Question

Consider a system of two particles In the xy plane: m_1 = 1.85 kg is at the location r_1 = (1.00i + 2.00J) m and has a velocity of (3.00i + 0.500j) m/s; m_2 = 2.90 kg Is at r_2 - (-4.00i - 3.00J) m and has velocity (3.00i - 2.00J) m/s. 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.) Find the position of the center of mass of the system and mark it on the grid. Determine the velocity of the center of mass and also show it on the diagram What is the total linear momentum of the system?

Answer 1

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