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

ID: 1422958 • Letter: F

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   7.8 kg -2.5 m -4.5 m 2.9 m/s -4.2 m/s

2    9.2 kg -3.4 m 3.4 m -5.1 m/s 5.1 m/s

3     8.6 kg 4.7 m -5.6 m -6 m/s 2 m/s

4     8.1 kg 5.6 m 2.5 m 3.9 m/s -2.9 m/s

1) What is the x position of the center of mass? in m.

2) What is the y position of the center of mass? in m.

3) What is the speed of the center of mass? in m/s.

4) When a fifth mass is placed at the origin, what happens to the horizontal (x) location of the center of mass?

It moves to the right.

It moves to the left.

It does not move.

It can not be determined unless you know the mass.

5) When a fifth mass is placed at the center of mass, what happens to the vertical (y) location of the center of mass?

It moves up.

It moves down.

It does not move.

It can not be determined unless you know the mass.

Explanation / Answer

here

Xcm = m1x1+ m2x2+ m3x3 + m4x4/(m1+m2+m3+m4)

Xcm = (-(7.8* 2.5)-(9.2* 3.4) + (8.6*4.7) + (8.1* 5.6))/(7.8 + 9.2+8.6 + 8.1)

Xcm = 1.038 m


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Ycm = m1y1+ m2y2+ m3y3 + m4y4/(m1+m2+m3+m4)

Ycm = (-(7.8* 4.5)+ (9.2* 3.4) + (8.6*5.6)- (8.1* 2.5))/(7.8 + 9.2+8.6 + 8.1)

Ycm = -0.94 m
-------------------------------


Vxcm = m1Vx1 + m2Vx2 + m3Vx3 + M4vx4 /(/(m1+m2+m3+m4)

Vxcm = m1Vx1 + m2Vx2 + m3Vx3 + M4vx4 /(/(m1+m2+m3+m4)

Vx cm = ((7.8* 2.9) - (9.2* 5.1) - (8.6*6)+(8.1* 3.9))/(7.8 + 9.2+8.6 + 8.1)

Vxcm = -1.31 m/s

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Vycm = m1Vx1 + m2Vx2 + m3Vx3 + M4vx4 /(/(m1+m2+m3+m4)

Vy cm = (-(7.8* 4.2) +(9.2* 5.1) + (8.6*2)-(8.1* 2.9))/(7.8 + 9.2+8.6 + 8.1)

Ycm = -4.39 m/s

---------------------------------

Vcm^2 = Vx^2 + Vy^2

Vcm^2 = (-1.31)^2 + (-4.39^2)

VCm = 4.58 m/s

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4. It does not move.

5. It does not move.

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