A 347-kg cannon, sliding freely on a frictionless horizontal plane at a speed of 8.07 m/s, shoots a 23.9-kg cannonball at an angle of 57.7° above the horizontal. The velocity of the ball relative to the cannon is such that when the shot occurs, the cannon stops cold. What is the velocity of the ball relative to the cannon?

Question

anonymous · Answer

X momentum same before and after so change is zero.

(m1 + m2) v1 = m2 V cos (57.7)

cannon + ball(?) start at 8.07m/s the ball emitted with horizontal velocity of V cos (57.7)

(347 + 23.9)(8.07) - (23.9) V cos(57.7) = 0

Solve that for V and you are done.

anonymous · Answer

thanks a lot! also when the ball is shoot there seems to be an additional momentum in y direction and i cant seem to understand how we could have an equation of momentum in y direction

anonymous · Answer

Yes,  y-momentum, (23.9) V sin (57.7) is absorbed by the springs and interaction with the ground of the cannon.