Question

Снаряд массой 100кг, летящий горизонтально вдоль железнодорожного пути со скоростью 500м/с, падает в вагон с песком массой 10т и застревает в нем. Найти скорость вагона, если он двигается со скоростью 36км/ч навстречу снаряду

Answer 1

Дано:
m1 = 100 кг
V1 = 500 м/с
m2 = 10 т = 10 000 кг
V2 = 36 км/ч = 10 м/с

U - ?


Импульс снаряда: 
p1 = m1*V1 = 100*500 = 50 000 кг*м/с

Импульс вагона до выстрела:
p2 = - m2*V2 = - 10 000*10 =  -100 000 кг*м/с  (знак "минус" означает, что вагон двигался навстречу снаряду)

Суммарный импульс до взаимодействия:

Po = P1 + P2 = 50 000 - 100 000 = - 50 000 кг*м/с

После взаимодействия:
P = (m1+m2)*U = 10 100*U

По закону сохранения импульса:

P = Po
10 100*U = - 50 000
U = - 50 000 / 10 100 ≈  - 4,95 ≈ - 5 м/с (знак "минус" говорит о том, что вагон продолжает движение в том же направлении, что и раньше, только со скоростью в 2 раза меньшей)

Answer 2

Problem Analysis

We are given a scenario where a projectile with a mass of 100 kg is flying horizontally along a railway track with a speed of 500 m/s. The projectile falls into a wagon filled with sand, which has a mass of 10 tons. The wagon is moving towards the projectile with a speed of 36 km/h. We need to find the velocity of the wagon after the projectile gets embedded in it.

Solution

To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

Let's denote the velocity of the wagon before the collision as v_wagon and the velocity of the projectile before the collision as v_projectile. After the collision, the projectile gets embedded in the wagon, so their masses combine.

The total momentum before the collision is given by the sum of the momentum of the wagon and the momentum of the projectile:

Total momentum before collision = momentum of wagon + momentum of projectile

The momentum of an object is given by the product of its mass and velocity:

Momentum = mass * velocity

Using this equation, we can write the equation for the total momentum before the collision as:

m_wagon * v_wagon + m_projectile * v_projectile = (m_wagon + m_projectile) * v_final

where: - m_wagon is the mass of the wagon, - m_projectile is the mass of the projectile, - v_projectile is the velocity of the projectile before the collision, - v_final is the velocity of the wagon after the collision.

We are given the values of m_wagon, m_projectile, v_projectile, and v_wagon. We need to solve for v_final.

Let's substitute the given values into the equation and solve for v_final.

Calculation

Given: - Mass of the wagon (m_wagon) = 10 tons = 10,000 kg - Mass of the projectile (m_projectile) = 100 kg - Velocity of the projectile before the collision (v_projectile) = 500 m/s - Velocity of the wagon before the collision (v_wagon) = 36 km/h = 36,000 m/3600 s = 10 m/s

Using the equation:

m_wagon * v_wagon + m_projectile * v_projectile = (m_wagon + m_projectile) * v_final

Substituting the given values:

10,000 kg * 10 m/s + 100 kg * 500 m/s = (10,000 kg + 100 kg) * v_final

Simplifying the equation:

100,000 kg*m/s + 50,000 kg*m/s = 10,100 kg * v_final

150,000 kg*m/s = 10,100 kg * v_final

Dividing both sides of the equation by 10,100 kg:

v_final = 150,000 kg*m/s / 10,100 kg

v_final ≈ 14.851 m/s

Answer

The velocity of the wagon after the projectile gets embedded in it is approximately 14.851 m/s.

Please let me know if I can help you with anything else.