A circus cannon, which has a mass M = 5000 kg, is tilted at θ = 50°. When it shoots a projectile at v0 = 40 m/s with respect to the cannon, the cannon recoils along a horizontal track at vcannon = 0.5 m/s with respect to the ground. a) At what angle to the horizontal does the projectile move with respect to the ground? b) What is the mass of the projectile? c) The cannon is now lowered to shoot horizontally. It fires the same projectile at the same speed relative to the cannon. With what speed does the cannon now recoil with respect to the ground? Relative Equations: Conservation of momentum I already figured out part a. The velocity of the projectile with respect to the ground is the velocity of the projectile with respect to the cannon + the velocity of the cannon with respect to the ground. I found the Velocity of the projectile with respect to the ground to 40.5 m/s. Then i did a ratio to find the angle with respect to the ground. (40m/s)/(50 degrees)=(40.5m/s)/(theta with respect to the ground) Theta equals 50.625 degrees. Its part b that I'm having trouble with. I know that the momentum of the cannon must be equal and opposite to that of the projectile in the x-direction. My equation looked like this: -(5000kg)(-.05m/s) = Mp(40.5cos(50.625)) From there i found the mass of the projectile to be 65.9444 kg, but this is not right. What am I overlooking. I've went over this 3 times now and I can't seem to find my mistake.