given that
m1 = 1.85 kg
r1 = (1i+2j )m
v1 = (3i+5j) m/s
m2 = 2.90 kg
r2 = (-4i-3j) m
v2 = (3i-2j) m/s
we know that
part(b)
center of mass of position is
Xcm = (m1*x1 + m2*x2 )/m1+m2
Xcm = (1.85*1+2.90*(-4))(1.85+2.90)
Xcm = -9.75/4.75
Xcm = -2.05 m
Ycm = (m1y1+m2y2)/m1+m2
Ycm = (1.85*2+2.90*(-3)) / (1.85+2.90)
Ycm = -5/4.75
Ycm = -1.05 m
Rcm = Xcm+Ycm
Rcm = (-2.05i-1.05j) m
part(c)
velocity of the center of mass
Vx,cm = (m1*V1x +m2*V2x)/m1+m2
Vx,cm = 1.85*3+2.90*3/1.85+2.90
Vx,cm = 3 m/s
Vy,cm = m1*V1y+m2*V2y /m1+m2
Vy,cm = 1.85*5+2.90*(-2) / ()1.85+2.90
Vy,cm = -0.72m/s
Vcm = Vx,cm+Vy,cm
Vcm = (3i -0.72 j) m/s
part(d)
total linear momentum
Px = m1*V1x+m2*V2x
Px = 1.85*3+2.90*3
Px =14.25 kg*m/s
Py = m1*V1y+m2*V2y
Py = 1.85*5+2.90*(-2)
Py = 9.25-5.8
Py =3.45 kg*m/s
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