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:
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
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