Four particles are in a 2-D plane with masses, x- and y- positions, and x- and y

Question

Four particles are in a 2-D plane with masses, x- and y- positions, and x- and y- velocities as given in the table below:

        m                x                y                Vx               Vy

1    8.2kg          -2.4m         -4.7m          2.9 m/s        -4.1 m/s

2    9.1kg          -3.6m          3.4m          -5 m/s           4.9 m/s

3    7.9kg          4.7m           -5.6m        -6.2 m/s         2 m/s

4    8.7kg          5.5m           2.7m          3.9 m/s        -3.2 m/s

1) What is the x position of the center of mass?

Explanation / Answer

1. x - position of Centre of mass =( m1* x1 + m2*x2 + m3*x3 + m4*x4 ) /( m1 +m2+m3+m4)
=(8.2*-2.4)+(9.1*-3.6)+(7.9*4.7)+(8.7*5.5)/(8.2+9.1+7.9+8.7)=0.959

2. y - position of Centre of mass =( m1* y1 + m2*y2 + m3*y3 + m4*y4 ) /( m1 +m2+m3+m4)
=(8.2*-4.7)+(9.1*3.4)+(7.9*-5.6)+(8.7*2.7)/(8.2+9.1+7.9+8.7)=-0.836

3. Vy of the centre of mass = ( m1* Vy1 + m2*Vy2 + m3*Vy3 + m4*Vy4 ) /( m1 +m2+m3+m4)
=(8.2*-4.1)+(9.1*4.9)+(7.9*2)+(8.7*-3.2)/(8.2+9.1+7.9+8.7)=-0.0315
Vx of the centre of mass = ( m1* Vx1 + m2*Vx2 + m3*Vx3 + m4*Vx4 ) /( m1 +m2+m3+m4)
=(8.2*2.9)+(9.1*-5)+(7.9*-6.2)+(8.7*3.9)/(8.2+9.1+7.9+8.7)=-1.084

therefore speed = sqrt (Vy^2 + Vx^2)=sqrt(1.176)=1.08
4. when a fifth particle is placed at the origin its x5 = 0, so in the formula m5*x5 = 0.. but its mass is not 0 . So the value of the denominator increase and the x position of the centre of mass decreases.

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