Question

Платформа массой 140 кг движется со скоростью 1 м\с. Спортсмен, который. бежит со скоростью 5 м\с, догоняет платформу и прыгает на нее. Определите скорость движения платформы после того, как на нее запрыгнул спортсмен. масса спортсмена 60 кг.

Answer 1

Закон сохранения импульса:
m1*v1 + m2v2 = (m1 + m2) * v
140*1 + 60*5 = (140 + 60) * v
440 = 200v
v = 2,2 м/с

Answer 2

Calculation of the Platform's Velocity

To determine the velocity of the platform after the athlete jumps onto it, we can use the principle of conservation of momentum. According to this principle, the total momentum before the jump is equal to the total momentum after the jump.

The momentum of an object is calculated by multiplying its mass by its velocity. Let's denote the mass of the platform as m1 (140 kg) and its velocity as v1 (1 m/s). The mass of the athlete is denoted as m2 (60 kg), and their velocity is v2 (5 m/s).

Before the jump, the total momentum is the sum of the momentum of the platform and the momentum of the athlete:

Total momentum before jump = m1 * v1 + m2 * v2 .

After the jump, the athlete and the platform move together with a common velocity, denoted as v. The total momentum after the jump is the sum of the momentum of the combined system:

Total momentum after jump = (m1 + m2) * v.

According to the principle of conservation of momentum, the total momentum before the jump is equal to the total momentum after the jump:

m1 * v1 + m2 * v2 = (m1 + m2) * v.

Now we can solve this equation to find the velocity v of the platform after the athlete jumps onto it.

Calculation:

Substituting the given values:

m1 = 140 kg (mass of the platform), v1 = 1 m/s (velocity of the platform), m2 = 60 kg (mass of the athlete), v2 = 5 m/s (velocity of the athlete).

We can solve the equation:

140 kg * 1 m/s + 60 kg * 5 m/s = (140 kg + 60 kg) * v.

Simplifying the equation:

140 kg + 300 kg = 200 kg * v.

440 kg = 200 kg * v.

Dividing both sides of the equation by 200 kg:

v = 440 kg / 200 kg.

v = 2.2 m/s.

Therefore, the velocity of the platform after the athlete jumps onto it is 2.2 m/s.

Please note that this calculation assumes an idealized scenario with no external forces or friction acting on the system.

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