This involves explosive problems. I posted the question under the description.

Question

The first hand held cell phone (Cellular phonis) m=1 kg is sitting in your hand when it suddenly explodes into 4 pieces. The first piece,

0.41 kg, moves with a velocity of 15 m/s towards 140°. The second piece, 0.06 kg, travels with a velocity of 25 m/s towards 253°. The

third piece, 0.22 kg, flies off with a velocity of 21 m/s towards 290°. Within seconds you look to see where the forth unknown piece is

headed and you see an Umbrella Cockatoo (Cacatua alba), m = 0.5 kg, traveling with a velocity of 14 m/s towards 125° unexpectedly

collide into it! The rattled cockatoo bounces off towards 70° going 10 m/s, while the 4 th piece bounces off in a different direction. Find

the velocity and direction of the last unknown piece of cell phone after the collision with the cockatoo.

Answer 1

net mass = 1kg

1) m = 0.41 kg with V = 15<cos(140),sin(140)>

2) m = 0.06 kg with V = 25<cos(253),sin(253)>

3)m = 0.22 kg with V = 21<cos(290),sin(290)>

4)m = 0.31 kg with V = 14<cos(125),sin(125)>

C) m=0.5 kg with V= 10<cos(70),sin(70)>

before collison

4 -> C at rest

after collision

X-> 4' + C' we need to find 4'

Momentum conserved X and Y

net 1/2 mv^2 before = net 1/2 mv^2 after

0.5*0.31*14<cos(125),sin(125)>^2 + 0.5*0.5*0 = 0.5*0.31*V<cos(x),sin(x)>^2 + 0.5*0.5*10<cos(70),sin(70)>^2

find V velocity and x angle

0.155*<64.48202,131.51797>= <0.155V^2*cos^2(x) + 2.9244, 0.155V^2*sin^2(x) + 22.07555>

eq1

0.155*64.48202 = 0.155V^2*cos^2(x) + 2.9244

7.0703131 = 0.155*V^2*cos^2(x)

45.61492/V^2 = cos^2(x)

eq2

0.155*131.51797 = 0.155V^2*sin^2(x) +22.07555

-1.69026465 = 0.155*V^2*(1-cos^2(x))

sub

-1.69026465 = 0.155*V^2*(1-45.61492/V^2)

V = 5.89151 m/s

solve for x the angle

M*vi = M*vf

14cos(125) = 5.89151cos(x)

x = -14.7681 degree or 345.2319 degree