How Do You Calculate the Third Piece's Velocity in a Grenade Explosion Problem?

[bgaspar00]

A grenade of mass 10 kg explodes into three pieces in the same plane, two of which, A (5.0 kg) and B (2.0 kg), move off as shown. Calculate the velocity of the 3.0 kg third piece, C.

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 here's 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!??

 

[neutrino]

I'm denied permission to view the file.

 

[chocokat]

I think you've mixed up your sines and cosines. The 'x' components should be cosine values, and the 'y' components should be sine values. To start you off, based on the picture:

Ax = - (500 cos 30)
Ay = 500 sin 30