Question

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.

Answer 1

`Position:(3.38m ,2.53m)`

`Velocity=(1.72m/s,1.22m/s)`

Step-by-step explanation:

From the question we are told that

Mass of particle 1

`M_1=2.0kg`

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

Velocity of Particle 1

`V_1= (3.1 m / s , 2.6 m / s )`

Mass of particle 2

`M_2=4.5kg`

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

Velocity of Particle 1

`V_2=( 1.1 m / s , 0.6 m / s )`

Generally the Position is mathematically given as

`X =( ((M_1 * Cx_1) + (M_2 *Cx_2))/((M_1 + M_2))`

`X=(2*2+4.5*4)/(2+4.5)`

`X=3.38`

`Y=((M_1* Cy_1 + M_2* Cy_2))/((M_1 + M_2))`

`Y =((2 * 6 + 4.5 * 1))/((2 + 4.5))`

`Y= 2.53 m`

Therefore the position is given as

`Position:(3.38m ,2.53m)`

Solving for Velocity

Generally the velocity of the system is mathematically Given as

`V_x =( (M_1 * vx_1 + M_2 * Vx_2))/((M_1 + M_2))`

`V_x=( (2 * 3.1 + 4.5 *1.1))/((2 + 4.5))`

`V_x=1.72m/s`

For Y

`V_y =((M_1* vy_1 + M_2* Vy_2))/((M_1 + M_2))`

`V_y=( (2 * 2.6) + (4.5*0.6))/((2 + 4.5))`

`V_y=1.22m/s`

Therefore Velocity

`V=(1.72m/s,1.22m/s)`