Simple recoil velocity question

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The discussion revolves around calculating the recoil velocity of a cannon after firing a shell, using the conservation of momentum. The initial attempt yielded a recoil velocity of -1.33 m/s, but the correct answer is 0.94 m/s. The error lies in not accounting for the angle of the shell's velocity; the component of the velocity at 45 degrees must be considered. Participants suggest focusing on the horizontal component of the velocity to arrive at the correct recoil calculation. The conversation emphasizes the importance of correctly applying vector components in momentum problems.
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Homework Statement


A cannon of mass 1500 kg fires a 10 kg shell with a velocity of 200 m/s at an angle of 45 degrees above the horizontal. Find the recoil velocity of the cannon across the level ground.


Homework Equations


Conservation of Momentum equation:
0=m1v1f + m2v2f
v1f = - (m2/m1)(v2f)


The Attempt at a Solution



Ok so I'm having trouble with this problem. I could've sworn I was doing it correctly, but here it goes. Because the cannon and the cannon ball are both at rest, initial velocities don't play a part in this question (I think?). So plugging in the numbers into the equation I got:

M1 = 1500
M2 = 10
V2f = 200

V1f = - (10/1500)(200) = -1.33 m/s

The anwser to this question is 0.94 m/s, and I can't figure out what I'm doing wrong for the life of me. Any push in the right direction would be awesome!
 
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the thing is ... use just m1v1+m2v2=0 and then find out the velocity of the cannon..
Now the question asked is to find the velocity of the cannon along the ground... so what you have to do is find the component of velocity in the angle given.

Anyway , your answer is partly correct.. you just got to do the component part.
 
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