
angular momentum of particle-2, [tex]\rho_2=3.1*3.6*2.8=31.248 kg.m^2/s[/tex]
Then, the net angular momentum is
[tex]\rho=\rho_1+\rho_2=21.45+31.248=52.69kg.m^2/s[/tex]
The direction of angular momentum of particle is always perpendicular to the motion of particle, as the particles are moving on xy-plane then the their direction will be perpendicular to xy-plane i.e. z-plane, it means out of the page.
To learn more about angular momentum refer here
https://brainly.com/question/14456632
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Your question is incomplete, here is the complete question.
In the instant shown in the diagram, two particles move in an xy-plane. Particle P1 has mass 6.5 kg and speed v1 = 2.2 m/s, and it is at distance d1 = 1.5 m from point O. Particle P2 has mass 3.1 kg and speed v2 = 3.6 m/s, and it is at distance d2 = 2.8 m from point O. What is the magnitude and direction of net angular momentum of the two particles about O?
A) 52.7 kg · m2/s out of the page B) 52.7 kg · m2/s into the page C) 21.5 kg · m2/s into the page D) 9.8 kg · m2/s into the page E) 9.8 kg · m2/s out of the page
