# Conservation of momentum in x and y directions

## Homework Statement

An m2 = 1.7 kg can of soup is thrown upward (north) with a velocity of v2 = 6 m/s. It
is immediately struck from the side by an m1 = 0.5 kg rock traveling at v1 = 7.8 m/s. (west)
The rock ricochets off at an angle of 46 degrees with a velocity of v3 = 5.6 m/s.
What is the angle of the can’s motion after the collision? Answer in units of degrees.

can someone look over my work? I'm not getting the right answer even after 2 attempts...

1. I broke up momentum into x and y components:
x-direction:
m1v1x + m2v2x = m1v1xf+ m2v2xf
m1v1x= m1v1xf + m2v2xf
I did 5.6cos46 to find V1xf (rock)
so plugging in m1= .5kg m2= 1.7kg and V1xf= 3.890086875m/s
i got V2xf of can= 1.149974449m/s

y-direction:
m1v1y + m2v2y = m1v1yf + m2v2yf
m2v2y= m1v1yf + m2v2yf
I did 5.6sin46 to find V1yf rock)
so plugging in m1= .5kg m2= 1.7kg and v1yf= 4.028302882 m/s
i got v2yf of can= 4.815205035 m/s

2. tantheta= V2yf/V2xf
so plugging in above numbers i got theta= 13.4318702 degrees

## Answers and Replies

LowlyPion
Homework Helper
I get 13.39° angle with the West axis. (The rock is traveling west)

What is the angle reference is it looking for?

it doesnt say...and how did you get 13.9? did you round any of the numbers?