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  

9.2 kg

-2.7 m

-4.7 m

3.2 m/s

-4.2 m/s

  2  

9.2 kg

-3.7 m

3.7 m

-5.2 m/s

5.2 m/s

  3  

9.3 kg

4.7 m

-5.7 m

-6.2 m/s

2.2 m/s

  4  

9.3 kg

5.7 m

2.7 m

4.2 m/s

-3.2 m/s

1)

What is the distance the center of mass is from the origin?

2)

What is the x-component of the velocity of the center of mass?

3)

What is the y-component of the velocity of the center of mass?

m

x

y

vx

vy

  1  

9.2 kg

-2.7 m

-4.7 m

3.2 m/s

-4.2 m/s

  2  

9.2 kg

-3.7 m

3.7 m

-5.2 m/s

5.2 m/s

  3  

9.3 kg

4.7 m

-5.7 m

-6.2 m/s

2.2 m/s

  4  

9.3 kg

5.7 m

2.7 m

4.2 m/s

-3.2 m/s

Explanation / Answer

1.

x - position of Centre of mass =( m1* x1 + m2*x2 + m3*x3 + m4*x4 ) /( m1 +m2+m3+m4)
= (9.2*-2.7)+(9.2*-3.7)+(9.3*4.7)+(9.3*5.7)/(9.2+9.2+9.3+9.3)

= 1.02 m

y - position of Centre of mass =( m1* y1 + m2*y2 + m3*y3 + m4*y4 ) /( m1 +m2+m3+m4)
= (9.2*-4.7)+(9.2*3.7)+(9.3*-5.7)+(9.3*2.7)/(9.2+9.2+9.3+9.3)

= -1.002 m

The distance from the origin to the center of mass is,

(xcm,ycm) = [1.02)2 + (-1.002)2] 1/2 = 1.43 m

2.

Vx of the centre of mass = ( m1* Vx1 + m2*Vx2 + m3*Vx3 + m4*Vx4 ) /( m1 +m2+m3+m4)
= (9.2*3.2)+(9.2*-5.2)+(9.3*-6.2)+(9.3*4.2)/37

= -1.0 m/s
3.
Vy of the centre of mass = ( m1* Vy1 + m2*Vy2 + m3*Vy3 + m4*Vy4 ) /( m1 +m2+m3+m4)

= (9.2*-4.2)+(9.2*5.2)+(9.3*2.2)+(9.3*-3.2)/37

= -0.0027 m/s

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