The gun, rigidly fixed on the railway platform, fires a shot along the railway bed at an angle a = 30 ° to the horizon.

The gun, rigidly fixed on the railway platform, fires a shot along the railway bed at an angle a = 30 ° to the horizon. Determine the speed u2 of the platform rollback if the projectile flies out with the speed u1 = 480m / s. The mass of the platform with the gun and shells m2 = 18 tons, the mass of the shell is t1 = 60kg.

∠α = 30 °.

m2 = 18 t = 18000 kg.

m1 = 60 kg.

u1 = 480 m / s.

u2 -?

Let us write the law of conservation of momentum for a closed-loop projectile-gun system in vector form: m1 * V1 + m2 * V2 = m1 * u1 + m2 * u2, where m1 * V1, m1 * u1 is the impulse of the projectile before and after the shot, m2 * V2, m2 * u2 – gun impulse before and after the shot.

Since before the shot, the projectile and the gun were at rest, then V1 = V2 = 0 m / s.

Let’s write the law of conservation of momentum for projections on the horizontal axis: 0 = m1 * u1 * cosα + m2 * u2.

u2 = – m1 * u1 * cosα / m2.

The “-” sign indicates that after firing the gun will move in the opposite direction of the projectile’s departure.

u2 = 60 kg * 480 m / s * cos30 ° / 18000 kg = 1.4 m / s.

Answer: the platform rollback speed will be u2 = 1.4 m / s.