A cannon weighing 800 kg shoots a core weighing 10 kg with an initial speed of 200 m / s relative to the Earth at an angle

A cannon weighing 800 kg shoots a core weighing 10 kg with an initial speed of 200 m / s relative to the Earth at an angle of 60 ° to the horizon. What is the recoil speed of the cannon?

M = 800 kg.

Vя “= 200 m / s.

m = 10 kg.

∠α = 60 °.

Vп “-?

For the cannon – core system, the law of conservation of momentum is valid in vector form: M * Vп + m * Vя = M * Vп “+ m * Vя”, where M * Vп, M * Vп “are the impulses of the gun before and after the shot, m * Vя, m * Vя “- the impulses of the nucleus before and after the shot.

Since the cannon and the cannonball were at rest, their impulses before the shot were equal to 0: M * Vp = m * Vя = 0.

M * Vp “+ m * Vя” = 0.

For projections on the horizontal axis, the law of conservation of momentum will be: M * Vp “= – m * Vя” * cosα

Vп “= – m * Vя” * cosα / M.

The “-” sign indicates that the speed of the cannon and the cannonball after the shot is directed in the opposite direction.

Vp “= 10 kg * 200 m / s * cos60 ° / 800 kg = 5 m / s.

Answer: the recoil speed of the gun is Vp “= 5 m / s.