Question

Part C (LINEAR MOMENTUM) Problem CI: Three spherical objects moving as shown in the diagram. m, 4kg. agram, m,- 4kg.m2 = 2mm 3m. , vz =10m/ v 5m/s 30° 60° V, 15m/s my Find the magnitude and direction of the total momentum of the spheres.

Answer 1

Momentum is given by,

P = m*v

here, m = mass

v = velocity

now total momentum will be,

Pt = P1 + P2 + P3

for mass m1,

m1 = 4 kg

v1 = 5 m/s in +x -axis = 5 i

So, P1 = 4*5 i = 20 i

for mass m2,

m2 = 2*m1 = 2*4 = 8 kg

v2 = 10 m/s at 30 deg to +x axis in anti-clockwise direction

v2 = 10*cos(30 deg) i + 10*sin(30 deg) j

So, P2 = 8*(10*cos(30 deg) i + 10*sin(30 deg) j)

P2 = 69.28 i + 40 j

for mass m3,

m3 = 3*m1 = 3*4 = 12 kg

v3 = 15 m/s at 60 deg to +x axis in clockwise direction

v3 = 15*cos(60 deg) i + 15*sin(60 deg) (-j)

So, P3 = 12*(15*cos(60 deg) i - 15*sin(60 deg) j)

P3 = 90 i - 155.88 j

therefore, Pt = 20 i + (69.28 i + 40 j) + (90 i - 155.88 j)

Pt = (20+69.28+90) i + (40 - 155.88) j

Pt = 179.28 i - 115.88 j

Magnitude of total momentum will be,

|Pt| = sqrt(179.28^2 + 115.88^2)

|Pt| = 213.47 kg*m/sec.

Direction of total momentum will be,

A = arctan(Pty/Ptx) = arctan(-115.88/179.28)

A = -32.88 deg

A = 32.88 deg from +x-axis in clockwise direction.

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