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I cannot show all of my work becuz it is basically just a sketch of what the diagram would look like, then drawing vectors and so on. However, i cannot get the right answer.

So heres the diagram it shows...

http://www.cbv.ns.ca/rv/physics/Physics12/BLM10-3.pdf

Δp = 0 where pi = 0

pix = pfx and piy = pfy

p1’

p1y’ = Cos30(5.0 kg (100.0 m/s))

= 433.013 kgm/s

p1x’ = Sin30(5.0(100m/s))

= 250

p2y’ = cos20 (2kg)(200 m/s)

= cos20 ( 400 kgm/s)

p2x’ = sin20 (400 kgm/s)

p1x’ + p2x’ + p3x’ = 0

500 + (- sin 20 (400 kgm/s)) + p3x’ = 0

p3x’ = 363.192 kgm/s

p1y’ + p2y’ + p3y’ = 0

500 kgm/s + (- cos 20(400 kgm/s)) + p3y’ = 0

p3y’ = 124.123kgm/s

p3 is found by resolving p3x’ and p3y’

p3x’

p3y’

p3

θ

p3^2 = [363.192kgm/s]2

+[124.123 kgm/s]2

|p3| = 383.816 kgm/s

θ = tan-1 [(124.123kgm/s)/(363.192kgm/s)]

= 19 degrees? wtf v3 = p3/m3

v3 = 383.816kgm/s/(3kg)

v3 = 128 m/s

This is completely wrong the answer is suppose to be V3= 128.92 m/s (S 82 E) or something....soo where did i go wrong? and is there an easier wayyyyy!!??