Conservation of momentum in x and y directions

  • Thread starter Maiia
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  • #1
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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

  • #2
LowlyPion
Homework Helper
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I get 13.39° angle with the West axis. (The rock is traveling west)

What is the angle reference is it looking for?
 
  • #3
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it doesnt say...and how did you get 13.9? did you round any of the numbers?
 

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