Consider a system consisting of three particles: m1 = 3 kg, vector v cm = m/s (c
ID: 1285644 • Letter: C
Question
Consider a system consisting of three particles: m1 = 3 kg, vector v cm = m/s (c) What is the total kinetic energy of this system? K tot = J (d) What is the translational kinetic energy of this system? K trans = J (e) What is the kinetic energy of this system relative to the center of mass? K rel = J One way to calculate K rel is to calculate the velocity of each particle relative to the center of mass, by subtracting the center-of-mass velocity from the particle's actual velocity to get the particle's velocity relative to the center of mass, then calculating the corresponding kinetic energy, then adding up the three relative kinetic energies. However, there is a much simpler way to determine the specified quantity, without having to do all those calculations; think about what you know about the relationships among the various kinds of kinetic energy in a multiparticle system. (If you wish, you can check your result by doing the complicated calculation just described.) vector p tot = kg A? m/s (b) What is the velocity of the center of mass of this system? vector v3 = m/s (a) What is the total momentum of this system? vector v2 = m/s m3 = 5 kg, vector v1 = m/s m2 = 4 kg,Explanation / Answer
total momentum will be
m1v1 + m2v2 + m3v3 = 3< 7, -4, 13 > + 4< -12, 7, -3 > + 5< -22, 35, 17 >
P = <-137, 191, 112>
velocity of centar of mass
Vcm = P/(m1+m2+m3) = <-137, 191, 112>/(3+4+5)
= <-11.41, 15.91, 9.33>
Ktotal =
m1 has KE = 1/2 m v^2 =
(.5)(3 kg)(7^2 + (-4)^2 + 13^2)(m^2/s^2) = 351 J
m2:
(.5)(4 kg)((-12)^2 + 7^2 + (-3)^2) = 404 J
m3:
(.5)(5 kg)((-22)^2 + 35^2 + 17^2) = 4995 J
Ktotal = 351 + 404 + 4995 = 5750 J
Ktrans = (Mass(total)*(Velocity of Ceneter of mass)^2)/2
= 12*0.5*<-11.41^2+15.91^2+9.33^2) = 2822.19 J
Krel = Ktotal - Ktrans = 5750 - 2822.19 = 2927.81 J