p.242 [illegible] I suppose the bullet completes its acceleration at some time when the muzzle is at the end of the barrel, and that it has a velocity v which is equal to the muzzle velocity. It follows from this that if the gun was originally at rest, the gun moves forward with a velocity u = mv/M when I have supposed that all momentum is conserved. A quantity (mv+Mv) must be constant, and if the bullet is not fired “uncompensated” so that it has some mass left in the tube behind it, then this equation will hold also in a slight modified form so that u = v(M+m)/(M+m) where m is the remaining mass of the bullet at end of t (approximately if acceleration is small). I presume therefore that an equal and opposite force is exerted on the gun in such time as to give it this velocity. The momentum of the bullet is then Mv = mv + mu, it has velocity v. If the gun was not originally stationary, but had some initial velocity U, then the final velocity would be V = (M+m)U + mv / (M+m). This can easily work out a proper relation to work out. p.243 (-1+2i) and 0.98 at 6 o'clock. 0.98 cm from the axis, which is not at all very different from the value of r.