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A system consists of two particles. Particle 1 with mass 2.0 k g 2.0kg is located at ( 2.0 m , 6.0 m ) (2.0m,6.0m) and has a velocity of ( 3.1 m / s , 2.6 m / s ) (3.1m/s,2.6m/s). Particle 2 with mass 4.5 k g 4.5kg is located at ( 4.0 m , 1.0 m ) (4.0m,1.0m) and has a velocity of ( 1.1 m / s , 0.6 m / s ) (1.1m/s,0.6m/s). Determine the position and the velocity of the center of mass of the system.

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okpalawalter8

Answer:

[tex]Position:(3.38m ,2.53m)[/tex]

[tex]Velocity=(1.72m/s,1.22m/s)[/tex]

Explanation:

From the question we are told that

Mass of particle 1  

 [tex]M_1=2.0kg[/tex]

Co-ordinate of particle 1 (2.0m,6.0m)

Velocity of Particle 1  

 [tex]V_1= (3.1 m / s , 2.6 m / s )[/tex]

Mass of particle 2  

[tex]M_2=4.5kg[/tex]

Co-ordinate of particle 2 (4.0m,1.0m)

Velocity of Particle 1  

[tex]V_2=( 1.1 m / s , 0.6 m / s )[/tex]

Generally the Position is mathematically given as  

[tex]X =\frac{ ((M_1 * Cx_1) + (M_2 *Cx_2)}{(M_1 + M_2)}[/tex]

 [tex]X=\frac{2*2+4.5*4}{2+4.5}[/tex]

[tex]X=3.38[/tex]

[tex]Y=\frac{(M_1* Cy_1 + M_2* Cy_2)}{(M_1 + M_2)}[/tex]

[tex]Y =\frac{(2 * 6 + 4.5 * 1)}{(2 + 4.5)}[/tex]

[tex]Y= 2.53 m[/tex]

Therefore the position is given as

[tex]Position:(3.38m ,2.53m)[/tex]

 Solving for Velocity

Generally the velocity of the system is mathematically Given as

[tex]V_x =\frac{ (M_1 * vx_1 + M_2 * Vx_2)}{(M_1 + M_2)}[/tex]

[tex]V_x=\frac{ (2 * 3.1 + 4.5 *1.1)}{(2 + 4.5)}[/tex]

[tex]V_x=1.72m/s[/tex]

For Y

[tex]V_y =\frac{(M_1* vy_1 + M_2* Vy_2)}{(M_1 + M_2)}[/tex]

[tex]V_y=\frac{ (2 * 2.6) + (4.5*0.6)}{(2 + 4.5)}[/tex]

[tex]V_y=1.22m/s[/tex]

Therefore Velocity

[tex]V=(1.72m/s,1.22m/s)[/tex]

 

 

 

 

 

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