Consider a system of two particles in the xy plane: m1 = 2.10 kg is at the locat
ID: 1458065 • Letter: C
Question
Consider a system of two particles in the xy plane:
m1 = 2.10 kg
is at the location
[r with arrow] 1 = (1.00 [i] + 2.00 [j] ) m
and has a velocity of (3.00 [i] + 0.500 [j] ) m/s;
m2 = 3.15 kg
is at
[r with arrow] 2 = (4.00 [i] 3.00 [j] ) m
and has velocity (3.00 [i] - 2.00 [j] ) m/s.
(a) Plot these particles on a grid or graph paper. Draw their position vectors and show their velocities. (Do this on paper. Your instructor may ask you to turn in this work.)
(b) Find the position of the center of mass of the system and mark it on the grid.
xCM__?___ m
yCM__?___ m
(c) Determine the velocity of the center of mass and also show it on the diagram.
vx,CM ___?___ m/s
vy,CM ___?___ m/s
(d) What is the total linear momentum of the system?
px___?__ kg · m/s
py ___?___ kg · m/s
Explanation / Answer
Given that
A system consisting of two particles of masses m1 =2.10kg
The location and velocity vectors for the first particle is r1 = (1.00 [i] + 2.00 [j] ) m
And has a velocity of (v1) = (3.00 [i] + 0.500 [j] ) m/s
Mass of the second particle is m2 =3.15kg
The location and velocity vectors for the second particle is
r2 = (4.00 [i] 3.00 [j] ) m
v2 = (3.00 [i] - 2.00 [j] ) m/s
Now the center of Xcm =m1x1+m2x2/(m1+m2) =2.10*1+3.15*-4/(2.10+3.15) = -2
Now the center of Ycm =m1y1+m2y2/(m1+m2) =2.10*2+3.15*-3/(2.10+3.15) = -1
Now the position is given by P =-2i-1j
The velocity of center of mass is given by
VXcm =m1v1+m2v2/(m1+m2) =2.10*3+3.15*3/(2.10+3.15) =3
VYcm =m1v1+m2v2/(m1+m2) =2.10*0.500+3.15*-2/(2.10+3.15) =-1
Hence the velocity is = 3i-1j
Now the total linear momentum is given by
P=mass*velocity =(2.10+3.15)(3i-1j) =(15.9i-5.35j)kg.m/s
The momentum along the x-axis is Px =15.9kgm/s
The momentum along the Y-axis is Py =-5.35Kgm/s