Directions. Answer the following problems on a graphing paper. Include illustrations and your complete solution. Identify important equations you used and box your final answer.
1. Particles A (m.-Z.0kg), 8 (ms-1.0 kg), and C (unknown) are placed in a 2-dimensional cartesian plane with the coordinates (1,0), (0,2), and (-3,0), respectively.
a. What should be the mass of Particle C so that the center of mass of the particles is located at the coordinate (1,1)? b. What should be the y coordinate of Particle C so that the
center of mass of the particles remains at its position?
2. A system of particles consists of four particles with masses 2 kg, 3 kg. 4 kg, and 5 kg located at positions (-1, -1), (1,-1), (1, 1), and (-1,
1) respectively.
a. What is the location of the center of mass of the system?
b. An additional particle with mass 6 kg is located at position (0,
0). What is the new location of the center of mass of the
system?
1. Particles A (m.-Z.0kg), 8 (ms-1.0 kg), and C (unknown) are placed in a 2-dimensional cartesian plane with the coordinates (1,0), (0,2), and (-3,0), respectively.
a. What should be the mass of Particle C so that the center of mass of the particles is located at the coordinate (1,1)? b. What should be the y coordinate of Particle C so that the
center of mass of the particles remains at its position?
2. A system of particles consists of four particles with masses 2 kg, 3 kg. 4 kg, and 5 kg located at positions (-1, -1), (1,-1), (1, 1), and (-1,
1) respectively.
a. What is the location of the center of mass of the system?
b. An additional particle with mass 6 kg is located at position (0,
0). What is the new location of the center of mass of the
system?
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Answer:
1. To solve this problem, we need to use the formula for the center of mass of two particles:
xcm = (m1x1 + m2x2)/(m1 + m2)
ycm = (m1y1 + m2y2)/(m1 + m2)
For three particles, we can use the same formula, but with three terms:
xcm = (m1x1 + m2x2 + m3x3)/(m1 + m2 + m3)
ycm = (m1y1 + m2y2 + m3y3)/(m1 + m2 + m3)
a. To find the mass of Particle C, we can use the fact that the center of mass is located at (1,1):
1 = (m1*1 + m2*0 + m3*(-3))/(m1 + m2 + m3)
1 = (m1 + 3m3)/(m1 + m2 + m3)
m1 + m2 + m3 = m1 + 3m3
m2 = 2m3
We also know that the total mass of the system is:
m1 + m2 + m3 = m
where m is the total mass of the system. Therefore, we can substitute m2 = 2m3 into this equation and simplify:
m1 + 3m3 = m
m3 = (m - m1)/3
Now we can use the x-coordinate formula to find m3:
1 = (m1*1 + m2*0 + m3*(-3))/m
1 = (m1 - 3(m - m1)/3)/m
1 = (2m1 - m)/m
m = 2m1 - 1
Substituting m3 = (m - m1)/3 and m2 = 2m3, we get:
m2 = 2(m - m1)/3
b. To find the y-coordinate of Particle C, we can use the fact that the center of mass remains at its position:
1 = (m1*1 + m2*0 + m3*(-3))/(m1 + m2 + m3)
1 = (m1 - 3m3)/(m1 + m2 + m3)
Substituting m2 = 2m3 and m3 = (m - m1)/3, we get:
1 = (m1 - 2(m - m1)/3)/(m1 + 2(m - m1)/3 + (m - m1)/3)
1 = m1/(2m - m1)
m1 = (2m - 1)/3
Now we can use the y-coordinate formula to find m3:
1 = (m1*0 + m2*2 + m3*0)/(m1 + m2 + m3)
1 = (4m3)/(m1 + 2m3 + m3)
Substituting m1 and m2 from above, we get:
1 = (4m3)/[(2m - 1)/3 + 4m3 + m3]
Solving for m3, we get:
m3 = (3/4)(2m - 1)/(5 - 3m)
Therefore, the y-coordinate of Particle C is:
y = (m1*0 + m2*0 + m3*y3)/(m1 + m2 + m3)
y = (2m3*(-3))/(m1 + 2m3 + m3)
y = -6m3/(2m - 1)
2. a. To find the location of the center of mass of the system, we can use the formula:
x_cm = (m1x1 + m2x2 + m3x3 + m4x4) / (m1 + m2 + m3 + m4)
y_cm = (m1y1 + m2y2 + m3y3 + m4y4) / (m1 + m2 + m3 + m4)
where m is the mass of each particle, x, and y are the coordinates of each particle.
Substituting the values, we get:
x_cm = (2*(-1) + 3*1 + 4*1 + 5*(-1)) / (2+3+4+5) = -0.2
y_cm = (2*(-1) + 3*(-1) + 4*1 + 5*1) / (2+3+4+5) = 0.2
Therefore, the center of mass of the system is located at (-0.2, 0.2).
b. To find the new location of the system's center of mass, we need to include the additional particle with a mass of 6 kg. Using the same formula as above, we get:
x_cm_new = (2*(-1) + 3*1 + 4*1 + 5*(-1) + 6*0) / (2+3+4+5+6) = 0
y_cm_new = (2*(-1) + 3*(-1) + 4*1 + 5*1 + 6*0) / (2+3+4+5+6) = 0
Therefore, the new location of the center of mass of the system is at (0, 0).