The cannon, standing on a smooth horizontal platform, shoots horizontally.

The cannon, standing on a smooth horizontal platform, shoots horizontally. The mass of the projectile is 20 kg, its muzzle velocity is 200 m / s. What speed does a cannon acquire when fired if its mass is 500 kg?

msn = 20 kg.

mp = 500 kg.

Vsn “= 200 m / s.

Vп “-?

Let us write down the law of conservation of momentum for a closed-loop system projectile-gun in vector form: mcn * Vcn + mp * Vp = mcn * Vcn “+ mp * Vp”, where msn, mp is the mass of the projectile and the gun, Vp is the speed of the gun before the shot, Vp ” is the speed of the gun after the shot, Vsn is the speed of the projectile before the shot, Vsn “is the speed of the projectile after the shot.

Since the projectile and the gun were at rest before the shot, their velocities were equal to 0: Vsn = Vp = 0 m / s.

The impulse conservation law will take the form: 0 = mсн * Vсн “+ mп * Vп”.

Vp “= – mсн * Vсн” / mp.

The “-” sign means that the speed of the cannon will be directed in the opposite direction of the projectile departure.

Vp “= 20 kg * 200 m / s / 500 kg = 8 m / s.

Answer: the speed of the gun after the shot will be Vp “= 8 m / s.