From the barrel of the gun, fixed on the platform, along the rails at an angle of 60 degrees to the horizon

From the barrel of the gun, fixed on the platform, along the rails at an angle of 60 degrees to the horizon, a projectile flew out at a speed of 600 m / s. The speed of the platform as a result of the shot is 0.3 m / s What is the ratio of the masses of the platform with the gun and the projectile?

∠α = 60 °.

Vsn “= 600 m / s.

Vp “= 0.3 m / s.

mp / msn -?

After the shot, the pulses of the platform Pp “and the projectile Pc” are equal in magnitude and oppositely directed: Pp “= – Pcn” – vector.

The “-” sign indicates the opposite direction of the impulses.

The momentum of the body P is the product of the velocity V and the mass of the body m: P = m * V.

mсн * Vсн “= mп * Vп” – vector.

For projections onto the horizontal axis, the impulse conservation law will have the form: mcn * Vcn “* cosα = mp * Vp”.

mп / mсн = Vсн “* cosα / Vп”.

mp / msn = 600 m / s * cos60 ° / 0.3 m / s = 1000.

Answer: the mass of the platform mp is 1000 times greater than the mass of the projectile mcn: mp / mcn = 1000.