Question

Пушка установлена на стоящей платформе. общая масса платформы равна 40 т. пушка выстреливает два раза вдоль пути. какую скорость будет иметь платформа после второго выстрела? масса снаряда равна 20 кг, скорость 1000м\с.

Answer 1

M<<M
2*m*V1=M*V
V=2*m*V/M=2*20*10^3/40*10^3=1 м/с

Answer 2

Calculation of the Platform's Velocity

To calculate the velocity of the platform after the second shot, we can use the principle of conservation of momentum. According to this principle, the total momentum before the shots is equal to the total momentum after the shots.

The momentum of an object is calculated by multiplying its mass by its velocity. In this case, the platform's momentum before the shots is zero since it is stationary. After the shots, the platform will have a non-zero velocity.

Let's denote the mass of the platform as m, the velocity of the platform after the second shot as v, the mass of the projectile as m_p, and the velocity of the projectile as v_p.

The total momentum before the shots is given by:

Total momentum before = 0

The total momentum after the shots is given by:

Total momentum after = (m + m_p) * v

According to the principle of conservation of momentum, the total momentum before and after the shots should be equal. Therefore, we can set up the following equation:

(m + m_p) * v = 0

Solving for v, we get:

v = 0 / (m + m_p)

Since the mass of the platform is given as 40 tons (or 40,000 kg) and the mass of the projectile is given as 20 kg, we can substitute these values into the equation to find the velocity of the platform after the second shot.

v = 0 / (40,000 + 20) = 0

Therefore, the velocity of the platform after the second shot is 0 m/s.

Please note that the platform's velocity after the second shot is zero because the total momentum before the shots is zero and the mass of the platform is much larger than the mass of the projectile.