At an instant when a particle of mass 80.0 grams has a velocity of 25 m/s in the positive y direction, a 75.0 grams particle has a velocity of 20.0 m/s in the positive x direction. What is the speed of the center of mass of this two-particle system at this instant?
Due to conservation of energy, no energy will be lost in this collision.....the vertical and horizontal components will remain the same BUT it will be a bigger mass.
yvel = 25 xvel = 20
Kintetic energy = 1/2mv^2
the y particle energy ye=1/2 (80)(25)^2 = 25000 The x particle energy xe = 1/2 (75)(20)^2 = 15000
The final particle has a mass of 155 with a kinetic energy of those two original particles COMBINED = 25000 + 15000 = 40000
since Kinetic Energy = 1/2mv^2 40000 = 1/2(155)v^2 Solving for v: v of the new particle is 22.72 m/s
A little more information you didn't ask for...ha
to find the resultant y component of the final particle's velocity 25000 = 1/2 mv^2 where m=155 then Vy = 17.960
to find the resultant x component of the final particle's velocity 15000 + 1/2 mv^2 Where m=155 then Vx = 13.912
The resultant angle of travel will be tan x = 17.960/13.912 or angle x = 52.23 degrees
~jc
Due to conservation of energy, no energy will be lost in this collision.....the energy of the vertical and horizontal components will remain the same BUT it will be a bigger mass.
yvel = 25 xvel = 20
Kintetic energy = 1/2mv^2
the y particle energy ye=1/2 (80)(25)^2 = 25000 The x particle energy xe = 1/2 (75)(20)^2 = 15000
The final particle has a mass of 155 with a kinetic energy of those two original particles COMBINED = 25000 + 15000 = 40000
since Kinetic Energy = 1/2mv^2 40000 = 1/2(155)v^2 Solving for v: v of the new particle is 22.72 m/s
A little more information you didn't ask for...ha
to find the resultant y component of the final particle's velocity 25000 = 1/2 mv^2 where m=155 then Vy = 17.960
to find the resultant x component of the final particle's velocity 15000 + 1/2 mv^2 Where m=155 then Vx = 13.912
The resultant angle of travel will be tan x = 17.960/13.912 or angle x = 52.23 degrees
~jc
