A cannon with a mass of mn = 200 kg is fired in a horizontal direction. What is the recoil velocity of the cannon

A cannon with a mass of mn = 200 kg is fired in a horizontal direction. What is the recoil velocity of the cannon after a shot if a 1 kg cannonball flew out at a speed of 400 m / s? How will the answer change if the nucleus flies out at an angle of 60 ° to the horizon?

mp = 200 kg.

mя = 1 kg.

Vя “= 400 m / s.

∠α = 60 °.

Vп “-?

mp * Vp + my * Vя = mp * Vp “+ my * Vя” – the law of conservation of momentum in vector form.

Since the cannon with the cannon ball was at rest before the shot, Vp = Vя = 0 m / s.

0 = mp * Vp “+ my * Vя”.

Horizontal shot.

For projections: mp * Vp “= – mя * Vя”.

Vp “= – mя * Vя” / mp.

The “-” sign shows that after the shot the cannon speeds Vp “and the nucleus Vя” will be directed in opposite directions.

Vp “= 1 kg * 400 m / s / 200 kg = 2 m / s.

Shot at an angle ∠α = 60 °.

0 = mп * Vп “+ mа * Vя” * cosα.

mp * Vp “= mя * Vя” * cosα.

Vp “= mя * Vя” ** cosα / mp.

Vp “= 1 kg * 400 m / s * cos60 ° / 200 kg = 1 m / s.

Answer: the recoil speed of the gun with a horizontal shot will be Vp “= 2 m / s, with a shot at an angle ∠α = 60 ° Vp” = 1 m / s.