# From the barrel of a cannon, fixed on a railway platform along the rails at an angle of 60 to the horizon

**From the barrel of a cannon, fixed on a railway platform along the rails at an angle of 60 to the horizon, a projectile weighing 10 kg flies out, the mass of the platform with a cannon is 10 tons, what is the ratio of the velocities of the projectile and the gun with which they will move after the shot.**

msn = 10 kg.

mp = 10 t = 10000 kg.

∠α = 60 °.

Vсн / Vп -?

When the projectile leaves the barrel of the gun, the law of conservation of momentum in vector form is valid: the impulse of the projectile mсн * Vсн is equal to the impulse of the gun with the platform mп * Vpsn.

msn * Vsn = mp * Vp.

Since the projectile flies out at an angle ∠α = 60 ° to the horizon, then for projections onto the horizontal axis the law of conservation of momentum will take the form: mсн * Vсн * cosα = mп * Vп.

Vsn / Vp = mp / msn * cosα.

Vsn / Vp = 10000 kg / 10 kg * cos60 ° = 2000.

Answer: the speed of the projectile will be 2000 times higher than the recoil speed of the gun: Vsn / Vp = 2000.