A particle of mass =6.50 kg moves with a velocity v1=x1i+y1j+z1k. A second particle of mass =3.50 kg moves with a velocity v2=x2i+y2j+z2k. And a third particle of mass =6.30 kg moves with a velocity v3=x3i+y3j+z3k. Here, (x1,y1,z1)=(-2.5,5.4,1.2), (x2,y2,z2)=(-2.3,0.1,-2.5) and (x3,y3,z3)=(3.3,-8.6,5.3).
What is the magnitude of the velocity of the center of mass for the system?
the sum of the masses.
then multiply each of the velocities by the fraction ofmass it contains, ie, on v1= 6.50/totalmass*(-2.5,5.4,1.2), or
v1=(-.997, 2.15, .479)
after scaling each of the three velocities with the mass fraction, then add the like components of each to get Vcm
To find the magnitude of the velocity of the center of mass for the system, we need to calculate the overall velocity of the center of mass using the principle of conservation of momentum.
The equation for the velocity of the center of mass is given by:
Vcm = (m1 * v1 + m2 * v2 + m3 * v3) / (m1 + m2 + m3)
Given the masses and velocities of the particles:
m1 = 6.50 kg, m2 = 3.50 kg, m3 = 6.30 kg
v1 = -2.5i + 5.4j + 1.2k
v2 = -2.3i + 0.1j - 2.5k
v3 = 3.3i - 8.6j + 5.3k
Substituting these values into the equation:
Vcm = (6.50 * (-2.5i + 5.4j + 1.2k) + 3.50 * (-2.3i + 0.1j - 2.5k) + 6.30 * (3.3i - 8.6j + 5.3k)) / (6.50 + 3.50 + 6.30)
Simplifying the equation, we get:
Vcm = (-16.25i + 35.1j + 7.8k - 8.05i + 0.35j - 8.75k + 20.79i - 54.18j + 33.39k) / 16.30
Combining like terms, we have:
Vcm = ((-16.25 - 8.05 + 20.79)i + (35.1 + 0.35 - 54.18)j + (7.8 - 8.75 + 33.39)k) / 16.30
Simplifying further:
Vcm = (-3.51i - 18.73j + 32.44k) / 16.30
Finally, we can find the magnitude of the velocity of the center of mass using the formula:
|Vcm| = sqrt((Vcm_x)² + (Vcm_y)² + (Vcm_z)²)
Substituting the values obtained above:
|Vcm| = sqrt((-3.51)² + (-18.73)² + (32.44)²)